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+\documentclass{report}
+
+\usepackage[T1]{fontenc}
+%\usepackage[a4paper]{geometry}
+\usepackage[leqno]{amsmath}
+\usepackage{amssymb}
+\usepackage{theorem}
+\usepackage[dvips]{graphicx}
+\usepackage{makeidx}
+
+\renewcommand{\th}{$^{\text{th}}$}
+\newcommand{\authorfont}[1]{\textsc{#1}}
+\renewcommand{\author}[2]{%
+ \def\tst{#1}
+ \ifx\tst\empty
+ \authorfont{#2}\index{#2}
+ \else
+ \authorfont{#2}\index{#1}
+ \fi}
+
+\theorembodyfont{\normalfont}
+\theoremheaderfont{\normalfont}
+\newtheorem{axiom}{Ax.}
+\newtheorem{theorem}{Th.}
+
+\makeindex
+
+\begin{document}
+\small
+\begin{verbatim}
+The Project Gutenberg EBook of The Algebra of Logic, by Louis Couturat
+
+This eBook is for the use of anyone anywhere at no cost and with
+almost no restrictions whatsoever. You may copy it, give it away or
+re-use it under the terms of the Project Gutenberg License included
+with this eBook or online at www.gutenberg.org
+
+
+Title: The Algebra of Logic
+
+Author: Louis Couturat
+
+Release Date: January 26, 2004 [EBook #10836]
+
+Language: English
+
+Character set encoding: TeX
+
+*** START OF THIS PROJECT GUTENBERG EBOOK THE ALGEBRA OF LOGIC ***
+
+
+
+
+Produced by David Starner, Arno Peters, Susan Skinner
+and the Online Distributed Proofreading Team.
+
+\end{verbatim}
+\normalsize
+\newpage
+
+\pagenumbering{roman}
+
+\begin{titlepage}
+ \begin{center}
+ {\LARGE\bfseries THE ALGEBRA OF LOGIC}\\[2.5cm]
+ BY\\[2.5cm]
+ {\Large LOUIS COUTURAT}\\[3cm]
+ AUTHORIZED ENGLISH TRANSLATION\\[2cm]
+ BY\\[2cm]
+ {\large LYDIA GILLINGHAM ROBINSON, B. A.}\\[1cm]
+ \textsc{With a Preface by PHILIP E. B. JOURDAIN. M. A. (Cantab.)}
+ \end{center}
+\end{titlepage}
+
+
+\section*{Preface}
+\addcontentsline{toc}{section}{\numberline{}Preface}
+
+Mathematical Logic is a necessary preliminary to logical
+Mathematics. ``Mathematical Logic'' is the name given by
+\author{}{Peano} to what is also known (after \author{}{Venn}) as ``Symbolic
+Logic''; and Symbolic Logic is, in essentials, the Logic of
+Aristotle,\index{Aristotle} given new life and power by being
+dressed up in the wonderful---almost magical---armour and
+accoutrements of Algebra. In less than seventy years, logic, to
+use an expression of \author{}{De Morgan's,} has so \emph{thriven}
+upon symbols and, in consequence, so grown and altered that the
+ancient logicians would not recognize it, and many old-fashioned
+logicians will not recognize it. The metaphor is not quite
+correct: Logic has neither grown nor altered, but we now see more
+\emph{of} it and more \emph{into} it.
+
+The primary significance of a symbolic calculus seems to lie in
+the economy of mental effort which it brings about, and to this is
+due the characteristic power and rapid development of mathematical
+knowledge. Attempts to treat the operations of formal logic in an
+analogous way had been made not infrequently by some of the more
+philosophical mathematicians, such as \author{}{Leibniz} and
+\author{}{Lambert}; but their labors remained little known, and it
+was \author{}{Boole} and \author{}{De Morgan,} about the middle of
+the nineteenth century, to whom a mathematical---though of course
+non-quantitative---way of regarding logic was due. By this, not
+only was the traditional or Aristotelian doctrine of logic
+reformed and completed, but out of it has developed, in course of
+time, an instrument which deals in a sure manner with the task of
+investigating the fundamental concepts of mathematics---a task
+which philosophers have repeatedly taken in hand, and in which
+they have as repeatedly failed.
+
+First of all, it is necessary to glance at the growth of
+symbolism in mathematics; where alone it first reached perfection.
+There have been three stages in the development
+of mathematical doctrines: first came propositions with particular
+numbers, like the one expressed, with signs subsequently
+invented, by ``$2 + 3 = 5$''; then came more general laws holding
+for all numbers and expressed by letters, such as
+\begin{displaymath}
+ \mbox{``}(a + b) c = a c + b c\mbox{''};
+\end{displaymath}
+lastly came the knowledge of more general laws of functions and
+the formation of the conception and expression ``function''. The
+origin of the symbols for particular whole numbers is very
+ancient, while the symbols now in use for the operations and
+relations of arithmetic mostly date from the sixteenth and
+seventeenth centuries; and these ``constant'' symbols together
+with the letters first used systematically by \author{}{Viète}
+(1540--1603) and \author{}{Descartes} (1596--1650), serve, by
+themselves, to express many propositions. It is not, then,
+surprising that \author{}{Descartes,} who was both a mathematician
+and a philosopher, should have had the idea of keeping the method
+of algebra while going beyond the material of traditional
+mathematics and embracing the general science of what thought
+finds, so that philosophy should become a kind of Universal
+Mathematics. This sort of generalization of the use of symbols for
+analogous theories is a characteristic of mathematics, and seems
+to be a reason lying deeper than the erroneous idea, arising from
+a simple confusion of thought, that algebraical symbols
+necessarily imply something quantitative, for the antagonism there
+used to be and is on the part of those logicians who were not and
+are not mathematicians, to symbolic logic. This idea of a
+universal mathematics was cultivated especially by
+\author{}{Gottfried Wilhelm Leibniz} (1646--1716).
+
+Though modern logic is really due to \author{}{Boole} and
+\author{}{De Morgan,} \author{}{Leibniz} was the first to have a
+really distinct plan of a system of mathematical logic. That this
+is so appears from research---much of which is quite recent---into
+\author{}{Leibniz's} unpublished work.
+
+The principles of the logic of \author{}{Leibniz,} and
+consequently of his whole philosophy, reduce to
+two\footnote{\author{}{Couturat,} \emph{La Logique de Leibniz
+d'après des documents inédits}, Paris, 1901, pp.~431--432, 48.}:
+(1) All our ideas are compounded of a very small number of simple
+ideas which
+form the ``alphabet of human thoughts'';%
+\index{Alphabet of human thought} (2) Complex ideas
+proceed from these simple ideas by a uniform and symmetrical
+combination which is analogous to arithmetical multiplication.
+With regard to the first principle, the number of simple ideas is
+much greater than \author{}{Leibniz} thought; and, with regard to the second
+principle, logic considers three operations---which we shall meet
+with in the following book under the names of logical multiplication,
+logical addition%
+\index{Addition!and multiplication!Logical} and negation---instead of only one.
+
+``Characters''\index{Characters} were, with \author{}{Leibniz,}
+any written signs, and ``real'' characters were those which---as
+in the Chinese ideography---represent ideas directly, and not the
+words for them. Among real characters, some simply serve to
+represent ideas, and some serve for reasoning. Egyptian and
+Chinese hieroglyphics and the symbols of astronomers and chemists
+belong to the first category, but
+\author{}{Leibniz} declared them to be imperfect, and desired the second
+category of characters for what he called his ``universal
+characteristic''.\footnote{\emph{Ibid}., p.~81.} It was not in the form of
+an algebra that \author{}{Leibniz} first conceived his characteristic,
+probably because he was then a novice in mathematics, but in the form of a
+universal language or script.\footnote{\emph{Ibid}., pp.~51, 78} It was in
+1676 that he first dreamed of a kind of algebra of thought,%
+\index{Algebra!of thought}\footnote{\emph{Ibid}., p.~61.} and it was the
+algebraic notation which then served as model for the
+characteristic.\footnote{\emph{Ibid}., p.~83.}
+
+\author{}{Leibniz} attached so much importance to the invention of proper
+symbols that he attributed to this alone the whole of his
+discoveries in mathematics.\footnote{\emph{Ibid}., p.~84.} And, in
+fact, his infinitesimal calculus\index{Calculus!Infinitesimal}
+affords a most brilliant example of the importance of, and
+\author{}{Leibniz'}s skill in devising, a suitable
+notation.\footnote{\emph{Ibid}., p.~84--87.}
+
+Now, it must be remembered that what is usually understood by the name
+``symbolic logic'', and which---though not its name---is chiefly due to
+\author{}{Boole,} is what \author{}{Leibniz} called a \emph{Calculus
+ ratiocinator},\index{Calculus!ratiocinator@\emph{ratiocinator}} and is
+only a part of the Universal Characteristic. In symbolic logic
+\author{}{Leibniz} enunciated the principal
+properties of what we now call logical multiplication, addition,%
+\index{Addition!and multiplication!Logical} negation, identity,
+class-inclusion, and the null-class; but the aim of
+\author{}{Leibniz's} researches was, as he said, to create ``a
+kind of general system of notation in which all the truths of
+reason should be reduced to a calculus. This could be, at the same
+time, a kind of universal written language, very different from
+all those which have been projected hitherto; for the characters
+and even the words would direct the reason, and the
+errors---excepting those of fact---would only be errors of
+calculation. It would be very difficult to invent this language or
+characteristic, but very easy to learn it without any
+dictionaries''. He fixed the time necessary to form it: ``I think
+that some chosen men could finish the matter within five years'';
+and finally remarked: ``And so I repeat, what I have often said,
+that a man who is neither a prophet nor a prince can never
+undertake any thing more conducive to the good of the human race
+and the glory of God''.
+
+In his last letters he remarked: ``If I had been less busy,
+or if I were younger or helped by well-intentioned young
+people, I would have hoped to have evolved a characteristic
+of this kind''; and: ``I have spoken of my general characteristic
+to the Marquis de l'Hôpital and others; but they paid no
+more attention than if I had been telling them a dream. It
+would be necessary to support it by some obvious use; but,
+for this purpose, it would be necessary to construct a part
+at least of my characteristic;---and this is not easy, above all
+to one situated as I am''.
+
+\author{}{Leibniz} thus formed projects of both what he called a
+\emph{characteristica universalis}, and what he called a \emph{calculus
+ ratiocinator};\index{Calculus!ratiocinator@\emph{ratiocinator}} it is not
+hard to see that these projects are interconnected, since a
+perfect universal characteristic would comprise, it seems, a
+logical calculus.\index{Calculus!Logical} \author{}{Leibniz} did
+not publish the incomplete results which he had obtained, and
+consequently his ideas had no continuators, with the exception of
+\author{}{Lambert} and some others, up to the time when
+\author{}{Boole,} \author{}{De Morgan,}
+\author{}{Schröder,} \author{}{MacColl,} and others rediscovered his
+theorems. But when the investigations of the principles of
+mathematics became the chief task of logical symbolism, the aspect
+of symbolic logic as a calculus ceased to be of such importance,
+as we see in the work of \author{}{Frege} and \author{}{Russell.}
+\author{}{Frege's} symbolism, though far better for logical analysis than
+\author{}{Boole's} or the more modern \author{}{Peano's,} for instance, is far
+inferior to \author{}{Peano's}---a symbolism in which the merits
+of internationality and power of expressing mathematical theorems
+are very satisfactorily attained---in practical convenience.
+\author{}{Russell,} especially in his later works, has used the ideas of
+\author{}{Frege,} many of which he discovered subsequently to, but
+independently of, \author{}{Frege,} and modified the symbolism of
+\author{}{Peano} as little as possible. Still, the complications
+thus introduced take away that simple character which seems
+necessary to a calculus, and which \author{}{Boole} and others
+reached by passing over certain distinctions which a subtler logic
+has shown us must ultimately be made.
+
+Let us dwell a little longer on the distinction pointed out by
+\author{}{Leibniz} between a \emph{calculus
+ ratiocinator}\index{Calculus!ratiocinator@\emph{ratiocinator}} and a
+\emph{characteristica universalis} or \emph{lingua characteristica}. The
+ambiguities of ordinary language are too well known for it to be necessary
+for us to give instances. The objects of a complete logical symbolism are:
+firstly, to avoid this disadvantage by providing an \emph{ideography}, in
+which the signs represent ideas and the relations between them
+\emph{directly} (without the intermediary of words), and secondly, so to
+manage that, from given premises, we can, in this ideography, draw all the
+logical conclusions which they imply by means of rules of transformation of
+formulas analogous to those of algebra,---in fact, in which we can replace
+reasoning by the almost mechanical process of calculation. This second
+requirement is the requirement of a \emph{calculus
+ ratiocinator}.\index{Calculus!ratiocinator@\emph{ratiocinator}} It is
+essential that the ideography should be complete, that only symbols with a
+well-defined meaning should be used---to avoid the same sort of ambiguities
+that words have---and, consequently,---that no suppositions should be
+introduced implicitly, as is commonly the case if the meaning of signs is
+not well defined. Whatever premises are necessary and sufficient%
+\index{Condition!Necessary and sufficient} for a conclusion should be
+stated explicitly.
+
+Besides this, it is of practical importance,---though it is
+theoretically irrelevant,---that the ideography should be concise,
+so that it is a sort of stenography.
+
+The merits of such an ideography are obvious: rigor of
+reasoning is ensured by the calculus character; we are
+sure of not introducing unintentionally any premise; and
+we can see exactly on what propositions any demonstration
+depends.
+
+We can shortly, but very fairly accurately, characterize the dual
+development of the theory of symbolic logic during the last sixty
+years as follows: The \emph{calculus
+ ratiocinator}\index{Calculus!ratiocinator@\emph{ratiocinator}}
+aspect of symbolic logic was developed by \author{}{Boole,}
+\author{}{De Morgan,} \author{}{Jevons,} \author{}{Venn,}
+\author{Peirce, C.~S.}{C. S. Peirce,} \author{}{Schröder,} Mrs.
+\author{}{Ladd-Franklin} and others; the \emph{lingua characteristica}
+aspect was developed by \author{}{Frege,} \author{}{Peano} and
+\author{}{Russell.} Of course there is no hard and fast boundary-line
+between the domains of these two parties. Thus \author{Peirce, C.~S.}{Peirce} and
+\author{}{Schröder} early began to work at the foundations of
+arithmetic with the help of the calculus of relations; and thus they
+did not consider the logical calculus\index{Calculus!Logical} merely
+as an interesting branch of algebra. Then \author{}{Peano} paid
+particular attention to the calculative aspect of his symbolism.
+\author{}{Frege} has remarked that his own symbolism is meant to be a
+\emph{calculus
+ ratiocinator}\index{Calculus!ratiocinator@\emph{ratiocinator}} as
+well as a \emph{lingua characteristica}, but the using of
+\author{}{Frege's} symbolism as a calculus would be rather like using
+a three-legged stand-camera for what is called ``snap-shot''
+photography, and one of the outwardly most noticeable things about
+\author{}{Russell's} work is his combination of the symbolisms of
+\author{}{Frege} and \author{}{Peano} in such a way as to preserve
+nearly all of the merits of each.
+
+The present work is concerned with the \emph{calculus
+ ratiocinator}\index{Calculus!ratiocinator@\emph{ratiocinator}} aspect,
+and shows, in an admirably succinct form, the beauty, symmetry and
+simplicity of the calculus of logic regarded as an algebra. In fact, it can
+hardly be doubted that some such form as the one in which
+\author{}{Schröder} left it is by far the best for exhibiting it from this
+point of view.\footnote{Cf.\ \author{Whitehead, A.~N.}{A.~N.\
+Whitehead,}
+ \emph{A Treatise on Universal Algebra with Applications}, Cambridge,
+ 1898.} The content of the
+present volume corresponds to the two first volumes of
+\author{}{Schröder's} great but rather prolix
+treatise.\footnote{\emph{Vorlesungen über die Algebra der Logik},
+ Vol.~I., Leipsic, 1890; Vol. II, 1891 and 1905. We may mention that
+ a much shorter \emph{Abriss} of the work has been prepared by
+ \author{Müller, Eugen}{Eugen Müller.} Vol.~III (1895) of
+ \author{}{Schröder's} work is on the logic of relatives founded by
+ \author{}{De Morgan} and \author{Peirce, C.~S.}{C. S. Peirce,}---a
+ branch of Logic that is only mentioned in the concluding sentences
+ of this volume.} Principally owing to the influence of
+\author{Peirce, C.~S.}{C. S. Peirce,} \author{}{Schröder} departed
+from the custom of \author{}{Boole, Jevons,} and himself (1877),
+which consisted in the making fundamental of the notion of
+\emph{equality}, and adopted the notion of \emph{subordination} or
+\emph{inclusion} as a primitive notion. A more orthodox
+\textsc{Boolian} exposition is that of
+\author{}{Venn,}\footnote{Symbolic Logic, London, 1881; 2nd ed.,
+ 1894.} which also contains many valuable historical notes.
+
+We will finally make two remarks.
+
+When \author{}{Boole} (cf.~\S\ref{ch:2} below) spoke of propositions determining
+a class of moments at which they are true, he really
+(as did \author{}{MacColl}) used the word ``proposition'' for what we
+now call a ``propositional function''. A ``proposition'' is a
+thing expressed by such a phrase as ``twice two are four'' or
+``twice two are five'', and is always true or always false. But
+we might seem to be stating a proposition when we say:
+``Mr. \author{Bryan, William Jennings}{William Jennings Bryan} is Candidate for the Presidency
+of the United States'', a statement which is sometimes true
+and sometimes false. But such a statement is like a mathematical
+\emph{function} in so far as it depends on a \emph{variable}---the
+time. Functions of this kind are conveniently distinguished
+from such entities as that expressed by the phrase ``twice
+two are four'' by calling the latter entities ``propositions'' and
+the former entities ``propositional functions'': when the variable
+in a propositional function is fixed, the function becomes a
+proposition. There is, of course, no sort of necessity why
+these special names should be used; the use of them is
+merely a question of convenience and convention.
+
+In the second place, it must be carefully observed that, in
+\S\ref{ch:13}, 0~and~1 are not \emph{defined} by expressions whose
+principal copulas are relations of inclusion. A definition is
+simply the convention that, for the sake of brevity or some other
+convenience, a certain new sign is to be used instead of a group
+of signs whose meaning is already known. Thus, it is the sign of
+\emph{equality} that forms the principal copula. The theory of
+definition has been most minutely studied, in modern times by
+\author{}{Frege} and \author{}{Peano.}
+\begin{quote}
+Philip E. B. Jourdain.
+\end{quote}
+Girton, Cambridge. England.
+
+\cleardoublepage
+{\renewcommand{\footnote}[1]{}
+\tableofcontents}
+
+\cleardoublepage
+\section*{Bibliography%
+ \footnote{This list contains only the works relating to the system of \author{}{Boole}
+ and \author{}{Schröder} explained in this work.}}
+\addcontentsline{toc}{section}{\numberline{}Bibliography}
+\bibliographystyle{none}
+\begin{description}
+\item[\textsc{George Boole}.] \emph{The Mathematical Analysis of Logic} (Cambridge
+ and London, 1847).
+
+\item[---] \emph{An Investigation of the Laws of Thought} (London and
+ Cambridge, 1854).
+
+\item[\textsc{W. Stanley Jevons}.] \emph{Pure Logic} (London, 1864).
+
+\item[---] ``On the Mechanical Performance of Logical Inference''
+ (\emph{Philosophical Transactions}, 1870).
+
+\item[\textsc{Ernst Schröder}.] \emph{Der Operationskreis des Logikkalkuls}
+ (Leipsic, 1877).
+
+\item[---] \emph{Vorlesungen über die Algebra der Logik}, Vol.~I (1890),
+ Vol.~II (1891), Vol.~III: \emph{Algebra und Logik der Relative}
+ (1895) (Leipsic).\footnote{\textsc{Eugen Müller} has prepared a part, and is preparing more, of
+ the publication of supplements to Vols.~II and~III, from the papers left
+ by \textsc{Schröder}.}
+
+\item[\textsc{Alexander MacFarlane}.] \emph{Principles of the Algebra of Logic,
+ with Examples} (Edinburgh, 1879).
+
+\item[\textsc{John Venn}.] \emph{Symbolic Logic}, 1881; 2nd. ed., 1894 (London).%
+ \footnote{A valuable work from the points of view of history and bibliography.}
+ \emph{Studies in Logic} by members of the Johns Hopkins University
+ (Boston, 1883): particularly Mrs. \textsc{Ladd-Franklin},
+ \textsc{O. Mitchell} and \textsc{C. S. Peirce}.
+
+\item[\textsc{A.~N.~Whitehead}.] \emph{A Treatise on Universal Algebra}. Vol. I
+ (Cambridge, 1898).
+
+\item[---] ``Memoir on the Algebra of Symbolic Logic'' (\emph{American
+ Journal of Mathematics}, Vol.~XXIII, 1901).
+
+\item[\textsc{Eugen Müller}.] \emph{Über die Algebra der Logik:}
+ I.~\emph{Die Grundlagen des Ge\-biete\-kalkuls;} II.~\emph{Das
+ Eliminationsproblem und die Syllogistik;} Programs of the Grand
+ Ducal Gymnasium of Tauberbischofsheim (Baden), 1900, 1901 (Leipsic).
+
+\item[\textsc{W.~E.~Johnson}.] ``Sur la théorie des
+ égalités logiques'' (\emph{Bibliothèque du Congrès international de
+ Philosophie}. Vol.~III, \emph{Logique et Histoire des Sciences;}
+ Paris, 1901).
+
+\item[\textsc{Platon Poretsky}.] \emph{Sept Lois fondamentales de la théorie des
+ égalités logiques} (Kazan, 1899).
+
+\item[---] \emph{Quelques lois ultérieures de la théorie des égalités logiques}
+ (Kazan, 1902).
+
+\item[---] ``Exposé élémentaire de la théorie des égalités logiques à
+ deux termes'' (\emph{Revue de Métaphysique et de Morale}. Vol. VIII,
+ 1900.)
+
+\item[---] ``Théorie des égalités logiques à trois termes'' (\emph{Bibliothèque
+ du Congrès international de Philosophie}). Vol. III. (\emph{Logique
+ et Histoire des Sciences}). (Paris, 1901, pp.~201--233).
+
+\item[---] \emph{Théorie des non-égalités logiques} (Kazan, 1904).
+
+\item[\textsc{E. V. Huntington}.] ``Sets of Independent Postulates for the
+ Algebra of Logic'' (\emph{Transactions of the American Mathematical
+ Society}, Vol. V, 1904).
+\end{description}
+
+\cleardoublepage
+\pagenumbering{arabic}
+\part*{THE ALGEBRA OF LOGIC.}
+
+\cleardoublepage
+\section{Introduction}\label{ch. 1}
+The algebra of logic was founded by
+\author{Boole}{George Boole} (1815--1864); it was developed and perfected
+by \author{Schröder}{Ernst Schröder} (1841--1902). The fundamental laws
+of this calculus were devised to express the principles of
+reasoning, the ``laws of thought''. But this calculus may be
+considered from the purely formal point of view, which is
+that of mathematics, as an algebra based upon certain principles
+arbitrarily laid down. It belongs to the realm of
+philosophy to decide whether, and in what measure, this
+calculus corresponds to the actual operations of the mind,
+and is adapted to translate or even to replace argument;
+we cannot discuss this point here. The formal value of this
+calculus and its interest for the mathematician are absolutely
+independent of the interpretation given it and of the application
+which can be made of it to logical problems. In
+short, we shall discuss it not as logic but as algebra.
+
+\section{The Two Interpretations of the Logical Calculus}
+\label{ch:2}\index{Calculus!Logical}
+There is one circumstance of particular interest,
+namely, that the algebra in question, like logic, is susceptible
+of two distinct interpretations, the parallelism between them
+being almost perfect, according as the letters represent concepts
+or propositions. Doubtless we can, with \author{}{Boole} and
+\author{}{Schröder,} reduce the two interpretations to one, by considering
+the concepts on the one hand and the propositions
+on the other as corresponding to \emph{assemblages} or \emph{classes}; since
+a concept determines the class of objects to which it is
+applied (and which in logic is called its \emph{extension}), and a
+proposition determines the class of the instances or moments
+of time in which it is true (and which by analogy can also
+be called its extension). Accordingly the calculus of concepts%
+\index{Concepts!Calculus of}\index{Calculus!of concepts} and the
+calculus of propositions become reduced to
+but one, the calculus of classes,%
+\index{Classes!Calculus of}\index{Calculus!of classes} or, as
+\author{}{Leibniz} called it, the theory of the whole and part, of that
+which contains and that which is contained. But as a matter of fact, the
+calculus of concepts and the calculus of propositions present certain
+differences, as we shall see, which prevent their complete identification
+from the formal point of view and consequently their reduction to a single
+``calculus of classes''.
+
+Accordingly we have in reality three distinct calculi, or,
+in the part common to all, three different interpretations of
+the same calculus. In any case the reader must not forget
+that the logical value and the deductive sequence of the
+formulas does not in the least depend upon the interpretations
+which may be given them, and, in order to
+make this necessary abstraction easier, we shall take care to
+place the symbols ``C.~I.'' (\emph{conceptual interpretation}) and ``P.~I.''
+(\emph{prepositional interpretation}) before all interpretative phrases.
+These interpretations shall serve only to render the formulas
+intelligible, to give them clearness and to make their meaning
+at once obvious, but never to justify them. They may
+be omitted without destroying the logical rigidity of the
+system.
+
+In order not to favor either interpretation we shall say
+that the letters represent \emph{terms}; these terms may be either
+concepts or propositions according to the case in hand.
+Hence we use the word \emph{term} only in the logical sense.
+When we wish to designate the ``terms'' of a sum we shall
+use the word \emph{summand} in order that the logical and mathematical
+meanings of the word may not be confused. A term
+may therefore be either a factor or a summand.
+
+\section{Relation of Inclusion}\label{ch:3}
+Like all deductive theories, the algebra of logic may be
+established on various systems of principles\footnote{See
+\author{}{Huntington,} ``Sets of Independent Postulates for the
+Algebra of Logic'', \emph{Transactions of the Am.\ Math.\ Soc.},
+Vol.~V, 1904, pp.~288--309. [Here he says: ``Any set of consistent
+postulates would give rise to a corresponding algebra, viz., the
+totality of propositions which follow from these postulates by
+logical deductions. Every set of postulates should be free from
+redundances, in other words, the postulates of each set should be
+\emph{independent}, no one of them deducible from the rest.'']};
+we shall choose the one which most nearly approaches the
+exposition of \author{}{Schröder} and current logical
+interpretation.
+
+The fundamental relation of this calculus is the binary
+(two-termed) relation which is called \emph{inclusion} (for
+classes), \emph{subsumption} (for concepts), or \emph{implication}
+(for propositions). We will adopt the first name as affecting
+alike the two logical interpretations, and we will represent this
+relation by the sign $<$ because it has formal properties
+analogous to those of the mathematical relation $<$ (``less
+than'') or more exactly $\leq$, especially the relation of not
+being symmetrical. Because of this analogy \author{}{Schröder}
+represents this relation by the sign $\in$ which we shall not
+employ because it is complex, whereas the relation of inclusion is
+a simple one.
+
+In the system of principles which we shall adopt, this
+relation is taken as a primitive idea and is consequently
+indefinable. The explanations which follow are not given
+for the purpose of \emph{defining} it but only to indicate its meaning
+according to each of the two interpretations.
+
+C.~I.: When $a$ and $b$ denote concepts, the relation $a < b$
+signifies that the concept $a$ is subsumed under the concept $b$;
+that is, it is a species with respect to the genus $b$. From
+the extensive point of view, it denotes that the class of $a$'s
+is contained in the class of $b$'s or makes a part of it; or,
+more concisely, that ``All $a$'s are $b$'s''. From the
+comprehensive point of view it means that the concept $b$ is contained
+in the concept $a$ or makes a part of it, so that consequently
+the character $a$ implies or involves the character $b$. Example:
+``All men are mortal''; ``Man implies mortal''; ``Who says
+man says mortal''; or, simply, ``Man, therefore mortal''.
+
+P.~I.: When~$a$ and $b$~denote propositions, the relation $a < b$
+signifies that the proposition~$a$ implies or involves the
+proposition $b$, which is often expressed by the hypothetical
+judgement, ``If~$a$ is true, $b$~is true''; or by~``$a$
+implies~$b$''; or more simply by~``$a$, therefore~$b$''. We see
+that in both interpretations the relation $<$ may be translated
+approximately by ``therefore''.
+
+\emph{Remark}.---Such a relation as ``$a < b$'' is a proposition,
+whatever may be the interpretation of the terms~$a$ and~$b$.
+Consequently, whenever a $<$~relation has two like relations
+(or even only one) for its members, it can receive only the
+propositional interpretation, that is to say, it can only denote
+an implication.
+
+A relation whose members are simple terms (letters) is
+called a \emph{primary} proposition; a relation whose members are
+primary propositions is called a \emph{secondary} proposition, and
+so on.
+
+From this it may be seen at once that the propositional
+interpretation is more homogeneous than the conceptual,
+since it alone makes it possible to give the same meaning
+to the copula~$<$ in both primary and secondary propositions.
+
+\section{Definition of Equality}\label{ch:4}
+There is a second copula
+that may be defined by means of the first; this is the
+copular~$=$ (``equal to''). By definition we have
+\begin{displaymath}
+ a = b,
+\end{displaymath}
+whenever
+\begin{displaymath}
+ a < b \text{ and } b < a
+\end{displaymath}
+are true at the same time, and then only. In other words,
+the single relation $a = b$ is equivalent to the two
+simultaneous relations $a < b$ and $b < a$.
+
+In both interpretations the meaning of the copula~$=$ is
+determined by its formal definition:
+
+C.~I.: $a = b$ means, ``All~$a$'s are $b$'s~and all~$b$'s are~$a$'s'';
+in other words, that the classes~$a$ and~$b$ coincide, that they are
+identical.\footnote{This does not mean that the concepts~$a$ and~$b$
+ have the same meaning. Examples: ``triangle'' and ``trilateral'',
+ ``equiangular triangle'' and ``equilateral triangle''.}
+
+P.~I.: $a = b$ means that~$a$ implies~$b$ and~$b$ implies~$a$; in
+other words, that the propositions~$a$ and~$b$ are equivalent,
+that is to say, either true or false at the same
+time.\footnote{This does not
+ mean that they have the same meaning. Example: ``The triangle ABC
+ has two equal sides'', and ``The triangle ABC has two equal
+ angles''.}
+
+\emph{Remark.}---The relation of equality is symmetrical by very
+reason of its definition: $a = b$ is equivalent to $b = a$. But
+the relation of inclusion is not symmetrical: $a < b$ is not
+equivalent to $b < a$, nor does it imply it. We might agree
+to consider the expression $a > b$ equivalent to $b < a$, but
+we prefer for the sake of clearness to preserve always the
+same sense for the copula~$<$. However, we might translate
+verbally the same inclusion $a < b$ sometimes by~``$a$ is contained
+in~$b$'', and sometimes by ``$b$ contains $a$''.
+
+In order not to favor either interpretation, we will call the first
+member of this relation the \emph{antecedent}\index{Antecedent} and
+the second the \emph{consequent}\index{Consequent}.
+
+C.~I.: The antecedent is the \emph{subject} and the consequent is
+the \emph{predicate} of a universal affirmative proposition.%
+\index{Affirmative propositions}
+
+P.~I.: The antecedent is the \emph{premise} or the
+\emph{cause},\index{Cause} and the consequent is the
+\emph{consequence}.\index{Consequence} When an implication is translated by
+a \emph{hypothetical} (or \emph{conditional}) judgment the antecedent is
+called the \emph{hypothesis} (or the \emph{condition}\index{Condition}) and
+the consequent is called the \emph{thesis}.
+
+When we shall have to demonstrate an equality we shall
+usually analyze it into two converse inclusions and demonstrate
+them separately. This analysis is sometimes made also
+when the equality is a datum (a \emph{premise}).
+
+When both members of the equality are propositions, it
+can be separated into two implications, of which one is
+called a \emph{theorem} and the other its \emph{reciprocal}. Thus whenever
+a theorem and its reciprocal are true we have an
+equality. A simple theorem gives rise to an implication
+whose antecedent is the \emph{hypothesis} and whose consequent is
+the \emph{thesis} of the theorem.
+
+It is often said that the hypothesis is the \emph{sufficient condition}%
+\index{Condition!Necessary and sufficient} of the thesis, and the
+thesis the \emph{necessary condition} of the hypothesis; that is
+to say, it is sufficient that the hypothesis be true for the
+thesis to be true; while it is necessary that the thesis be true
+for the hypothesis to be true also. When a theorem and its
+reciprocal are true we say that its hypothesis
+is the necessary and sufficient condition%
+\index{Condition!Necessary and sufficient} of the thesis; that is to say,
+that it is at the same time both cause and consequence.
+
+\section{Principle of Identity}\label{ch:5}
+The first principle or axiom
+of the algebra of logic is the \emph{principle of identity}, which is
+formulated thus:
+\begin{axiom}\index{Axioms}
+ \begin{displaymath}
+ a < a,
+ \end{displaymath}
+ whatever the term $a$ may be.
+\end{axiom}
+
+C.~I.: ``All $a$'s are $a$'s'', \emph{i.e.}, any class whatsoever
+is contained in itself.
+
+P.~I.: ``$a$ implies $a$'', \emph{i.e.}, any proposition
+whatsoever implies itself.
+
+This is the primitive formula of the principle of identity.
+By means of the definition of equality, we may deduce from
+it another formula which is often wrongly taken as the expression
+of this principle:
+\begin{displaymath}
+ a = a,
+\end{displaymath}
+whatever~$a$ may be; for when we have
+\begin{displaymath}
+ a < a, a < a,
+\end{displaymath}
+we have as a direct result,
+\begin{displaymath}
+ a = a.
+\end{displaymath}
+
+C.~I.: The class~$a$ is identical with itself.
+
+P.~I.: The proposition~$a$ is equivalent to itself.
+
+\section{Principle of the Syllogism}\label{ch:6}
+
+Another principle of the algebra of logic is the principle of the
+\emph{syllogism}, which may be formulated as follows:
+\begin{axiom}\index{Axioms}
+ \begin{displaymath}
+ (a < b) (b < c) < (a < c).
+ \end{displaymath}
+\end{axiom}
+
+C.~I.: ``If all~$a$'s are~$b$'s, and if all~$b$'s are~$c$'s, then all~$a$'s
+are~$c$'s''. This is the principle of the \emph{categorical
+ syllogism}.\index{Categorical syllogism}
+
+P.~I.: ``If~$a$ implies~$b$, and if~$b$ implies~$c$, $a$~implies~$c$.''
+This is the principle of the \emph{hypothetical syllogism}.
+
+We see that in this formula the principal copula has always
+the sense of implication because the proposition is a
+secondary one.
+
+By the definition of equality the consequences%
+\index{Consequences!of the syllogism} of the
+principle of the syllogism may be stated in the following
+formulas\footnote{Strictly speaking, these formulas presuppose the laws of multiplication
+which will be established further on; but it is fitting to cite
+them here in order to compare them with the principle of the syllogism
+from which they are derived.}:
+\begin{alignat*}{2}
+ (a &< b) &\quad (b = c) &< (a < c), \\
+ (a &= b) &\quad (b < c) &< (a < c), \\
+ (a &= b) &\quad (b - c) &< (a = c).
+\end{alignat*}
+
+The conclusion is an equality only when both premises are equalities.
+
+The preceding formulas can be generalized as follows:
+\begin{alignat*}{3}
+ (a &< b) &\quad (b &< c) &\quad (c < d) &< (a < d), \\
+ (a &= b) &\quad (b &= c) &\quad (c = d) &< (a = d).
+\end{alignat*}
+
+Here we have the two chief formulas of the \emph{sorites}. Many
+other combinations may be easily imagined, but we can have
+an equality for a conclusion only when all the premises are
+equalities. This statement is of great practical value. In a
+succession of deductions we must pay close attention to see
+if the transition from one proposition to the other takes place
+by means of an equivalence or only of an implication. There
+is no equivalence between two extreme propositions unless
+all intermediate deductions are equivalences; in other words,
+if there is one single implication in the chain, the relation
+of the two extreme propositions is only that of implication.
+
+\section{Multiplication and Addition}\label{ch:7}
+The algebra of logic admits of three operations, \emph{logical
+ multiplication}, \emph{logical addition},%
+\index{Addition!and multiplication!Logical} and \emph{negation}.
+The two former are binary operations, that is to say, combinations
+of two terms having as a consequent a third term which may or may
+not be different from each of them. The existence of the logical
+product and logical sum of two terms must necessarily answer the
+purpose of a double postulate, for simply to define an entity is
+not enough for it to exist. The two postulates may be formulated
+thus:
+\begin{axiom}\index{Axioms}
+ Given any two terms,~$a$ and~$b$, then there is a
+ term~$p$ such that
+ \begin{displaymath}
+ p < a, p < b,
+ \end{displaymath}
+ and that for every value of~$x$ for which
+ \begin{displaymath}
+ x < a, x < b,
+ \end{displaymath}
+ we have also
+ \begin{displaymath}
+ x < p.
+ \end{displaymath}
+\end{axiom}
+\begin{axiom}\index{Axioms}
+ Given any two terms,~$a$ and~$b$, there exists
+ a term~$s$ such that
+ \begin{displaymath}
+ a < s, b < s,
+ \end{displaymath}
+ we have also
+ \begin{displaymath}
+ s < x.
+ \end{displaymath}
+\end{axiom}
+It is easily proved that the terms~$p$ and~$s$ determined by
+the given conditions are unique, and accordingly we can
+define \emph{the} product $ab$ and \emph{the} sum $a + b$ as being respectively
+the terms~$p$ and~$s$.
+
+C.~I.: 1. The product of two classes is a class~$p$ which
+is contained in each of them and which contains every
+(other) class contained in each of them;
+
+2. The sum of two classes~$a$ and~$b$ is a class~$s$ which
+contains each of them and which is contained in every (other)
+class which contains each of them.
+
+Taking the words ``less than'' and ``greater than'' in a metaphorical sense
+which the analogy of the relation~$<$ with the mathematical relation of
+inequality suggests, it may be said that the product of two classes is the
+greatest class contained in both, and the sum of two classes is the
+smallest class which contains both.\footnote{According to another analogy
+ \author{}{Dedekind} designated the logical sum and product by the same
+ signs as the least common multiple and greatest common divisor (\emph{Was
+ sind und was sollen die Zahlen?} Nos.~8 and~17, 1887. [Cf.\ English
+ translation entitled \emph{Essays on Number} (Chicago, Open Court
+ Publishing Co.\ 1901, pp.~46 and 48)] \author{Cantor, Georg}{Georg
+ Cantor} originally gave them the same designation (\emph{Mathematische
+ Annalen}, Vol.~XVII, 1880).} Consequently the product of two
+classes is the part that is common to each (the class of their
+common elements) and the sum of two classes is the class of all
+the elements which belong to at least one of them.
+
+P.~I.: 1. The product of two propositions is a proposition
+which implies each of them and which is implied by every
+proposition which implies both:
+
+2. The sum of two propositions is the proposition which
+is implied by each of them and which implies every proposition
+implied by both.
+
+Therefore we can say that the product of two propositions
+is their weakest common cause, and that their sum is their
+strongest common consequence, strong and weak being used
+in a sense that every proposition which implies another is
+stronger than the latter and the latter is weaker than the
+one which implies it. Thus it is easily seen that the product
+of two propositions consists in their \emph{simultaneous affirmation}:
+``$a$~and $b$~are true'', or simply~``$a$ and~$b$''; and that their
+sum consists in their \emph{alternative affirmation},%
+\index{Alternative!affirmation} ``either~$a$ or~$b$
+is true'', or simply~``$a$ or~$b$''.
+
+\emph{Remark}.---Logical addition%
+\index{Addition!Logical, not disjunctive} thus defined is not disjunctive;%
+\footnote{\author{}{Boole,} closely following analogy with
+ordinary mathematics, premised, as a necessary condition to the
+definition of ``$x + y$'', that~$x$ and~$y$ were mutually
+exclusive. \author{}{Jevons,} and practically all mathematical
+logicians after him, advocated, on various grounds, the definition
+of ``logical addition'' in a form which does not necessitate
+mutual exclusiveness.} that is to say, it does not presuppose that
+the two summands have no element in common.
+
+\section{Principles of Simplification and Composition}\label{ch:8}
+The
+two preceding definitions, or rather the postulates which
+precede and justify them, yield directly the following formulas:
+
+\begin{gather}
+ ab < a, \qquad ab < b, \label{eq:simplification1}\\
+ (x < a) (x < b) < (x < ab), \label{eq:composition1}\\
+ a < a + b, \qquad b < a + b, \label{eq:simplification2}\\
+ (a < x) (b < x) < (a + b < x). \label{eq:composition2}
+\end{gather}
+
+Formulas~(\ref{eq:simplification1}) and~(\ref{eq:simplification2})
+bear the name of the \emph{principle of simplification} because by
+means of them the premises of an argument may be simplified by
+deducing therefrom weaker propositions, either by deducing one of
+the factors from a
+product, or by deducing from a proposition a sum (alternative)%
+\index{Alternative}
+of which it is a summand.
+
+Formulas (\ref{eq:composition1}) and (\ref{eq:composition2}) are
+called the \emph{principle of composition},%
+\index{Composition!Principle of} because by means of them two inclusions of
+the same antecedent or the same consequent may be combined
+(\emph{composed}). In the first case we have the product of the
+consequents, in the second, the sum of the antecedents.
+
+The formulas of the principle of composition can be transformed
+into equalities by means of the principles of the
+syllogism and of simplification. Thus we have
+\begin{gather*}
+ \tag*{1 (Syll.)} (x < ab) (ab < a) < (x < a), \\
+ \tag{Syll.} (x < ab) (ab < b) < (x < b).\\
+ \intertext{Therefore}
+ \tag{Comp.} (x < ab) < (x < a) (x < b).\\
+ \tag*{2 (Syll.)} (a < a + b) (a + b < x) < (a < x),\\
+ \tag{Syll.} (b < a + b) (a + b < x) < (b < x).\\
+ \intertext{Therefore}
+ \tag{Comp.} (a + b < x) < (a < x) (b < x).
+\end{gather*}
+
+If we compare the new formulas with those preceding, which are their
+converse propositions, we may write
+\begin{gather*}
+ (x < ab) = (x < a) (x < b),\\
+ (a + b < x) = (a < x) (b < x).
+\end{gather*}
+
+Thus, to say that~$x$'s contained in~$ab$ is equivalent to
+saying that it is contained at the same time in both~$a$ and~$b$;
+and to say that~$x$ contains $a + b$ is equivalent to saying
+that it contains at the same time both~$a$ and~$b$.
+
+\section{The Laws of Tautology and of Absorption}\label{ch:9}
+
+Since the definitions of the logical sum and product do not
+imply any order among the terms added or multiplied,
+logical addition and multiplication%
+\index{Addition!and multiplication!Logical} evidently possess commutative
+and associative properties which may be expressed in
+the formulas
+\begin{displaymath}
+ \begin{aligned}
+ ab &= ba,\\
+ (ab) c &= a (bc),\\
+ \end{aligned} \qquad
+ \begin{aligned}
+ a + b &= b + a,\\
+ (a + b) + c &= a + (b + c).\\
+ \end{aligned}
+\end{displaymath}
+
+Moreover they possess a special property which is expressed
+in the \emph{law of tautology:}
+\begin{displaymath}
+ a = aa, \qquad a = a + a.
+\end{displaymath}
+
+\emph{Demonstration:}
+\begin{gather*}
+ \tag*{1 (Simpl.)} aa < a,\\
+ \tag*{(Comp.)} (a < a) (a < a) = (a < aa)\\
+ \intertext{whence, by the definition of equality,}
+ (aa < a) (a < aa) = (a - aa)
+\end{gather*}
+
+In the same way:
+\begin{gather*}
+ \tag*{2 (Simpl.)} a < a + a,\\
+ \tag*{(Comp.)} (a < a) (a < a) = (a + a < a),\\
+ \intertext{whence}
+ (a < a + a) (a + a < a) = (a = a + a).
+\end{gather*}
+
+From this law it follows that the sum or product of any
+number whatever of equal (identical) terms is equal to one
+single term. Therefore in the algebra of logic there are
+neither multiples nor powers, in which respect it is very
+much simpler than numerical algebra.%
+\index{Algebra!of logic compared to mathematical algebra}
+
+Finally, logical addition and multiplication%
+\index{Addition!and multiplication!Logical} posses a
+remarkable property which also serves greatly to simplify
+calculations, and which is expressed by the \emph{law of absorption:}%
+\index{Absorption!Law of}
+\begin{displaymath}
+ a + ab = a, \qquad a (a + b) = a.
+\end{displaymath}
+
+\emph{Demonstration:}
+\begin{gather*}
+ \tag*{1 (Comp.)} (a < a) (ab < a) < (a + ab < a),\\
+ \tag*{(Simpl.)} a < a + ab,\\
+ \intertext{whence, by the definition of equality,}
+ (a + ab < a) (a < a + ab) = (a + ab = a).
+\end{gather*}
+
+In the same way:
+\begin{align*}
+ \tag*{1 (Comp.)} (a < a) (a < a + b) < [a < a (a + b)],\\
+ \tag*{(Simpl.)} a (a + b) < a,\\
+ \intertext{whence}
+ [a < a (a + b)] [a (a + b) < a] = [a (a + b) = a].
+\end{align*}
+Thus a term~$(a)$ \emph{absorbs} a summand~$(ab)$ of which it is a
+factor, or a factor $(a + b)$ of which it is a summand.
+
+\section{Theorems on Multiplication and Addition}\label{ch:10}
+We
+can now establish two theorems with regard to the combination
+of inclusions and equalities by addition and multiplication:%
+\index{Addition!and multiplication!Theorems on}
+\begin{theorem}
+ \begin{displaymath}
+ (a < b) < (ac < bc), \qquad (a < b) < (a + c < b + c).
+ \end{displaymath}
+\end{theorem}
+
+\emph{Demonstration:}
+\begin{gather*}
+ \tag*{1 (Simpl.)} ac < c, \\
+ \tag*{(Syll.)} (ac < a) (a < b) < (ac < b), \\
+ \tag*{(Comp.)} (ac < b) (ac < c) < (ac < bc). \\
+ \tag*{2 (Simpl.)} c < b + c, \\
+ \tag*{(Syll.)} (a < b) (b < b + c) < (a < b + c). \\
+ \tag*{(Comp.)} (a < b + c) (a < b + c) < (a + c < b + c).
+\end{gather*}
+
+This theorem may be easily extended to the case of
+equalities:
+\begin{displaymath}
+ (a = b) < (ac = bc), \qquad (a = b) < (a + c = b + c).
+\end{displaymath}
+\begin{theorem}
+ \begin{gather*}
+ (a < b) (c < d) < (ac < bd), \\
+ (a < b) (c < d) < (a + c < b + d).\\
+ \end{gather*}
+\end{theorem}
+
+\emph{Demonstration:}
+\begin{align*}
+ \tag*{1 (Syll.)} (ac < a) (a < b) < (ac < b),\\
+ \tag*{(Syll.)} (ac < c) (c < d) < (ac < a),\\
+ \tag*{(Comp.)} (ac < b) (ac < d) < (ac < bd).\\
+ \tag*{2 (Syll.)} (a < b) (b < b + d) < (a < b + d),\\
+ \tag*{(Syll.)} (c < d) (d < b + d) < (c < b + d),\\
+ \tag*{(Comp.)} (a < b + d) (c < b + d) < (a + c < b + d).
+\end{align*}
+
+This theorem may easily be extended to the case in which
+one of the two inclusions is replaced by an equality:
+\begin{gather*}
+ (a = b) (c < d) < (ac < bd),\\
+ (a = b) (c < d) < (a + c < b + d).
+\end{gather*}
+
+When both are replaced by equalities the result is an
+equality:
+\begin{align*}
+ (a = b) (c = d) &< (ac = bd),\\
+ (a = b) (c = d) &< (a + c = b + d).
+\end{align*}
+
+To sum up, two or more inclusions or equalities can be
+added or multiplied together member by member; the result
+will not be an equality unless all the propositions combined are equalities.
+
+\section{The First Formula for Transforming Inclusions
+into Equalities}\label{ch:11}
+We can now demonstrate an important
+formula by which an inclusion may be transformed into an
+equality, or \emph{vice versa}:
+\begin{displaymath}
+ (a < b) = (a = ab) \qquad (a < b) = (a + b = b)
+\end{displaymath}
+
+\emph{Demonstration:}
+\begin{displaymath}
+ (a < b) < (a = ab),\qquad (a < b) < (a + b = b).\tag*{1.}
+\end{displaymath}
+
+For
+\begin{gather*}
+ \tag{Comp.} (a < a) (a < b) < (a < ab),\\
+ (a < b) (b < b) < (a + b < b).
+\end{gather*}
+
+On the other hand, we have
+\begin{gather*}
+ \tag{Simpl.} ab < a, b < a + b,\\
+ \tag{Def. =} (a < ab) (ab < a) = (a = ab)\\
+ (a + b < b) (b < a + b) = (a + b = b).
+\end{gather*}
+
+\begin{displaymath}
+ \tag*{2.} (a = ab) < (a < b),\qquad (a + b = b) < (a < b).
+\end{displaymath}
+
+For
+\begin{gather*}
+ (a - ab) (ab < b) < (a < b),\\
+ (a < a + b) (a + b = b) < (a < b).
+\end{gather*}
+
+\emph{Remark}.---If we take the relation of equality as a primitive
+idea (one not defined) we shall be able to define the relation of
+inclusion by means of one of the two preceding formulas.\footnote{See
+ \author{}{Huntington,} \emph{op.~cit.}, \S\ref{ch:1}.} We shall then be able
+to demonstrate the principle of the syllogism.\footnote{This can be
+ demonstrated as follows: By definition we have $(a < b) = (a =
+ ab)$, and $(b < c) = (b = bc)$. If in the first equality we
+ substitute for~$b$ its value derived from the second equality, then
+ $a = abc$. Substitute for~$a$ its equivalent $ab$, then $ab = abc$.
+ This equality is equivalent to the inclusion, $ab < c$.
+ Conversely substitute~$a$ for $ab$; whence we have $a < c$.
+ Q.E.D.}
+
+From the preceding formulas may be derived an interesting result:
+\begin{displaymath}
+ (a = b) = (ab = a + b).
+\end{displaymath}
+
+For
+\begin{gather*}
+ \tag*{1.} (a = b) = (a < b) (b < a),\\
+ (a < b) = (a = ab), (b < a) = (a + b = a),\\
+ \tag{Syll.} (a = ab) (a + b = a) < (ab = a + b).
+\end{gather*}
+
+\begin{gather*}
+ \tag*{2.} (ab=a+b) < (a + b < ab),\\
+ \tag{Comp.} (a+b < ab) = (a < ab) (b < ab),\\
+ (a < ab) (ab < a) = (a = ab) = (a < b),\\
+ (b < ab) (ab < b) = (b=ab) = (b < a),
+\end{gather*}
+
+Hence
+\begin{displaymath}
+ (ab = a + b) < (a < b) (b < a) = (a = b).
+\end{displaymath}
+
+\section{The Distributive Law}\label{ch:12}
+The principles previously
+stated make it possible to demonstrate the \emph{converse distributive
+law}, both of multiplication with respect to addition, and of
+addition with respect to multiplication,
+\begin{displaymath}
+ ac+bc < (a + b) c,\qquad ab+c < (a+c)(b+c).
+\end{displaymath}
+
+\emph{Demonstration:}
+\begin{gather*}
+ (a < a+b) < [ac < (a + b)c],\\
+ (b < a+b) < [bc < (a + b)c];\\
+ \intertext{whence, by composition,}
+ [ac < (a+b)c] [bc < (a+b)c] < [ac+bc < (a+b)c]
+\end{gather*}
+
+\begin{gather*}
+ \tag*{2.} (ab < a) < (ab+c < a+c), \\
+ (ab < b) < (ab+c < b+c),
+\end{gather*}
+whence, by composition,
+\begin{displaymath}
+ (ab+c < a+c)(ab+c < b+c) < [ab+c < (a+c)(b+c)].
+\end{displaymath}
+
+But these principles are not sufficient to demonstrate the
+\emph{direct distributive law}
+\begin{displaymath}
+ (a+b)c < ac+bc,\qquad (a+c)(b+c) < ab+c,
+\end{displaymath}
+and we are obliged to postulate one of these formulas or
+some simpler one from which they can be derived. For
+greater convenience we shall postulate the formula
+\begin{axiom}\index{Axioms}
+ \begin{displaymath}
+ (a + b) c < a c + b c.
+ \end{displaymath}
+\end{axiom}
+
+This, combined with the converse formula, produces the equality
+\begin{displaymath}
+ (a+b)c = ac+bc
+\end{displaymath}
+which we shall call briefly the \emph{distributive law.}
+
+From this may be directly deduced the formula
+\begin{displaymath}
+ (a + b)(c + d) = ac + bc + ad + bd,
+\end{displaymath}
+and consequently the second formula of the distributive law,
+\begin{displaymath}
+ (a + c) (b + c) = ab + c.
+\end{displaymath}
+For
+\begin{displaymath}
+ (a + c) (b + c) = ab + ac + bc + c,
+\end{displaymath}
+and, by the law of absorption,
+\begin{displaymath}
+ ac + bc + c = c.
+\end{displaymath}
+
+This second formula implies the inclusion cited above,
+\begin{displaymath}
+ (a + c) (b + c) < ab + c,
+\end{displaymath}
+which thus is shown to be proved.
+
+\emph{Corollary}.---We have the equality
+\begin{displaymath}
+ ab + ac + bc = (a + b) (a + c) (b + c),
+\end{displaymath}
+for
+\begin{displaymath}
+ (a + b) (a + c) (b + c) = (a + bc) (b + c) = ab + ac + bc.
+\end{displaymath}
+
+It will be noted that the two members of this equality
+differ only in having the signs of multiplication and addition
+transposed (compare \S\ref{ch:14}).
+
+\section{Definition of 0 and 1}\label{ch:13}
+
+We shall now define and introduce into the logical calculus two
+special terms which we shall designate by 0 and by 1, because of some
+formal analogies that they present with the zero and unity of
+arithmetic. These two terms are formally defined by the two following
+principles which affirm or postulate their existence.
+
+\begin{axiom}\index{Axioms}
+ There is a term~0 such that whatever value
+ may be given to the term~$x$, we have
+ \begin{displaymath}
+ 0 < x.
+ \end{displaymath}
+\end{axiom}
+
+\begin{axiom}\index{Axioms}
+ There is a term~1 such that whatever value
+ may be given to the term~$x$, we have
+ \begin{displaymath}
+ x < 1.
+ \end{displaymath}
+\end{axiom}
+
+It may be shown that each of the terms thus defined is
+unique; that is to say, if a second term possesses the same
+property it is equal to (identical with) the first.
+
+The two interpretations of these terms give rise to paradoxes which we
+shall not stop to elucidate here, but which will be justified by the
+conclusions of the theory.\footnote{Compare the
+ author's\index{Couturat} \emph{Manuel de Logistique}, Chap.~I., \S
+ 8, Paris, 1905 [This work, however, did not appear].}
+
+C.~I.: 0~denotes the class contained in every class; hence it is
+the ``null'' or ``void'' class which contains no element (Nothing
+or Naught), 1~denotes the class which contains all classes; hence
+it is the totality of the elements which are contained within it.
+It is called, after \author{}{Boole,} the ``universe of
+discourse'' or simply the ``whole''.
+
+P.~I.: 0~denotes the proposition which implies every proposition;
+it is the ``false'' or the ``absurd'', for it implies
+notably all pairs of contradictory propositions,%
+\index{Contradictory!propositions} 1~denotes
+the proposition which is implied in every proposition; it is
+the ``true'', for the false may imply the true whereas the true
+can imply only the true.
+
+By definition we have the following inclusions
+\begin{displaymath}
+ 0 < 0,\quad 0 < 1,\quad 1 < 1,
+\end{displaymath}
+the first and last of which, moreover, result from the principle of
+identity. It is important to bear the second in mind.
+
+C.~I.: The null class is contained in the \emph{whole}.\footnote{The
+ rendering ``Nothing is everything'' must be avoided.}
+
+P.~I.: The false implies the true.
+
+By the definitions of~0 and~1 we have the equivalences
+\begin{displaymath}
+ (a < 0) = (a = 0),\quad (1 < a) = (a = 1),
+\end{displaymath}
+since we have
+\begin{displaymath}
+ 0 < a,\quad a < 1
+\end{displaymath}
+whatever the value of~$a$.
+
+Consequently the principle of composition%
+\index{Composition!Principle of} gives rise to
+the two following corollaries:
+\begin{gather*}
+ (a = 0) (b = 0) = (a + b = 0),\\
+ (a = 1) (b = 1) = (ab = 1).
+\end{gather*}
+
+Thus we can combine two equalities having~0 for a second member by
+adding their first members, and two equalities having~1 for a
+second member by multiplying their first members.
+
+Conversely, to say that a sum is ``null'' [zero] is to say that
+each of the summands is null; to say that a product is equal
+to 1 is to say that each of its factors is equal to 1.
+
+Thus we have
+\begin{align*}
+ (a + b = 0) &< (a = 0),\\
+ (ab = 1) &< (a = 1),
+\end{align*}
+and more generally (by the principle of the syllogism)
+\begin{align*}
+ (a < b) (b = 0) &< (a = 0),\\
+ (a < b) (a = 1) &< (b = 1).
+\end{align*}
+
+It will be noted that we can not conclude from these the
+equalities $ab = 0$ and $a + b = 1$. And indeed in the conceptual
+interpretation the first equality denotes that the part
+common to the classes~$a$ and~$b$ is null; it by no means
+follows that either one or the other of these classes is null.
+The second denotes that these two classes combined form
+the whole; it by no means follows that either one or the
+other is equal to the whole.
+
+The following formulas comprising the rules for the calculus
+of~0 and~1, can be demonstrated:
+\begin{gather*}
+ a \times 0 = 0, \quad a + 1 = 1,\\
+ a + 0 = a, \quad a \times 1 = a.
+\end{gather*}
+
+For
+\begin{align*}
+ (0 < a) &= (0 = 0 \times a) = (a + 0 = a),\\
+ (a < 1) &= (a = a \times 1) = (a + 1 = 1).
+\end{align*}
+
+Accordingly it does not change a term to add~0 to it or to multiply it
+by~1. We express this fact by saying that 0~is the \emph{modulus} of
+addition and~1 the \emph{modulus} of multiplication.%
+\index{Addition!and multiplication!Modulus of}
+
+On the other hand, the product of any term whatever by~0 is 0~and the
+sum of any term whatever with 1~is~1.
+
+These formulas justify the following interpretation of the
+two terms:
+
+C.~I.: The part common to any class whatever and to the
+null class is the null class; the sum of any class whatever
+and of the whole is the whole. The sum of the null class and
+of any class whatever is equal to the latter; the part common
+to the whole and any class whatever is equal to the latter.
+
+P.~I.: The simultaneous affirmation of any proposition
+whatever and of a false proposition is equivalent to the latter
+(\emph{i.e.}, it is false); while their alternative affirmation%
+\index{Alternative!affirmation} is equal to the former. The
+simultaneous affirmation of any proposition whatever and of a true
+proposition is equivalent to the former; while their alternative
+affirmation is equivalent to the latter (\emph{i.e.}, it is true).
+
+\emph{Remark.}---If we accept the four preceding formulas as
+axioms, because of the proof afforded by the double interpretation,
+we may deduce from them the paradoxical formulas
+\begin{displaymath}
+ 0 < x, \text{ and } x < 1,
+\end{displaymath}
+by means of the equivalences established above,
+\begin{displaymath}
+ (a - ab) = (a < b) = (a + b = b).
+\end{displaymath}
+
+\section{The Law of Duality}\label{ch:14}
+We have proved that a perfect
+symmetry exists between the formulas relating to multiplication
+and those relating to addition. We can pass from one class
+to the other by interchanging the signs of addition and
+multiplication, on condition that we also interchange the
+terms~0 and~1 and reverse the meaning of the sign~< (or
+transpose the two members of an inclusion). This symmetry, or
+\emph{duality} as it is called, which exists in principles and definitions,
+must also exist in all the formulas deduced from them as
+long as no principle or definition is introduced which would
+overthrow them. Hence a true formula may be deduced
+from another true formula by transforming it by the principle
+of duality; that is, by following the rule given above. In its
+application the \emph{law of duality} makes it possible to replace
+two demonstrations by one. It is well to note that this law
+is derived from the definitions of addition and multiplication
+(the formulas for which are reciprocal by duality)
+and not, as is often thought%
+\footnote{\label{fn:boole}\author{}{Boole} thus derives it
+(\emph{Laws of Thought}, London 1854, Chap.~III, Prop.~IV).}, from
+the laws of negation which have not yet been stated. We shall see
+that these laws possess the same property and consequently
+preserve the duality, but they do not originate it; and duality
+would exist even if the idea of negation were not introduced. For
+instance, the equality (\S\ref{ch:12})
+\begin{displaymath}
+ ab + ac + bc = (a + b) (a + c) (b + c)
+\end{displaymath}
+is its own reciprocal by duality, for its two members are
+transformed into each other by duality.
+
+It is worth remarking that the law of duality is only
+applicable to primary propositions. We call [after \author{}{Boole}]
+those propositions \emph{primary} which contain but one copula
+($<$ or $=$). We call those propositions \emph{secondary} of which
+both members (connected by the copula $<$ or $=$) are primary
+propositions, and so on. For instance, the principle of
+identity and the principle of simplification are primary propositions,
+while the principle of the syllogism and the principle
+of composition are secondary propositions.
+
+\section{Definition of Negation}\label{ch:15}
+
+The introduction of the terms 0 and 1 makes it possible for us to
+define \emph{negation}. This is a ``uni-nary'' operation which
+transforms a single term into another term called its
+\emph{negative}.\footnote{[In French] the same word \emph{negation}
+ denotes both the operation and its result, which becomes equivocal.
+ The result ought to be denoted by another word, like [the English]
+ ``negative''. Some authors say, ``supplementary'' or ``supplement'',
+ [\emph{e.g.} \author{}{Boole} and \author{}{Huntington}], Classical logic
+ makes use of the term ``contradictory'' especially for
+ propositions.} The negative of $a$ is called not-$a$ and is written
+$a'$.\footnote{We adopt here the notation of \author{}{MacColl;}
+ \author{}{Schröder} indicates not-$a$ by $a_1$ which prevents the
+ use of indices and obliges us to express them as exponents. The
+ notation $a'$ has the advantage of excluding neither indices nor
+ exponents. The notation $\bar{a}$ employed by many authors is
+ inconvenient for typographical reasons. When the negative affects a
+ proposition written in an explicit form (with a copula) it is
+ applied to the copula $<$ or $=$) by a vertical bar ($\nless$) or
+ $\not=$). The accent can be considered as the indication of a
+ vertical bar applied to letters.} Its formal definition implies the
+following postulate of existence\footnote{\author{}{Boole} follows
+ Aristotle\index{Aristotle} in usually calling the law of duality the
+principle of contradiction%
+\index{Contradiction!Principle of} ``which affirms that it is
+impossible for any being to possess a quality and at the same time
+not to possess it''. He writes it in the form of an equation of
+the second degree, $x - x^{2} = 0$, or $x (1 - x) = 0$ in which $1
+- x$ expresses the universe less $x$, or not $x$. Thus he regards
+the law of duality as derived from negation as stated in
+note~\ref{fn:boole} above.}:
+\begin{axiom}\index{Axioms}
+ Whatever the term~$a$ may be, there is also a
+ term~$a'$ such that we have at the same time
+ \begin{displaymath}
+ aa' = 0, \quad a + a' = 1.
+ \end{displaymath}
+
+ It can be proved by means of the following \emph{lemma} that if
+ a term so defined exists it is unique:
+
+ If at the same time
+ \begin{displaymath}
+ ac = bc, \quad a + c = b + c,
+ \end{displaymath}
+ then
+ \begin{displaymath}
+ a = b.
+ \end{displaymath}
+\end{axiom}
+
+\emph{Demonstration.}---Multiplying both members of the second
+premise by~$a$, we have
+\begin{displaymath}
+ a + ac = ab + ac.
+\end{displaymath}
+
+Multiplying both members by~$b$,
+\begin{displaymath}
+ ab + bc = b + bc.
+\end{displaymath}
+
+By the first premise,
+\begin{displaymath}
+ ab + ac = ab + bc.
+\end{displaymath}
+
+Hence
+\begin{displaymath}
+ a + ac = b + bc,
+\end{displaymath}
+which by the law of absorption may be reduced to
+\begin{displaymath}
+ a = b.
+\end{displaymath}
+
+\emph{Remark.}---This demonstration rests upon the direct distributive
+law. This law cannot, then, be demonstrated by means
+of negation, at least in the system of principles which we are
+adopting, without reasoning in a circle.
+
+This lemma being established, let us suppose that the same
+term~$a$ has two negatives; in other words, let $a'_{1}$ and
+$a'_{2}$ be two terms each of which by itself satisfies the
+conditions of the definition. We will prove that they are equal.
+Since, by hypothesis,
+\begin{gather*}
+ aa'_1 = 0, \quad a + a'_1 = 1,\\
+ aa'_2 = 0, \quad a + a'_2 = 1,
+\end{gather*}
+we have
+\begin{displaymath}
+ aa'_1 = a a'_2, \quad a + a'_1 = a + a'_2;
+\end{displaymath}
+whence we conclude, by the preceding lemma, that
+\begin{displaymath}
+ a'_1 = a'_2.
+\end{displaymath}
+
+We can now speak of \emph{the} negative of a term as of a unique
+and well-defined term.
+
+The \emph{uniformity} of the operation of negation may be expressed
+in the following manner:
+
+If $a = b$, then also $a' = b'$. By this proposition, both
+members of an equality in the logical calculus may be
+``denied''.
+
+\section{The Principles of Contradiction and of Excluded Middle}
+\label{ch:16}\index{Contradiction!Principle of|(}
+By definition, a term and its negative verify the
+two formulas
+\begin{displaymath}
+ aa' = 0, \quad a + a' = 1,
+\end{displaymath}
+which represent respectively the \emph{principle of contradiction} and
+the \emph{principle of excluded middle}.%
+\footnote{As \author{}{Mrs. Ladd-Franklin} has truly remarked
+(\author{}{Baldwin,}\index{Baldwin} \emph{Dictionary of Philosophy
+and Psychology}, article ``Laws of Thought''), the principle of
+\emph{contradiction} is not sufficient to define
+\emph{contradictories}; the principle of excluded middle must be
+added which equally deserves the name of principle of
+contradiction. This is why \author{}{Mrs. Ladd-Franklin} proposes
+to call them respectively the \emph{principle of exclusion} and
+the \emph{principle of
+exhaustion}, inasmuch as, according to the first, two contradictory terms%
+\index{Contradictory!terms}
+are \emph{exclusive} (the one of the other); and, according to the second, they
+are \emph{exhaustive} (of the universe of discourse).}
+
+C.~I.: 1. The classes~$a$ and~$a'$ have nothing in common;
+in other words, no element can be at the same time both~$a$
+and not-$a$.
+
+2. The classes~$a$ and~$a'$ combined form the whole; in
+other words, every element is either~$a$ or not-$a$.
+
+P.~I.: 1. The simultaneous affirmation of the propositions
+$a$ and not-$a$ is false; in other words, these two propositions
+cannot both be true at the same time.
+
+2. The alternative affirmation%
+\index{Alternative!affirmation} of the propositions~$a$ and
+not-$a$ is true; in other words, one of these two propositions
+must be true.
+
+Two propositions are said to be \emph{contradictory} when one is
+the negative of the other; they cannot both be true or false
+at the same time. If one is true the other is false; if one
+is false the other is true.
+
+This is in agreement with the fact that the terms 0 and 1
+are the negatives of each other; thus we have
+\begin{displaymath}
+ 0 \times 1 = 0, \quad 0 + 1 = 1.
+\end{displaymath}
+
+Generally speaking, we say that two terms are \emph{contradictory}
+when one is the negative of the other.%
+\index{Contradiction!Principle of|)}
+
+\section{Law of Double Negation}\label{ch:17}
+Moreover this reciprocity is general: if a term~$b$ is the negative of the
+term~$a$, then the term~$a$ is the negative of the term~$b$. These two
+statements are expressed by the same formulas
+\begin{displaymath}
+ ab = 0, \quad a + b = 1,
+\end{displaymath}
+and, while they unequivocally determine $b$ in terms of $a$, they likewise
+determine $a$ in terms of $b$. This is due to the symmetry of these
+relations, that is to say, to the commutativity\index{Commutativity} of
+multiplication and addition. This reciprocity is expressed by the \emph{law
+ of double negation}
+\begin{displaymath}
+ (a')' = a,
+\end{displaymath}
+which may be formally proved as follows: $a'$ being by hypothesis
+the negative of $a$, we have
+\begin{displaymath}
+ a a' = 0, \quad a + a' = 1.
+\end{displaymath}
+
+On the other hand, let $a''$ be the negative of $a'$; we have,
+in the same way,
+\begin{displaymath}
+ a' a'' = 0, \quad a' + a'' = 1.
+\end{displaymath}
+
+But, by the preceding lemma, these four equalities involve
+the equality
+\begin{displaymath}
+ a = a''.
+\end{displaymath}
+Q. E. D.
+
+This law may be expressed in the following manner:
+
+If $b = a'$, we have $a = b'$, and conversely, by symmetry.
+
+This proposition makes it possible, in calculations, to
+transpose the negative from one member of an equality to
+the other.
+
+The law of double negation makes it possible to conclude
+the equality of two terms from the equality of their negatives
+(if $a' = b'$ then $a = b$), and therefore to cancel the negation
+of both members of an equality.
+
+From the characteristic formulas of negation together with
+the fundamental properties of~0 and~1, it results that every
+product which contains two contradictory factors is null, and
+that every sum which contains two contradictory summands
+is equal to~1.
+
+In particular, we have the following formulas:
+\begin{displaymath}
+ a = ab + ab', \quad a = (a + b) (a + b'),
+\end{displaymath}
+which may be demonstrated as follows by means of the
+distributive law:
+\begin{gather*}
+ a = a \times 1 = a (b + b') = ab + ab', \\
+ a = a + 0 = a + bb' = (a + b) (a + b').
+\end{gather*}
+
+These formulas indicate the principle of the method of
+development which we shall explain in detail later (\S\S\ref{ch:21} sqq.)
+
+\section{Second Formulas for Transforming Inclusions
+into Equalities}\label{ch:18}
+We can now establish two very important
+equivalences between inclusions and equalities:
+\begin{displaymath}
+ (a < b) = (ab' = 0), \quad (a < b) = (a' + b = 1).
+\end{displaymath}
+
+\emph{Demonstration}.---1. If we multiply the two members of the
+inclusion $a < b$ by $b'$ we have
+\begin{displaymath}
+ (ab' < bb') = (ab' < 0) = (ab' = 0).
+\end{displaymath}
+
+2. Again, we know that
+\begin{displaymath}
+ a = ab + ab'.
+\end{displaymath}
+
+Now if $ab' = 0$,
+\begin{displaymath}
+ a = ab + 0 = ab.
+\end{displaymath}
+
+On the other hand: 1. Add $a'$ to each of the two members
+of the inclusion $a < b$; we have
+\begin{displaymath}
+ (a' + a < a' + b) = (1 < a' + b) = a' + b = 1).
+\end{displaymath}
+
+2. We know that
+\begin{displaymath}
+ b = (a + b)(a' + b).
+\end{displaymath}
+
+Now, if $a' + b = 1$,
+\begin{displaymath}
+ b = (a + b) \times 1 = a + b.
+\end{displaymath}
+
+By the preceding formulas, an inclusion can be transformed
+at will into an equality whose second member is either 0 or 1.
+Any equality may also be transformed into an equality of
+this form by means of the following formulas:
+\begin{displaymath}
+ (a = b) = (ab' + a'b = 0), \quad (a = b) = [(a + b')(a' + b) = 1].
+\end{displaymath}
+
+\emph{Demonstration:}
+\begin{gather*}
+ (a = b) = (a < b)(b < a) = (ab' = 0)(a'b = 0) = (ab' + a'b = 0),\\
+ (a = b) = (a < b)(b < a) = (a' + b = 1)(a + b' = 1) =
+ [(a' + b')(a' + b) = 1].
+\end{gather*}
+
+Again, we have the two formulas
+\begin{displaymath}
+ (a = b) = [(a + b)(a' + b') = 0], \quad (a = b) = (ab + a'b' = 1),
+\end{displaymath}
+which can be deduced from the preceding formulas by performing
+the indicated multiplications (or the indicated additions)
+by means of the distributive law.
+
+\section{The Law of Contraposition}
+\label{ch:19}
+
+We are now able to demonstrate the \emph{law of contraposition},%
+\index{Contraposition!Law of}
+\begin{displaymath}
+ (a < b) = (b' < a').
+\end{displaymath}
+
+\emph{Demonstration}.---By the preceding formulas, we have
+\begin{displaymath}
+ (a < b) = (ab = 0) = (b' < a').
+\end{displaymath}
+
+Again, the law of contraposition may take the form
+\begin{displaymath}
+ (a < b') = (b < a'),
+\end{displaymath}
+which presupposes the law of double negation. It may be
+expressed verbally as follows: ``Two members of an inclusion
+may be interchanged on condition that both are denied''.
+
+C.~I.: ``If all $a$ is $b$, then all not-$b$ is not-$a$, and conversely''.
+
+P.~I.: ``If $a$ implies $b$, not-$b$ implies not-$a$ and conversely'';
+in other words, ``If $a$ is true $b$ is true'', is equivalent to
+saying, ``If $b$ is false, $a$ is false''.
+
+This equivalence is the principle of the \emph{reductio ad absurdum}
+(see hypothetical arguments, \emph{modus tollens}, \S\ref{ch:58}).
+
+\section{Postulate of Existence}\label{ch:20}
+One final axiom may be
+formulated here, which we will call the \emph{postulate of existence}:
+\begin{axiom}\label{axiom:IX}\index{Axioms}
+ \begin{displaymath}
+ 1 \nless 0
+ \end{displaymath}
+\end{axiom}
+whence may be also deduced $1 \neq 0$.
+
+In the conceptual interpretation (C.~I.) this axiom means
+that the universe of discourse is not null, that is to say, that
+it contains some elements, at least one. If it contains but
+one, there are only two classes possible, $1$ and $0$. But even
+then they would be distinct, and the preceding axiom would
+be verified.
+
+In the propositional interpretation (P.~I.) this axiom signifies
+that the true and false are distinct; in this case, it bears
+the mark of evidence and necessity. The contrary
+proposition, $1 = 0$, is, consequently, the type of \emph{absurdity}%
+\index{Absurdity!Type of}
+(of the formally false proposition) while the propositions $0 = 0$,
+and $1 = 1$ are types of \emph{identity} (of the formally true proposition).
+Accordingly we put
+\begin{displaymath}
+ (1 = 0) = 0, \quad (0 = 0) = (l = 1) = 1.
+\end{displaymath}
+
+More generally, every equality of the form
+\begin{displaymath}
+ x = x
+\end{displaymath}
+is equivalent to one of the identity-types; for, if we reduce
+this equality so that its second member will be~$0$ or~$1$, we find
+\begin{displaymath}
+ (xx' + xx' = 0) = (0 = 0), \quad (xx + x'x' = 1) = (1 = 1).
+\end{displaymath}
+
+On the other hand, every equality of the form
+\begin{displaymath}
+ x = x'
+\end{displaymath}
+is equivalent to the absurdity-type, for we find by the same
+process,
+\begin{displaymath}
+ (xx + x'x' = 0) = (1 = 0), \quad (xx' + xx' = 1) = (0 = 1).
+\end{displaymath}
+
+\section{The Development of~0 and of~1}\label{ch:21}
+Hitherto we
+have met only such formulas as directly express customary
+modes of reasoning and consequently offer direct evidence.
+
+We shall now expound theories and methods which depart
+from the usual modes of thought and which constitute more
+particularly the algebra of logic in so far as it is a formal
+and, so to speak, automatic method of an absolute universality
+and an infallible certainty, replacing reasoning by calculation.
+
+The fundamental process of this method is \emph{development}.
+Given the terms $a, b, c \ldots$ (to any finite number), we can
+develop 0 or 1 with respect to these terms (and their negatives)
+by the following formulas derived from the distributive law:
+\begin{align*}
+ 0 &= aa',\\
+ 0 &= aa' + bb' = (a + b) (a + b') (a' + b) (a' + b'),\\
+ 0 &= aa' + bb' + cc' =
+ \begin{aligned}[t]
+ (a &+ b + c) (a + b + c') (a + b' + c)\\
+ &\times (a + b' + c') (a' + b + c)\\
+ &\times (a' + b + c') (a' + b' + c) (a' + b' + c');\\
+ \end{aligned}\\
+ 1 &= a + a',\\
+ 1 &= (a + a') (b + b') = ab + ab' + a'b + a'b',\\
+ 1 &= (a + a') (b + b') (c + c')
+ \begin{aligned}[t]
+ &= abc + abc' + ab'c + ab'c'\\
+ &+ a'bc + a'bc' + a'b'c + a'b'c';\\
+ \end{aligned}
+\end{align*}
+and so on. In general, for any number~$n$ of simple terms;
+0~will be developed in a product containing $2^n$ factors, and
+1~in a sum containing $2^n$ summands. The factors of zero
+comprise all possible additive combinations, and the summands
+of~1 all possible multiplicative combinations of the~$n$ given
+terms and their negatives, each combination comprising~$n$
+different terms and never containing a term and its negative
+at the same time.
+
+The summands of the development of~1 are what \author{}{Boole} called
+the \emph{constituents}\index{Constituents} (of the universe of
+discourse). We may equally well call them, with
+\author{}{Poretsky,}\footnote{See the Bibliography, page xiv.} the
+\emph{minima} of discourse, because they are the smallest classes
+into which the universe of discourse is divided with reference to
+the~$n$ given terms. In the same way we shall call the factors of
+the development of~0 the \emph{maxima} of discourse, because they
+are the largest classes that can be determined in the universe of
+discourse by means of the $n$ given terms.
+
+\section{Properties of the Constituents}
+\label{ch:22}\index{Constituents!Properties of} The constituents
+or minima of discourse possess two properties characteristic of
+contradictory terms (of which they are a generalization); they are
+\emph{mutually exclusive}, \emph{i.e.}, the product of any two of
+them is~0; and they are \emph{collectively exhaustive},
+\emph{i.e.}, the sum of all ``exhausts'' the universe of
+discourse. The latter property is evident from the preceding
+formulas. The other results from the fact that any two
+constituents differ at least in the ``sign'' of one of the terms
+which serve as factors, \emph{i.e.}, one contains this term as a
+factor and the other the negative of this term. This is enough, as
+we know, to ensure that their product be null.
+
+The maxima of discourse possess analogous and correlative
+properties; their combined product is equal to~0, as we have
+seen; and the sum of any two of them is equal to~1, inasmuch
+as they differ in the sign of at least one of the terms which
+enter into them as summands.
+
+For the sake of simplicity, we shall confine ourselves, with
+\author{}{Boole} and \author{}{Schröder,} to the study of the constituents or
+minima of discourse, \emph{i.e.}, the developments of~1. We shall
+leave to the reader the task of finding and demonstrating the
+corresponding theorems which concern the maxima of discourse or
+the developments of~0.
+
+\section{Logical Functions}\label{ch:23}
+We shall call a \emph{logical function}
+any term whose expression is complex, that is, formed of
+letters which denote simple terms together with the signs of
+the three logical operations.\footnote{In this algebra the logical function is analogous to the \emph{integral
+function} of ordinary algebra, except that it has no powers beyond
+the first.}
+
+A logical function may be considered as a function of all
+the terms of discourse, or only of some of them which may
+be regarded as unknown or variable and which in this case
+are denoted by the letters $x, y, z$. We shall represent a
+function of the variables or unknown quantities, $x, y, z$, by
+the symbol $f(x, y, z)$ or by other analogous symbols, as in
+ordinary algebra. Once for all, a logical function may be
+considered as a function of any term of the universe of discourse,
+whether or not the term appears in the explicit expression
+of the function.
+
+\section{The Law of Development}\label{ch:24}
+This being established,
+we shall proceed to develop a function $f(x)$ with respect to~$x$.
+Suppose the problem solved, and let
+\begin{displaymath}
+ ax + bx'
+\end{displaymath}
+be the development sought. By hypothesis we have the
+equality
+\begin{displaymath}
+ f(x) = ax + bx'
+\end{displaymath}
+for all possible values of~$x$. Make $x = 1$ and consequently
+$x' = 0$. We have
+\begin{displaymath}
+ f(1) = a.
+\end{displaymath}
+
+Then put $x = 0$ and $x' = 1$; we have
+\begin{displaymath}
+ f(0) = b.
+\end{displaymath}
+
+These two equalities determine the coefficients~$a$ and~$b$ of
+the development which may then be written as follows:
+\begin{displaymath}
+ f(x) = f(1)x + f(0)x',
+\end{displaymath}
+in which $f(1)$, $f(0)$ represent the value of the function $f(x)$
+when we let $x = 1$ and $x = 0$ respectively.
+
+\emph{Corollary.}---Multiplying both members of the preceding
+equalities by~$x$ and~$x'$ in turn, we have the following pairs
+of equalities (\author{}{MacColl}):
+\begin{alignat*}{2}
+ xf(x) &= ax &\quad x'f(x) &= bx'\\
+ xf(x) &= xf(1), &\quad x'f(x) &= x'f(0).
+\end{alignat*}
+
+Now let a function of two (or more) variables be developed with
+respect to the two variables~$x$ and~$y$. Developing $f(x, y)$
+first with respect to~$x$, we find
+\begin{displaymath}
+ f(x, y) = f(1, y)x + f(0, y)x'.
+\end{displaymath}
+
+Then, developing the second member with respect to~$y$,
+we have
+\begin{displaymath}
+ f(x, y) = f(1, 1)xy + f(1, 0)xy' + f(0, 1)x'y + f(0, 0)x'y'
+\end{displaymath}
+
+This result is symmetrical with respect to the two variables,
+and therefore independent of the order in which the developments
+with respect to each of them are performed.
+
+In the same way we can obtain progressively the development
+of a function of $3, 4, \ldots\ldots$, variables.
+
+The general law of these developments is as follows:
+
+To develop a function with respect to $n$ variables, form all
+the constituents of these $n$ variables and multiply each of
+them by the value assumed by the function when each of
+the simple factors of the corresponding constituent is equated
+to~1 (which is the same thing as equating to~0 those factors
+whose negatives appear in the constituent).
+
+When a variable with respect to which the development is
+made, $y$~for instance, does not appear explicitly in the
+function ($f(x)$ for instance), we have, according to the
+general law,
+\begin{displaymath}
+ f(x) = f(x)y + f(x)y'.
+\end{displaymath}
+
+In particular, if $a$ is a constant term, independent of the
+variables with respect to which the development is made,
+we have for its successive developments,
+\begin{align*}
+ a &= ax + ax',\\
+ a &= axy + axy' + ax'y + ax'y',\\
+ a &= \begin{aligned}[t]
+ axyz &+ axyz' + axy'z + axy'z' + ax'yz + ax'yz' + ax'y'z\\
+ &+ ax'y'z'.\footnotemark\index{Classification of dichotomy}
+ \end{aligned}
+\end{align*}
+\footnotetext{These formulas express the method of classification
+ by dichotomy.} and so on. Moreover these formulas may be directly
+obtained by multiplying by~$a$ both members of each development
+of~1.
+
+\emph{Cor}. 1. We have the equivalence
+\begin{displaymath}
+ (a + x') (b + x) = ax + bx + ab = ax + bx'.
+\end{displaymath}
+
+For, if we develop with respect to $x$, we have
+\begin{displaymath}
+ ax + bx' + abx + abx' = (a + ab)x + (b + ab)x' = ax + bx'.
+\end{displaymath}
+
+\emph{Cor}. 2. We have the equivalence
+\begin{displaymath}
+ ax + bx' + c = (a + c)x + (b + c)x'.
+\end{displaymath}
+
+For if we develop the term $c$ with respect to $x$, we find
+\begin{displaymath}
+ ax + bx' + cx + cx' = (a + c)x + (b + c)x'.
+\end{displaymath}
+
+Thus, when a function contains terms (whose sum is
+represented by~$c$) independent of~$x$, we can always reduce it
+to the developed form $ax + bx'$ by adding~$c$ to the coefficients
+of both~$x$ and~$x'$. Therefore we can always consider a
+function to be reduced to this form.
+
+In practice, we perform the development by multiplying
+each term which does not contain a certain letter ($x$ for
+instance) by $(x + x')$ and by developing the product according
+to the distributive law. Then, when desired, like terms may
+be reduced to a single term.
+
+\section{The Formulas of De Morgan}
+\label{ch:25}\index{De Morgan!Formulas of|(}
+
+\emph{In any development of 1, the sum of a certain number of
+ constituents is the negative of the sum of all the others.}
+
+For, by hypothesis, the sum of these two sums is equal to~1, and their
+product is equal to~0, since the product of two different constituents
+is zero.
+
+From this proposition may be deduced the formulas of
+\author{}{De Morgan:}
+\begin{displaymath}
+ (a + b)' = a'b', \quad (ab)' = a' + b'.
+\end{displaymath}
+
+\emph{Demonstration}.---Let us develop the sum $(a + b)$:
+\begin{displaymath}
+ a + b = ab + ab' + ab + a'b = ab + ab' + a'b.
+\end{displaymath}
+
+Now the development of~1 with respect to~$a$ and~$b$ contains
+the three terms of this development plus a fourth term $a'b'$.
+This fourth term, therefore, is the negative of the sum of the
+other three.
+
+We can demonstrate the second formula either by a correlative
+argument (\emph{i.e.}, considering the development of 0 by
+factors) or by observing that the development of $(a' + b')$,
+\begin{displaymath}
+ a'b + ab' + a'b',
+\end{displaymath}
+differs from the development of 1 only by the summand $ab$.
+
+How \author{}{De Morgan's} formulas may be generalized is now
+clear; for instance we have for a sum of three terms,
+\begin{displaymath}
+ a + b + c = abc + abc' + ab'c + ab'c'+ a'bc + a'bc' + a'b'c.
+\end{displaymath}
+
+This development differs from the development of 1 only
+by the term $a'b'c'$. Thus we can demonstrate the formulas
+\begin{displaymath}
+ (a + b + c)' = a'b'c', \quad (abc)' = a' + b' + c',
+\end{displaymath}
+which are generalizations of \author{}{De Morgan's} formulas.
+
+The formulas of \author{}{De Morgan} are in very frequent use in
+calculation, for they make it possible to perform the negation
+of a sum or a product by transferring the negation to the
+simple terms: the negative of a sum is the product of the
+negatives of its summands; the negative of a product is the
+sum of the negatives of its factors.
+
+These formulas, again, make it possible to pass from a primary
+proposition to its correlative proposition by duality, and to
+demonstrate their equivalence. For this purpose it is only necessary
+to apply the law of contraposition%
+\index{Contraposition!Law of} to the given proposition, and then
+to perform the negation of both members.
+
+\emph{Example:}
+\begin{displaymath}
+ ab + ac + bc = (a + b) (a + c) (b + c).
+\end{displaymath}
+
+\emph{Demonstration:}
+\begin{align*}
+ (ab + ac + bc)' &= [(a + b) (a + c) (b + c)], \\
+ (ab)'(ac)'(bc)' &= (a + b)'+(a + c)' + (b + c)', \\
+ (a' + b') (a'+ c') (b' + c') &= a'b' + a'c' + b'c'.
+\end{align*}
+
+Since the simple terms, $a, b, c$, may be any terms, we may
+suppress the sign of negation by which they are affected, and
+obtain the given formula.
+
+Thus \author{}{De Morgan's} formulas furnish a means by which to
+find or to demonstrate the formula correlative to another; but, as
+we have said above (\S\ref{ch:14}), they are not the basis of
+this correlation.%
+\index{De Morgan!Formulas of|)}
+
+\section{Disjunctive Sums}\label{ch:26}
+By means of development we can transform any sum into a
+\emph{disjunctive} sum, \emph{i.e.}, one in which each product of
+its summands taken two by two is zero. For, let $(a + b + c)$ be a
+sum of which we do not know whether or not the three terms are
+disjunctive; let us assume that they are not. Developing, we have:
+\begin{displaymath}
+ a + b + c = abc + abc' + ab'c + ab'c' + a'bc + a'bc' + a'b'c.
+\end{displaymath}
+
+Now, the first four terms of this development constitute
+the development of a with respect to $b$ and $c$; the two
+following are the development of $a'b$ with respect to $c$. The
+above sum, therefore, reduces to
+\begin{displaymath}
+ a + a'b + a'b'c,
+\end{displaymath}
+and the terms of this sum are disjunctive like those of the
+preceding, as may be verified. This process is general and,
+moreover, obvious. To enumerate without repetition all the
+$a$'s, all the $b$'s, and all the $c$'s, etc., it is clearly sufficient to
+enumerate all the $a$'s, then all the $b$'s which are not $a$'s, and
+then all the $c$'s which are neither $a$'s nor $b$'s, and so on.
+
+It will be noted that the expression thus obtained is not
+symmetrical, since it depends on the order assigned to the
+original summands. Thus the same sum may be written:
+\begin{displaymath}
+ b + ab' + a'b'c, \quad c + ac' + a'bc', \ldots.
+\end{displaymath}
+
+Conversely, in order to simplify the expression of a sum,
+we may suppress as factors in each of the summands (arranged
+in any suitable order) the negatives of each preceding summand.
+Thus, we may find a symmetrical expression for a
+sum. For instance,
+\begin{displaymath}
+ a + a'b = b + ab' = a + b.
+\end{displaymath}
+
+\section{Properties of Developed Functions}\label{ch:27}
+The practical
+utility of the process of development in the algebra of logic
+lies in the fact that developed functions possess the following
+property:
+
+The sum or the product of two functions developed with respect to
+the same letters is obtained simply by finding the sum or the
+product of their coefficients. The negative of a developed
+function is obtained simply by replacing the coefficients of its
+development by their negatives.
+
+We shall now demonstrate these propositions in the case
+of two variables; this demonstration will of course be of
+universal application.
+
+Let the developed functions be
+\begin{gather*}
+ a_1 xy + b_1 xy' + c_1 x'y + d_1 x'y',\\
+ a_2 xy + b_2 xy' + c_2 x'y + d_2 x'y'.
+\end{gather*}
+
+1. I say that their sum is
+\begin{displaymath}
+ (a_1 + a_2) xy + (b_1 + b_2) xy' + (c_1 + c_2) x'y + (d_1 + d_2) x'y'.
+\end{displaymath}
+
+This result is derived directly from the distributive law.
+
+2. I say that their product is
+\begin{displaymath}
+ a_1 a_2 xy + b_1 b_2 xy' + c_1 c_2 x'y + d_1 d_2 x'y',
+\end{displaymath}
+for if we find their product according to the general rule
+(by applying the distributive law), the products of two terms
+of different constituents will be zero; therefore there will remain
+only the products of the terms of the same constituent, and,
+as (by the law of tautology) the product of this constituent
+multiplied by itself is equal to itself, it is only necessary to
+obtain the product of the coefficients.
+
+3. Finally, I say that the negative of
+\begin{displaymath}
+ axy + bxy' + cx'y + dx'y'
+\end{displaymath}
+is
+\begin{displaymath}
+ a'xy + b'xy' + c'x'y + d'x'y'.
+\end{displaymath}
+
+In order to verify this statement, it is sufficient to prove
+that the product of these two functions is zero and that their
+sum is equal to 1. Thus
+\begin{gather*}
+ \begin{split}
+ (axy &+ bxy' + cx'y + dx'y') (a'xy + b'xy' + c'x'y + d'x'y')\\
+ &= (aa'xy + bb'xy' + cc'x'y + dd'x'y')\\
+ &= (0 \cdot xy + 0 \cdot xy' + 0 \cdot x'y + 0 \cdot x'y') = 0\\
+ (axy &+ bxy' + cx'y + dx'y') + (a'xy + b'xy' + c'x'y + d'x'y')\\
+ &= [(a + a') xy + (b + b') xy' + (c + c') x'y + (d + d') x'y']\\
+ &= (1xy + 1xy' + 1x'y + 1x'y') = 1.
+ \end{split}
+\end{gather*}
+
+\emph{Special Case}.---We have the equalities:
+\begin{align*}
+ (ab + a'b')' &= ab' + a'b,\\
+ (ab'+ a'b')' &= ab + a'b',
+\end{align*}
+which may easily be demonstrated in many ways; for instance, by
+observing that the two sums $(ab + a'b')$ and $(ab'+a'b)$ combined
+form the development of~1; or again by \emph{performing} the
+negation $(ab + a'b')'$ by means of \author{}{De Morgan's}
+formulas~(\S\ref{ch:25}).
+
+From these equalities we can deduce the following equality:
+\begin{displaymath}
+ (ab'+ a b = 0) = (ab + a'b'= 1),
+\end{displaymath}
+which result might also have been obtained in another way
+by observing that~(\S\ref{ch:18})
+\begin{displaymath}
+ (a = b) = (ab'+ a'b = 0) = [(a + b') (a'+ b) = 1],
+\end{displaymath}
+and by performing the multiplication indicated in the last
+equality.
+
+\textsc{Theorem}.---\emph{We have the following equivalences:}%
+\footnote{\author{}{W. Stanley Jevons,} \emph{Pure Logic}, 1864,
+p.~61.}
+\begin{displaymath}
+ (a = bc' + b'c) = (b = ac' + a'c) = (c = ab'+ a'b).
+\end{displaymath}
+
+For, reducing the first of these equalities so that its second
+member will be~0,
+\begin{align*}
+ a(bc + b'c') + a' (bc'+ b'c) &= 0,\\
+ abc + ab'c' + a'bc' + a'b'c &= 0.
+\end{align*}
+
+Now it is clear that the first member of this equality is
+symmetrical with respect to the three terms $a, b, c$. We may
+therefore conclude that, if the two other equalities which differ
+from the first only in the permutation of these three letters
+be similarly transformed, the same result will be obtained,
+which proves the proposed equivalence.
+
+\emph{Corollary}.---If we have at the same time the three inclusions:
+\begin{displaymath}
+ a < bc'+b'c, \quad b< ac'+a'c, \quad c< ab'+a'b.
+\end{displaymath}
+we have also the converse inclusion, an therefore the
+corresponding equalities
+\begin{displaymath}
+ a=bc'+b'c, \quad b=ac'+a'c, \quad c=ab'+a'b.
+\end{displaymath}
+
+For if we transform the given inclusions into equalities, we
+shall have
+\begin{displaymath}
+ abc + ab'c' = 0, \quad abc + a'bc' = 0, \quad abc + a'b'c = 0,
+\end{displaymath}
+whence, by combining them into a single equality,
+\begin{displaymath}
+ abc + ab'c' + a'bc' + a'b'c = 0.
+\end{displaymath}
+
+Now this equality, as we see, is equivalent to any one of
+the three equalities to be demonstrated.
+
+\section{The Limits of a Function}\label{ch:28}
+A term~$x$ is said to be
+\emph{comprised} between two given terms,~$a$ and~$b$, when it contains
+one and is contained in the other; that is to say, if we have,
+for instance,
+\begin{displaymath}
+ a < x, \quad x < b,
+\end{displaymath}
+which we may write more briefly as
+\begin{displaymath}
+ a < x < b.
+\end{displaymath}
+
+Such a formula is called a \emph{double inclusion}. When the
+term~$x$ is variable and always comprised between two
+constant terms $a$ and $b$, these terms are called the \emph{limits}
+of $x$. The first (contained in $x$) is called \emph{inferior limit}; the
+second (which contains $x$) is called the \emph{superior limit}.
+
+\author{}{Theorem.}---\emph{A developed function is comprised between the sum
+and the product of its coefficients.}
+
+We shall first demonstrate this theorem for a function of
+one variable,
+\begin{displaymath}
+ ax + bx'.
+\end{displaymath}
+
+We have, on the one hand,
+\begin{align*}
+ (ab < a) &< (abx < ax),\\
+ (ab < b) &< (abx' < bx').
+\end{align*}
+
+Therefore
+\begin{displaymath}
+ abx + abx' < ax + bx',
+\end{displaymath}
+or
+\begin{displaymath}
+ ab < ax + bx'.
+\end{displaymath}
+
+On the other hand,
+\begin{align*}
+ (a < a + b) &< [ax < (a + b)x],\\
+ (b < a + b) &< [bx' < (a + b)x'].
+\end{align*}
+
+Therefore
+\begin{displaymath}
+ ax + bx' < (a + b) (x + x'),
+\end{displaymath}
+or
+\begin{displaymath}
+ ax + bx' < a + b.
+\end{displaymath}
+
+To sum up,
+\begin{displaymath}
+ ab < ax + bx' < a + b.
+\end{displaymath}
+Q. E. D.
+
+\emph{Remark} 1. This double inclusion may be expressed in the
+following form:%
+\footnote{\author{}{Eugen Müller,} \emph{Aus der Algebra der
+Logik}, Art. II.}
+\begin{displaymath}
+ f(b) < f(x) < f(a).
+\end{displaymath}
+
+For
+\begin{gather*}
+f(a) = aa + ba' = a + b,\\
+f(b) = ab + bb' = ab.
+\end{gather*}
+
+But this form, pertaining as it does to an equation of one
+unknown quantity, does not appear susceptible of generalization,
+whereas the other one does so appear, for it is readily seen
+that the former demonstration is of general application.
+Whatever the number of variables $n$ (and consequently the
+number of constituents $2^{n}$) it may be demonstrated in exactly
+the same manner that the function contains the product of
+its coefficients and is contained in their sum. Hence the
+theorem is of general application.
+
+\emph{Remark} 2.---This theorem assumes that all the constituents
+appear in the development, consequently those that are wanting
+must really be present with the coefficient~0. In this case,
+the product of all the coefficients is evidently~0. Likewise
+when one coefficient has the value~1, the sum of all the
+coefficients is equal to~1.
+
+It will be shown later (\S\ref{ch:38}) that a function may reach
+both its limits, and consequently that they are its extreme
+values. As yet, however, we know only that it is always
+comprised between them.
+
+\section{Formula of Poretsky.%
+\protect\footnote{\author{}{Poretsky,} ``Sur les méthodes pour
+résoudre les égalités logiques''. (\emph{Bull. de la Soc.
+phys.-math. de Kazan}, Vol. II, 1884).}}\label{ch:29}
+
+We have the equivalence
+\begin{displaymath}
+ (x = ax + bx') = (b < x < a).
+\end{displaymath}
+
+\emph{Demonstration.}---First multiplying by~$x$ both members of
+the given equality [which is the first member of the entire
+secondary equality], we have
+\begin{displaymath}
+ x = ax,
+\end{displaymath}
+which, as we know, is equivalent to the inclusion
+\begin{displaymath}
+ x < a.
+\end{displaymath}
+
+Now multiplying both members by $x'$, we have
+\begin{displaymath}
+ 0 = bx',
+\end{displaymath}
+which, as we know, is equivalent to the inclusion
+\begin{displaymath}
+ b < x.
+\end{displaymath}
+
+Summing up, we have
+\begin{displaymath}
+ (x = ax + bx') < (b < x < a).
+\end{displaymath}
+
+Conversely,
+\begin{displaymath}
+ (b < x < a) < (x = ax + bx').
+\end{displaymath}
+
+For
+\begin{align*}
+ (x < a) &= (x = ax),\\
+ (b < x) &= (bx' = 0).
+\end{align*}
+
+Adding these two equalities member to member [the second
+members of the two larger equalities],
+\begin{displaymath}
+ (x = ax) (o = bx) < (x = ax + bx').
+\end{displaymath}
+
+Therefore
+\begin{displaymath}
+ (b < x < a) < (x = ax + bx')
+\end{displaymath}
+and thus the equivalence is proved.
+
+\section{Schröder's Theorem.%
+\protect\footnote{\author{}{Schröder,} \emph{Operationskreis des Logikkalküls} (1877), Theorem 20.}}%
+\label{ch:30}
+
+The equality
+\begin{displaymath}
+ ax + bx' = 0
+\end{displaymath}
+signifies that~$x$ lies between~$a'$ and~$b$.
+
+\emph{Demonstration:}
+\begin{align*}
+ (ax + bx' = 0) &= (ax = 0) (bx' = 0),\\
+ (ax = 0) &= (x < a'),\\
+ (bx' = 0) &= (b < x).
+\end{align*}
+Hence
+\begin{displaymath}
+ (ax + bx' = 0) = (b < x < a').
+\end{displaymath}
+
+Comparing this theorem with the formula of \author{}{Poretsky,} we
+obtain at once the equality
+\begin{displaymath}
+ (ax + bx' = 0) = (x = a' x + bx'),
+\end{displaymath}
+which may be directly proved by reducing the formula of
+\author{}{Poretsky} to an equality whose second member is~0, thus:
+\begin{displaymath}
+ (x = a'x + bx') = [x (ax + b'x') + x' (a'x + bx') = 0] = (ax + bx' = 0).
+\end{displaymath}
+
+If we consider the given equality as an \emph{equation} in which
+$x$~is the unknown quantity, \author{}{Poretsky's} formula will be
+its solution.
+
+From the double inclusion
+\begin{displaymath}
+ b < x < a'
+\end{displaymath}
+we conclude, by the principle of the syllogism, that
+\begin{displaymath}
+ b < a'
+\end{displaymath}
+
+This is a consequence of the given equality and is independent
+of the unknown quantity~$x$. It is called the
+\emph{resultant of the elimination} of~$x$ in the given equation. It is
+equivalent to the equality
+\begin{displaymath}
+ ab = 0.
+\end{displaymath}
+
+Therefore we have the implication
+\begin{displaymath}
+ (ax + bx' = 0) < (ab = 0).
+\end{displaymath}
+
+Taking this consequence into consideration, the solution
+may be simplified, for
+\begin{displaymath}
+ (ab = 0) = (b = a'b).
+\end{displaymath}
+
+Therefore
+\begin{displaymath}
+ \begin{split}
+ x &= a'x + bx' = a'x + a'bx'\\
+ &= a'bx + a'b'x + a'bx' = a'b + a'b'x\\
+ & = b + a'b'x + b + a'x.\\
+ \end{split}
+\end{displaymath}
+
+This form of the solution conforms most closely to common sense:
+since~$x'$ contains~$b$ and is contained in~$a'$, it is natural
+that~$x$ should be equal to the sum of~$b$ and a part of~$a'$
+(that is to say, the part common to~$a'$ and~$x$). The solution is
+generally indeterminate (between the limits~$a'$ and~$b$); it is
+determinate only when the limits are equal,
+\begin{displaymath}
+ a' = b,
+\end{displaymath}
+for then
+\begin{displaymath}
+ x = b + a'x = b + bx = b = a'.
+\end{displaymath}
+
+Then the equation assumes the form
+\begin{displaymath}
+ (ax + a'x' = 0) = (a' = x)
+\end{displaymath}
+and is equivalent to the double inclusion
+\begin{displaymath}
+ (a' < x < a') = (x = a').
+\end{displaymath}
+
+\section{The Resultant of Elimination}\label{ch:31}
+When $ab$ is not zero, the equation is impossible (always false), because
+it has a false consequence. It is for this reason that \author{}{Schröder}
+considers the resultant of the elimination as a \emph{condition} of the
+equation. But we must not be misled by this equivocal word. The resultant
+of the elimination of $x$ is not a \emph{cause}\index{Cause} of
+the equation, it is a \emph{consequence}\index{Consequence} of it; it is not a \emph{sufficient}%
+\index{Condition!Necessary but not sufficient} but a \emph{necessary}
+condition.
+
+The same conclusion may be reached by observing that
+$ab$ is the inferior limit of the function $ax + bx'$, and that
+consequently the function can not vanish unless this limit is~0.
+\begin{displaymath}
+ (ab < ax + bx') (ax + bx' = 0) < (ab = 0).
+\end{displaymath}
+
+We can express the resultant of elimination in other equivalent
+forms; for instance, if we write the equation in the form
+\begin{displaymath}
+ (a + x') (b + x) = 0,
+\end{displaymath}
+we observe that the resultant
+\begin{displaymath}
+ ab = 0
+\end{displaymath}
+is obtained simply by dropping the unknown quantity (by
+suppressing the terms~$x$ and~$x'$). Again the equation may be
+written:
+\begin{displaymath}
+ a'x + b'x' = 1
+\end{displaymath}
+and the resultant of elimination:
+\begin{displaymath}
+ a' + b' = 1.
+\end{displaymath}
+
+Here again it is obtained simply by dropping the unknown
+quantity.%
+\footnote{This is the method of elimination of Mrs.
+\author{}{Ladd-Franklin} and Mr. \author{}{Mitchell,} but this
+rule is deceptive in its apparent simplicity, for it cannot be
+applied to the same equation when put in either of the forms
+\begin{displaymath}
+ ax + bx' = 0, \quad (a' + x') (b' +x) = 1.
+\end{displaymath}
+
+Now, on the other hand, as we shall see (\S\ref{ch:54}), for inequalities it
+may be applied to the forms
+\begin{displaymath}
+ ax + bx' \neq 0, \quad (a' + x') (b' + x) \neq 1.
+\end{displaymath}
+and not to the equivalent forms
+\begin{displaymath}
+ (a + x') (b + x) \neq 0, \quad a'x + b'x' \neq 1.
+\end{displaymath}
+
+Consequently, it has not the mnemonic property attributed to it, for, to
+use it correctly, it is necessary to recall to which forms it is applicable.}
+
+\emph{Remark}. If in the equation
+\begin{displaymath}
+ ax + bx' = 0
+\end{displaymath}
+we substitute for the unknown quantity~$x$ its value derived
+from the equations,
+\begin{displaymath}
+ x = a'x + bx', \quad x' = ax + b'x',
+\end{displaymath}
+we find
+\begin{displaymath}
+ (abx + abx' = 0) = (ab = 0),
+\end{displaymath}
+that is to say, the resultant of the elimination of~$x$ which, as
+we have seen, is a consequence of the equation itself. Thus we are
+assured that the value of~$x$ verifies this equation. Therefore we
+can, with \author{}{Voigt,} define the solution of an equation as
+that value which, when substituted for~$x$ in the equation,
+reduces it to the resultant of the elimination of~$x$.
+
+\emph{Special Case}.---When the equation contains a term
+independent of~$x$, \emph{i.e.}, when it is of the form
+\begin{displaymath}
+ ax + bx' + c = 0
+\end{displaymath}
+it is equivalent to
+\begin{displaymath}
+ (a+c)x + (b+c)x' = 0,
+\end{displaymath}
+and the resultant of elimination is
+\begin{displaymath}
+ (a + c) (b + c) = ab + c = 0,
+\end{displaymath}
+whence we derive this practical rule: To obtain the resultant of
+the elimination of~$x$ in this case, it is sufficient to equate to
+zero the product of the coefficients of~$x$ and~$x'$, and add to
+them the term independent of~$x$.
+
+\section{The Case of Indetermination}\label{ch:32}
+Just as the resultant
+\begin{displaymath}
+ ab = 0
+\end{displaymath}
+corresponds to the case when the equation is possible, so the
+equality
+\begin{displaymath}
+ a + b =0
+\end{displaymath}
+corresponds to the case of \emph{absolute indetermination}. For in
+this case the equation both of whose coefficients are zero
+$(a = 0)$, $(b = 0)$, is reduced to an identity $(0 = 0)$, and
+therefore is ``identically'' verified, whatever the value of~$x$ may
+be; it does not determine the value of~$x$ at all, since the
+double inclusion
+\begin{displaymath}
+ b < x < a'
+\end{displaymath}
+then becomes
+\begin{displaymath}
+ 0 < x < 1
+\end{displaymath}
+which does not limit in any way the variability of~$x$. In this
+case we say that the equation is \emph{ indeterminate}.
+
+We shall reach the same conclusion if we observe that
+$(a + b)$ is the superior limit of the function $ax + bx$ and that,
+if this limit is 0, the function is necessarily zero for all
+values of $x$,
+\begin{displaymath}
+ (ax + bx' < a + b) (a + b = 0) < (ax + bx' = 0).
+\end{displaymath}
+
+\emph{Special Case}.---When the equation contains a term independent of~$x$,
+\begin{displaymath}
+ ax + bx' + c = 0,
+\end{displaymath}
+the condition of absolute indetermination takes the form
+\begin{displaymath}
+ a + b + c = 0.
+\end{displaymath}
+For
+\begin{align*}
+ ax + bx' + c &= (a + c)x + (b + c)x', \\
+ (a + c) + (b + c) &= a + b + c = 0.
+\end{align*}
+
+\section{Sums and Products of Functions}\label{ch:33}
+It is desirable
+at this point to introduce a notation borrowed from mathematics,
+which is very useful in the algebra of logic. Let $f(x)$
+be an expression containing one variable; suppose that the
+class of all the possible values of $x$ is determined; then the
+class of all the values which the function $f(x)$ can assume
+in consequence will also be determined. Their sum will be
+represented by $\sum_{x} f(x)$ and their product by $\prod_{x}f(x)$ This
+is a new notation and not a new notion, for it is merely the
+idea of sum and product applied to the values of a function.
+
+When the symbols $\sum$ and $\prod$ are applied to propositions,
+they assume an interesting significance:
+\begin{displaymath}
+ \prod_{x} [f(x) = 0]
+\end{displaymath}
+means that $f(x) = 0$ is true for \emph{every} value of $x$; and
+\begin{displaymath}
+ \sum _{x} [f(x) = 0]
+\end{displaymath}
+that $f(x) = 0$ is true for \emph{some} value of $x$. For, in
+order that a product may be equal to~1 (\emph{i.e.}, be true), all
+its factors must be equal to~1 (\emph{i.e.}, be true); but, in
+order that a sum may be equal to~1 (\emph{i.e.}, be true), it is
+sufficient that only one of its summands be equal to $1$
+(\emph{i.e.}, be true). Thus we have a means of expressing
+universal and particular propositions when they are applied to
+variables, especially those in the form: ``For every value of~$x$
+such and such a proposition is true'', and ``For some value
+of~$x$, such and such a proposition is true'', etc.
+
+For instance, the equivalence
+\begin{displaymath}
+ (a = b) = (ac = bc) (a + c = b + c)
+\end{displaymath}
+is somewhat paradoxical because the second member contains
+a term~($c$) which does not appear in the first. This equivalence
+is independent of~$c$, so that we can write it as follows,
+considering~$c$ as a variable~$x$
+\begin{displaymath}
+ \prod_{x} [(a= b) = (ax = bx) (a + x = b + x)],
+\end{displaymath}
+or, the first member being independent of~$x$,
+\begin{displaymath}
+ (a = b) = \prod_{x} [(ax = bx) (a + x = b + x)].
+\end{displaymath}
+
+In general, when a proposition contains a variable term,
+great care is necessary to distinguish the case in which it is
+true for \emph{every} value of the variable, from the case in which
+it is true only for some value of the variable.%
+\footnote{This is the same as the distinction made in mathematics between
+\emph{identities} and \emph{equations}, except that an equation may not be verified by
+any value of the variable.} This is the
+purpose that the symbols $\prod$ and $\sum$ serve.
+
+Thus when we say for instance that the equation
+\begin{displaymath}
+ ax + bx' = 0
+\end{displaymath}
+is possible, we are stating that it can be verified by some
+value of~$x$; that is to say,
+\begin{displaymath}
+ \sum_{x} (ax + bx' = 0),
+\end{displaymath}
+and, since the necessary and sufficient condition%
+\index{Condition!Necessary and sufficient} for this is that the resultant
+$(ab = 0)$ is true, we must write
+\begin{displaymath}
+ \sum_{x} (ax + bx = 0) = (ab = 0),
+\end{displaymath}
+although we have only the implication
+\begin{displaymath}
+ (ax + bx = 0) < (ab = 0).
+\end{displaymath}
+
+On the other hand, the necessary and sufficient condition%
+\index{Condition!Necessary and sufficient} for the equation to be verified
+by every value of~$x$ is that
+\begin{displaymath}
+ a + b = 0.
+\end{displaymath}
+
+\emph{Demonstration}.---1. The condition is sufficient, for if
+\begin{displaymath}
+ (a + b = 0) = (a = 0) (b = 0),
+\end{displaymath}
+we obviously have
+\begin{displaymath}
+ ax + bx' = 0
+\end{displaymath}
+whatever the value of $x$; that is to say,
+\begin{displaymath}
+ \prod_{x} (ax+ bx' = 0).
+\end{displaymath}
+
+2. The condition is necessary, for if
+\begin{displaymath}
+ \prod_{x} (ax + bx') = 0,
+\end{displaymath}
+the equation is true, in particular, for the value $x = a$; hence
+\begin{displaymath}
+ a + b = 0.
+\end{displaymath}
+
+Therefore the equivalence
+\begin{displaymath}
+ \prod_{x} (ax + bx' = 0) = (a + b = 0)
+\end{displaymath}
+is proved.%
+\footnote{\author{}{Eugen Müller,} \emph{op.~cit}.} In this
+instance, the equation reduces to an identity: its first member is
+``identically'' null.
+
+\section{The Expression of an Inclusion by Means of an
+Indeterminate}\label{ch:34}
+The foregoing notation is indispensable in
+almost every case where variables or indeterminates occur in
+one member of an equivalence, which are not present in the
+other. For instance, certain authors predicate the two following
+equivalences
+\begin{displaymath}
+ (a < b) = (a = bu) = (a + v = b),
+\end{displaymath}
+in which $u$, $v$ are two ``indeterminates''. Now, each of the
+two equalities has the inclusion ($a < b$) as its consequence,
+as we may assure ourselves by eliminating $u$ and $v$ respectively
+from the following equalities:
+
+\begin{displaymath}
+ \tag*{1.} [a (b' + u') + a'bu = 0] = [(ab' + a'b) u + au' = 0].
+\end{displaymath}
+
+Resultant:
+\begin{displaymath}
+ [(ab' + a'b) a = 0] = (ab' = 0) = (a < b).
+\end{displaymath}
+\begin{displaymath}
+ \tag*{2.} [(a + v) b' + a'bv = 0] = [b'v + (ab' + a'b) v' = 0].
+\end{displaymath}
+
+Resultant:
+\begin{displaymath}
+ [b' (ab' + a'b) = 0] = (ab' = 0) + (a < b).
+\end{displaymath}
+
+But we cannot say, conversely, that the inclusion implies
+the two equalities for \emph{any values} of~$u$ and~$v$; and, in fact, we
+restrict ourselves to the proof that this implication holds for
+some value of~$u$ and~$v$, namely for the particular values
+\begin{displaymath}
+ u = a, \quad b = v;
+\end{displaymath}
+for we have
+\begin{displaymath}
+ (a = ab) = (a < b) = (a + b = b).
+\end{displaymath}
+
+But we cannot conclude, from the fact that the implication
+(and therefore also the equivalence) is true for \emph{some} value of
+the indeterminates, that it is true for \emph{all}; in particular, it is
+not true for the values
+\begin{displaymath}
+ u = 1, \quad v = 0,
+\end{displaymath}
+for then $(a = bu)$ and $(a + v = b)$ become $(a=b)$, which
+obviously asserts more than the given inclusion $(a < b)$.%
+\footnote{Likewise if we make
+\begin{displaymath}
+ u = 0, \quad v = 1,
+\end{displaymath}
+we obtain the equalities
+\begin{displaymath}
+ (a = 0), \quad (b = 1),
+\end{displaymath}
+which assert still more than the given inclusion.}
+
+Therefore we can write only the equivalences
+\begin{displaymath}
+ (a < b) = \sum_u (a = bu) = \sum_v (a + v = b),
+\end{displaymath}
+but the three expressions
+\begin{displaymath}
+ (a < b), \quad \prod_u (a = bu), \quad \prod_v (a + v = b)
+\end{displaymath}
+are not equivalent.%
+\footnote{According to the remark in the preceding note, it is clear that we have
+\begin{displaymath}
+ \prod_v (a = bu) = (a = b = 0), \quad \prod_v (a + v = b) = (a = b = 1),
+\end{displaymath}
+since the equalities affected by the sign $\prod$ may be likewise verified
+by the values
+\begin{displaymath}
+ u = 0, \quad u = 1 \quad \text{and} \quad v = 0, \quad v = 1.
+\end{displaymath}
+If we wish to know within what limits the indeterminates $u$ and $v$ are
+variable, it is sufficient to solve with respect to them the equations
+\begin{displaymath}
+ (a < b) = (a = bu), \quad (a < b) = (a + v = b),
+\end{displaymath}
+or
+\begin{displaymath}
+ (ab' = a'bu + ab' + au', \quad ab' = ab' + b'v + a'bv',
+\end{displaymath}
+or
+\begin{displaymath}
+ a'bu + abu' = 0, \quad a'b'v + a'bv' = 0,
+\end{displaymath}
+from which (by a formula to be demonstrated later on) we derive the
+solutions
+\begin{displaymath}
+ u = ab + w (a + b'), \quad v = a'b + w (a + b),
+\end{displaymath}
+or simply
+\begin{displaymath}
+ u = ab + wb', \quad v = a'b + wa,
+\end{displaymath}
+$w$ being absolutely indeterminate. We would arrive at these solutions
+simply by asking: By what term must we multiply $b$ in order to obtain
+$a$? By a term which contains $ab$ plus any part of $b'$. What term must
+we add to $a$ in order to obtain $b$? A term which contains $a'b$ plus
+any part of $a$. In short, $u$ can vary between $ab$ and $a + b'$, $v$ between
+$a'b$ and $a + b$.}
+
+\section{The Expression of a Double Inclusion by Means
+of an Indeterminate}\label{ch:35}
+\textsc{Theorem}. \emph{The double inclusion
+\begin{displaymath}
+ b < x < a
+\end{displaymath}
+is equivalent to the equality $x = au + bu'$ together with the
+condition ($b < a$), $u$ being a term absolutely indeterminate.}
+
+\emph{Demonstration}.---Let us develop an equality in question,
+\begin{align*}
+ x(a'u + b'u') + x'(au + bu') &= 0, \\
+ (a'x + ax')u + (b'x + bx')u' &= 0.
+\end{align*}
+
+Eliminating $u$ from it,
+\begin{displaymath}
+ a'b'x + abx' = 0.
+\end{displaymath}
+
+This equality is equivalent to the double inclusion
+\begin{displaymath}
+ ab < x < a + b.
+\end{displaymath}
+
+But, by hypothesis, we have
+\begin{displaymath}
+ (b < a) = (ab = b) = (a + b = a).
+\end{displaymath}
+
+The double inclusion is therefore reduced to
+\begin{displaymath}
+ b < x < a.
+\end{displaymath}
+
+So, whatever the value of~$u$, the equality under consideration
+involves the double inclusion. Conversely, the double inclusion
+involves the equality, whatever the value of $x$ may be,
+for it is equivalent to
+\begin{displaymath}
+ a'x + bx' = 0,
+\end{displaymath}
+and then the equality is simplified and reduced to
+\begin{displaymath}
+ ax'u + b'xu' = 0.
+\end{displaymath}
+
+We can always derive from this the value of $u$ in terms
+of $x$, for the resultant $(ab'xx' = 0)$ is identically verified.
+The solution is given by the double inclusion
+\begin{displaymath}
+ b' x < u < a' + x.
+\end{displaymath}
+
+\emph{Remark}.---There is no contradiction between this result,
+which shows that the value of~$u$ lies between certain limits,
+and the previous assertion that~$u$ is absolutely indeterminate;
+for the latter assumes that~$x$ is any value that will verify the
+double inclusion, while when we evaluate~$u$ in terms of~$x$ the
+value of $x$ is supposed to be determinate, and it is with
+respect to this particular value of~$x$ that the value of~$u$ is
+subjected to limits.%
+\footnote{Moreover, if we substitute for~$x$ its inferior limit~$b$ in the inferior
+limit of~$u$, this limit becomes $bb' = 0$; and, if we substitute for~$x$ its
+superior limit~$a$ in the superior limit of~$u$, this limit becomes $a + a' = 1$.}
+
+In order that the value of~$u$ should be completely determined,
+it is necessary and sufficient that we should have
+\begin{displaymath}
+ b'x = a' + x,
+\end{displaymath}
+that is to say,
+\begin{displaymath}
+ b' xax' + (b + x') (a'+ x) = 0
+\end{displaymath}
+or
+\begin{displaymath}
+ bx + a' x' =0.
+\end{displaymath}
+
+Now, by hypothesis, we already have
+\begin{displaymath}
+ a' x + bx' = 0.
+\end{displaymath}
+
+If we combine these two equalities, we find
+\begin{displaymath}
+ (a + b = 0) = (a = 1) (b = 0).
+\end{displaymath}
+
+This is the case when the value of~$x$ is absolutely indeterminate,
+since it lies between the limits~0 and~1.
+
+In this case we have
+\begin{displaymath}
+ u = b'x = a + x = x.
+\end{displaymath}
+
+In order that the value of $u$ be absolutely indeterminate,
+it is necessary and sufficient that we have at the same time
+\begin{displaymath}
+ b'x = 0, \quad a'+x = 1,
+\end{displaymath}
+or
+\begin{displaymath}
+ b'x + ax' = 0,
+\end{displaymath}
+that is
+\begin{displaymath}
+ a < x < b.
+\end{displaymath}
+
+Now we already have, by hypothesis,
+\begin{displaymath}
+ b < x < a;
+\end{displaymath}
+so we may infer
+\begin{displaymath}
+ b = x = a.
+\end{displaymath}
+
+This is the case in which the value of $x$ is completely
+determinate.
+
+\section{Solution of an Equation Involving One Unknown
+Quantity}\label{ch:36}
+The solution of the equation
+\begin{displaymath}
+ ax + bx' = 0
+\end{displaymath}
+may be expressed in the form
+\begin{displaymath}
+ x = a'u + bu',
+\end{displaymath}
+$u$~being an indeterminate, on condition that the resultant of
+the equation be verified; for we can prove that this equality
+implies the equality
+\begin{displaymath}
+ ab'x + a'bx' = 0,
+\end{displaymath}
+which is equivalent to the double inclusion
+\begin{displaymath}
+ a'b < x < a' + b.
+\end{displaymath}
+
+Now, by hypothesis, we have
+\begin{displaymath}
+ (ab = 0) = (a'b = b) = (a' + b = a').
+\end{displaymath}
+
+Therefore, in this hypothesis, the proposed solution implies
+the double inclusion
+\begin{displaymath}
+ b < x < a';
+\end{displaymath}
+which is equivalent to the given equation.
+
+\emph{Remark}.---In the same hypothesis in which we have
+\begin{displaymath}
+ (ab=0)=(b < a'),
+\end{displaymath}
+we can always put this solution in the simpler but less symmetrical
+forms
+\begin{displaymath}
+ x = b + a'u, \quad x = a'(b+u).
+\end{displaymath}
+For
+
+1. We have identically
+\begin{displaymath}
+ b = bu + bu'.
+\end{displaymath}
+Now
+\begin{displaymath}
+ (b < a') < (bu < a'u).
+\end{displaymath}
+Therefore
+\begin{displaymath}
+ (x = bu' + a'u) = (x = b + a'u).
+\end{displaymath}
+
+2. Let us now demonstrate the formula
+\begin{displaymath}
+ x = a'b + a'u.
+\end{displaymath}
+Now
+\begin{displaymath}
+ a'b = b.
+\end{displaymath}
+Therefore
+\begin{displaymath}
+ x = b + a'u
+\end{displaymath}
+which may be reduced to the preceding form.
+
+Again, we can put the same solution in the form
+\begin{displaymath}
+ x = a'b + u(ab + a'b'),
+\end{displaymath}
+which follows from the equation put in the form
+\begin{displaymath}
+ ab'x + a'bx' = 0,
+\end{displaymath}
+if we note that
+\begin{displaymath}
+ a'+ b = ab + a'b + a'b'
+\end{displaymath}
+and that
+\begin{displaymath}
+ ua'b < a'b.
+\end{displaymath}
+
+This last form is needlessly complicated, since, by hypothesis,
+\begin{displaymath}
+ ab = 0.
+\end{displaymath}
+Therefore there remains
+\begin{displaymath}
+ x = a'b + ua'b'
+\end{displaymath}
+which again is equivalent to
+\begin{displaymath}
+ x = b + ua',
+\end{displaymath}
+since
+\begin{displaymath}
+ a'b = b \quad\text{and}\quad a' = a'b + a'b'.
+\end{displaymath}
+
+Whatever form we give to the solution, the parameter~$u$ in it is
+absolutely indeterminate, \emph{i.e.}, it can receive all possible
+values, including~0 and~1; for when $u = 0$ we have
+\begin{displaymath}
+ x = b,
+\end{displaymath}
+and when $u = 1$ we have
+\begin{displaymath}
+ x = a',
+\end{displaymath}
+and these are the two extreme values of~$x$.
+
+Now we understand that~$x$ is determinate in the particular
+case in which $a' = b$, and that, on the other hand, it is
+absolutely indeterminate when
+\begin{displaymath}
+ b = 0, \quad a' = 1, \quad (\text{or } a = 0).
+\end{displaymath}
+
+Summing up, the formula
+\begin{displaymath}
+ x = a'u + bu'
+\end{displaymath}
+replaces the ``limited'' variable~$x$ (lying between the limits~$a'$
+and~$b$) by the ``unlimited'' variable~$u$ which can receive all
+possible values, including~0 and~1.
+
+\emph{Remark}.%
+\footnote{\author{}{Poretsky.} \emph{Sept lois}, Chaps.~XXXIII and
+XXXIV.}---The formula of solution
+\begin{displaymath}
+ x = a'x + bx'
+\end{displaymath}
+is indeed equivalent to the given equation, but not so the
+formula of solution
+\begin{displaymath}
+ x = a'u + bu'
+\end{displaymath}
+as a function of the indeterminate~$u$. For if we develop the
+latter we find
+\begin{displaymath}
+ ab'x + a'bx' + ab(xu + x'u') + a'b'(xu' + x'u) = 0,
+\end{displaymath}
+and if we compare it with the developed equation
+\begin{displaymath}
+ ab + ab'x + a'bx' = 0,
+\end{displaymath}
+we ascertain that it contains, besides the solution, the equality
+\begin{displaymath}
+ ab(xu' + x'u) = 0,
+\end{displaymath}
+and lacks of the same solution the equality
+\begin{displaymath}
+ a'b'(xu' + x'u) = 0.
+\end{displaymath}
+
+Moreover these two terms disappear if we make
+\begin{displaymath}
+ u = x
+\end{displaymath}
+and this reduces the formula to
+\begin{displaymath}
+ x = a'x + bx'.
+\end{displaymath}
+
+From this remark, \author{}{Poretsky} concluded that, in general, the
+solution of an equation is neither a consequence nor a cause
+of the equation. It is a cause of it in the particular case in which
+\begin{displaymath}
+ ab = 0,
+\end{displaymath}
+and it is a consequence of it in the particular case in which
+\begin{displaymath}
+ (a'b' = 0) = (a + b = i).
+\end{displaymath}
+
+But if $ab$ is not equal to~0, the equation is unsolvable and
+the formula of solution absurd, which fact explains the
+preceding paradox. If we have at the same time
+\begin{displaymath}
+ ab = 0 \quad\text{and}\quad a + b = 1,
+\end{displaymath}
+the solution is both consequence and cause at the same time,
+that is to say, it is equivalent to the equation. For when
+$a = b$ the equation is determinate and has only the one
+solution
+\begin{displaymath}
+ x = a' = b.
+\end{displaymath}
+
+Thus, whenever an equation is solvable, its solution is one of its
+causes; and, in fact, the problem consists in finding a value of
+$x$ which will verify it, \emph{i.e.}, which is a cause of it.
+
+To sum up, we have the following equivalence:
+\begin{displaymath}
+ (ax + bx' = 0) = (ab = 0) \sum_u (x = a'u + bu')
+\end{displaymath}
+which includes the following implications:
+\begin{gather*}
+ (ax + bx' = 0) < (ab = 0), \\
+ (ax + bx' = 0) < \sum_u (x = a'u + bu'), \\
+ (ab = 0) \sum_u (x = a'u + bu') < (ax + bx' = 0).
+\end{gather*}
+
+\section{Elimination of Several Unknown Quantities}\label{ch:37}
+We shall now consider an equation involving several unknown
+quantities and suppose it reduced to the normal form, \emph{i.e.},
+its first member developed with respect to the unknown quantities,
+and its second member zero. Let us first concern ourselves with
+the problem of elimination. We can eliminate the unknown
+quantities either one by one or all at once.
+
+For instance, let
+\begin{equation}\label{eq:phi}
+ \begin{split}
+ \phi(x,y,z) &= axyz + bxyz' + cxy'z + dxy'z'\\
+ &+ fx'yz + gx'yz' + hx'y'z + kx'y'z' = 0\\
+ \end{split}
+\end{equation}
+be an equation involving three unknown quantities.
+
+We can eliminate $z$ by considering it as the only unknown
+quantity, and we obtain as resultant
+\begin{displaymath}
+ (axy + cxy' + fx'y + hx'y') (bxy + dxy' + gx'y + kx'y') = 0
+\end{displaymath}
+or
+\begin{equation}\label{eq:phi2}
+ abxy + cdxy' + fgx'y + hkx'y' = 0.
+\end{equation}
+
+If equation (\ref{eq:phi}) is possible, equation (\ref{eq:phi2}) is possible as well;
+that is, it is verified by some values of~$x$ and~$y$. Accordingly
+we can eliminate~$y$ from the equation by considering it as
+the only unknown quantity, and we obtain as resultant
+\begin{displaymath}
+ (abx + fgx') (cdx + hkx') = 0
+\end{displaymath}
+or
+\begin{equation}\label{eq:phi3}
+ abcdx + fghkx' = 0.
+\end{equation}
+
+If equation~(\ref{eq:phi}) is possible, equation~(\ref{eq:phi3}) is also possible;.
+that is, it is verified by some values of~$x$. Hence we can
+eliminate~$x$ from it and obtain as the final resultant,
+\begin{displaymath}
+ abcd \cdot fghk = 0
+\end{displaymath}
+which is a consequence of~(\ref{eq:phi}), independent of the unknown
+quantities. It is evident, by the principle of symmetry, that
+the same resultant would be obtained if we were to eliminate
+the unknown quantities in a different order. Moreover this
+result might have been foreseen, for since we have (\S\ref{ch:28})
+\begin{displaymath}
+ abcdfghk < \phi(x,y,z),
+\end{displaymath}
+$\phi(x,y,z)$ can vanish only if the product of its coefficients
+is zero:
+\begin{displaymath}
+ \left[\phi(x,y,z) = 0\right] < (abcdfghk = 0).
+\end{displaymath}
+
+Hence we can eliminate all the unknown quantities at once
+by equating to 0 the product of the coefficients of the
+function developed with respect to all these unknown quantities.
+
+We can also eliminate some only of the unknown quantities at one
+time. To do this, it is sufficient to develop the first member
+with respect to these unknown quantities and to equate the product
+of the coefficients of this development to~0. This product will
+generally contain the other unknown quantities. Thus the resultant
+of the elimination of~$z$ alone, as we have seen, is
+\begin{displaymath}
+ abxy + cdxy' + fgx'y + hkx'y' = 0
+\end{displaymath}
+and the resultant of the elimination of~$y$ and~$z$ is
+\begin{displaymath}
+ abcdx + fghkx' = 0.
+\end{displaymath}
+
+These partial resultants can be obtained by means of the
+following practical rule: Form the constituents relating to the
+unknown quantities to be retained; give each of them, for a
+coefficient, the product of the coefficients of the constituents
+of the general development of which it is a factor, and equate
+the sum to~0.
+
+\section{Theorem Concerning the Values of a Function}\label{ch:38}
+\emph{All the values which can be assumed by a function of any number
+of variables $f(x, y, z \ldots)$ are given by the formula
+\begin{displaymath}
+ abc \ldots k + u(a + b + c + \ldots + k),
+\end{displaymath}
+in which $u$ is absolutely indeterminate, and $a, b, c \ldots, k$ are
+the coefficients of the development of~$f$.}
+
+\emph{Demonstration}.---It is sufficient to prove that in the equality
+\begin{displaymath}
+ f(x, y, z \ldots) = abc \ldots k + u(a + b + c + \ldots + k)
+\end{displaymath}
+$u$ can assume all possible values, that is to say, that this
+equality, considered as an equation in terms of $u$, is indeterminate.
+
+In the first place, for the sake of greater homogeneity, we
+may put the second member in the form
+\begin{displaymath}
+ u'abc \ldots k + u(a + b + c + \ldots + k),
+\end{displaymath}
+for
+\begin{displaymath}
+ abc \ldots k = uabc \ldots k + u'abc \ldots k,
+\end{displaymath}
+and
+\begin{displaymath}
+ uabc \ldots k < u(a + b + c + \ldots + k).
+\end{displaymath}
+
+Reducing the second member to~0 (assuming there are
+only three variables $x, y, z$)
+\begin{displaymath}
+ \begin{split}
+ (axyz &+ bxyz' + cxy'z + \ldots + kx'y'z')\\
+ &\times [ua'b'c' \ldots k' + u'(a' + b' + c' + \ldots + k')]\\
+ &+ (a'xyz + b'xyz' + c'xy'z + \ldots + k'x'y'z')\\
+ &\times [u(a + b + c + \ldots + k) + u'abc \ldots k] = 0,\\
+ \end{split}
+\end{displaymath}
+or more simply
+\begin{displaymath}
+ \begin{split}
+ u(a &+ b + c + \ldots + k) (a'xyz + b'xyz + c'xy'z + \ldots + k'x'y'z')\\
+ &+ u'(a' + b' + c' + \ldots+ k') (axyz + bxyz +\ldots + kx'y z) = 0.\\
+ \end{split}
+\end{displaymath}
+
+If we eliminate all the variables $x, y, z$, but not the indeterminate
+$u$, we get the resultant
+\begin{displaymath}
+ \begin{split}
+ u(a &+ b + c + \ldots + k) a'b'c' \ldots k'\\
+ &+ u'(a' + b' + c'+\ldots + k')abc \ldots k = 0.\\
+ \end{split}
+\end{displaymath}
+
+Now the two coefficients of~$u$ and~$u'$ are identically zero; it
+follows that~$u$ is absolutely indeterminate, which was to be
+proved.\footnote{\author{Whitehead, A.~N.}{Whitehead,}
+\emph{Universal Algebra}, Vol.~I, \S 33 (4).}
+
+From this theorem follows the very important consequence
+that a function of any number of variables can be changed
+into a function of a single variable without diminishing or
+altering its ``variability''.
+
+\emph{Corollary}.---A function of any number of variables can
+become equal to either of its limits.
+
+For, if this function is expressed in the equivalent form
+\begin{displaymath}
+ abc \ldots k + u(a + b + c + \ldots + k),
+\end{displaymath}
+it will be equal to its minimum $(abc \ldots k)$ when $u = 0$, and
+to its maximum $(a + b + c + \ldots + k)$ when $u = 1$.
+
+Moreover we can verify this proposition on the primitive
+form of the function by giving suitable values to the
+variables.
+
+Thus a function can assume all values comprised between
+its two limits, including the limits themselves. Consequently,
+it is absolutely indeterminate when
+\begin{displaymath}
+ a b c \ldots k = 0 \quad\text{and}\quad a + b + c + \ldots + k = 1
+\end{displaymath}
+at the same time, or
+\begin{displaymath}
+ a b c \ldots k = 0 = a' b' c' \ldots k'.
+\end{displaymath}
+
+\section{Conditions of Impossibility and Indetermination}\label{ch:39}
+
+The preceding theorem enables us to find the conditions
+under which an equation of several unknown quantities is
+impossible or indeterminate. Let $f(x, y, z \ldots)$ be the first
+member supposed to be developed, and $a, b, c \ldots, k$ its
+coefficients. The necessary and sufficient condition for the
+equation to be possible is
+\begin{displaymath}
+ abc \ldots k = 0.
+\end{displaymath}
+
+For, (1) if~$f$ vanishes for some value of the unknowns,
+its inferior limit $abc \ldots k$ must be zero; (2) if $abc \ldots k$ is zero,
+$f$~may become equal to it, and therefore may vanish for certain
+values of the unknowns.
+
+The necessary and sufficient condition%
+\index{Condition!Necessary and sufficient} for the equation to
+be indeterminate (identically verified)%
+\index{Condition!of impossibility and indetermination} is
+\begin{displaymath}
+ a + b + c \ldots + k = 0.
+\end{displaymath}
+
+For, (1) if $a + b + c + \ldots + k$ is zero, since it is the
+superior limit of $f$, this function will always and necessarily
+be zero; (2) if $f$ is zero for all values of the unknowns,
+$a + b + c + \ldots + k$ will be zero, for it is one of the values
+of $f$.
+
+Summing up, therefore, we have the two equivalences
+\begin{gather*}
+ \sum [f(x, y, z, \ldots) = 0] = (a b c \ldots k = 0).\\
+ \prod[f(x, y, z, \ldots) = 0] = (a + b + c \ldots + k = 0).
+\end{gather*}
+
+The equality $a b c \ldots k = 0$ is, as we know, the resultant
+of the elimination of all the unknowns; it is the consequence
+that can be derived from the equation (assumed to be verified)
+independently of all the unknowns.
+
+\section{Solution of Equations Containing Several Unknown
+Quantities}\label{ch:40}
+On the other hand, let us see how
+we can solve an equation with respect to its various unknowns,
+and, to this end, we shall limit ourselves to the
+case of two unknowns
+\begin{displaymath}
+ axy + bxy' + cx'y + dx'y' = 0.
+\end{displaymath}
+First solving with respect to $x$,
+\begin{displaymath}
+ x = (a'y + b'y')x + (cy + dy')x'.
+\end{displaymath}
+
+The resultant of the elimination of $x$ is
+\begin{displaymath}
+ acy + bdy' = 0.
+\end{displaymath}
+If the given equation is true, this resultant is true.
+
+Now it is an equation involving $y$ only; solving it,
+\begin{displaymath}
+ y = (a' + c')y + bdy'.
+\end{displaymath}
+
+Had we eliminated~$y$ first and then~$x$, we would have
+obtained the solution
+\begin{displaymath}
+ y = (a'x + c'x')y + (bx + dx')y'
+\end{displaymath}
+and the equation in~$x$
+\begin{displaymath}
+ abx + cdx' = 0,
+\end{displaymath}
+whence the solution
+\begin{displaymath}
+ x = (a' + b')x + cdx'.
+\end{displaymath}
+
+We see that the solution of an equation involving two
+unknown quantities is not symmetrical with respect to these
+unknowns; according to the order in which they were eliminated,
+we have the solution
+\begin{align*}
+ x &= (a'y + b'y')x + (cy + dy')x',\\
+ y &= (a' + c')y + bdy',
+\end{align*}
+or the solution
+\begin{align*}
+ x &= (a' + b')x + cdx,\\
+ y &= (a'x + c'x')y + (bx + dx')y'.
+\end{align*}
+
+If we replace the terms $x, y$, in the second members by
+indeterminates $u, v$, one of the unknowns will depend on only
+one indeterminate, while the other will depend on two. We
+shall have a symmetrical solution by combining the two formulas,
+\begin{align*}
+ x &= (a' + b')u + cdu',\\
+ y &= (a' + c')v + bdv',
+\end{align*}
+but the two indeterminates~$u$ and $v$~will no longer be independent
+of each other. For if we bring these solutions into
+the given equation, it becomes
+\begin{displaymath}
+ abcd + ab'c'uv + a'bd'uv' + a'cd'u'v + b'c'du'v' = 0
+\end{displaymath}
+or since, by hypothesis, the resultant $abcd = 0$ is verified,
+\begin{displaymath}
+ ab'c'uv + a'bd'uv' + a'cdu'v + b'c'du'v' = 0.
+\end{displaymath}
+
+This is an ``equation of condition'' which the indeterminates
+$u$~and $v$~must verify; it can always be verified, since its
+resultant is identically true,
+\begin{displaymath}
+ ab'c' \cdot a'bd' \cdot a'cd' \cdot b'c'd = aa' \cdot bb' \cdot cc' \cdot dd' = 0,
+\end{displaymath}
+but it is not verified by any pair of values attributed to~$u$
+and~$v$.
+
+Some general symmetrical solutions, \emph{i.e.}, symmetrical
+solutions in which the unknowns are expressed in terms of several
+independent indeterminates, can however be found.
+This problem has been treated by \author{}{Schröder}%
+\footnote{\emph{Algebra der Logik}, Vol.~I, \S 24.},
+by \author{Whitehead, A.~N.}{Whitehead}%
+\footnote{\emph{Universal Algebra}, Vol.~I, \S\S 35--37.}
+and by \author{}{Johnson.}%
+\footnote{``Sur la théorie des égalités logiques'', \emph{Bibl. du Cong. intern. de Phil.},
+Vol. III, p.~185 (Paris, 1901).}
+
+This investigation has only a purely technical interest; for,
+from the practical point of view, we either wish to eliminate
+one or more unknown quantities (or even all), or else we seek
+to solve the equation with respect to one particular unknown.
+In the first case, we develop the first member with respect
+to the unknowns to be eliminated and equate the product of
+its coefficients to~0. In the second case we develop with
+respect to the unknown that is to be extricated and apply
+the formula for the solution of the equation of one unknown
+quantity. If it is desired to have the solution in terms of
+some unknown quantities or in terms of the known only, the
+other unknowns (or all the unknowns) must first be eliminated
+before performing the solution.
+
+\section{The Problem of Boole}\label{ch:41}\index{Boole!Problem of|(}
+
+According to \author{}{Boole} the most general problem of the algebra
+of logic is the following\footnote{\emph{Laws of Thought}, Chap.~IX,
+ \S 8.}:
+
+Given any equation (which is assumed to be possible)
+\begin{displaymath}
+ f(x, y, z,\ldots) = 0,
+\end{displaymath}
+and, on the other hand, the expression of a term~$t$ in terms
+of the variables contained in the preceding equation
+\begin{displaymath}
+ t = \varphi(x,y,z,\ldots)
+\end{displaymath}
+to determine the expression of $t$ in terms of the constants
+contained in $f$ and in $\varphi$.
+
+Suppose $f$ and $\varphi$ developed with respect to the variables
+$x, y, z \ldots$ and let $p_1, p_2, p_3, \ldots$ be their constituents:
+\begin{align*}
+ f(x, y, z, \ldots) &= A p_1 + B p_2 + C p_3 + \ldots,\\
+ \phi (x, y, z, \ldots) &= ap_1 + bp_2 + cp_3 + \ldots.
+\end{align*}
+
+Then reduce the equation which expresses~$t$ so that its
+second member will be~0:
+\begin{displaymath}
+ \begin{split}
+ (t \phi' + t' \phi = 0) &= [(a' p_1 + b' p_2 + c' p_3 + \ldots) t\\
+ &\qquad + (a p_1 + b p_2 + c p_3 + \ldots) t' = 0].\\
+ \end{split}
+\end{displaymath}
+
+Combining the two equations into a single equation and
+developing it with respect to~$t$:
+\begin{multline*}
+ [(A + a') p_1 + (B + b') p_2 + (C+ c') p_3 + \ldots] t\\
+ + [(A + a) p_1 + (B + b) p_2 + (C + c) p_3 + \ldots]t' = 0.
+\end{multline*}
+
+This is the equation which gives the desired expression
+of $t$. Eliminating $t$, we obtain the resultant
+\begin{displaymath}
+ A p_1 + B p_2 + C p_3 + \ldots = 0,
+\end{displaymath}
+as we might expect. If, on the other hand, we wish to eliminate
+$x, y, z,\ldots$ (\emph{i.e.}, the constituents $p_1 , p_2 , p_3
+\ldots$), we put the equation in the form
+\begin{displaymath}
+ (A + a't + at')p_1 + (B + b't + bt')p_2 + (C + c't + ct') p_3 + \ldots = 0,
+\end{displaymath}
+and the resultant will be
+\begin{displaymath}
+ (A + a't + at') (B + b't + bt')(C + c't + ct')\ldots = 0,
+\end{displaymath}
+an equation that contains only the unknown quantity~$t$ and
+the constants of the problem (the coefficients of~$f$ and of~$\varphi$).
+From this may be derived the expression of~$t$ in terms of
+these constants. Developing the first member of this equation
+\begin{displaymath}
+ (A + a') (B + b') (C + c') \ldots \times t + (A + a) (B + b) (C + c) \ldots \times t' = 0.
+\end{displaymath}
+
+The solution is
+\begin{displaymath}
+ t = (A + a) (B + b) (C + c) \ldots + u(A'a + B'b + C'c + \ldots).
+\end{displaymath}
+
+The resultant is verified by hypothesis since it is
+\begin{displaymath}
+ ABC \ldots = 0,
+\end{displaymath}
+which is the resultant of the given equation
+\begin{displaymath}
+ f(x, y, z, \ldots) = 0.
+\end{displaymath}
+
+We can see how this equation contributes to restrict the
+variability of $t$. Since $t$ was defined only by the function $\varphi$,
+it was determined by the double inclusion
+\begin{displaymath}
+ abc\ldots < t < a + b + c + \ldots.
+\end{displaymath}
+
+Now that we take into account the condition $f=0$, $t$~is
+determined by the double inclusion
+\begin{displaymath}
+ (A + a) (B + b) (C + c) \ldots < t < (A'a + B'b + C'c + \ldots).%
+ \footnote{\author{Whitehead, A.~N.}{Whitehead,} \emph{Universal Algebra}, p.~63.}
+\end{displaymath}
+
+The inferior limit can only have increased and the superior
+limit diminished, for
+\begin{displaymath}
+ abc \ldots < (A + a) (B + b) (C + c) \ldots
+\end{displaymath}
+and
+\begin{displaymath}
+ A'a + B'b + C'c \ldots < a + b + c \ldots.
+\end{displaymath}
+
+The limits do not change if $A = B = C = \ldots = 0$, that
+is, if the equation $f = 0$ is reduced to an identity, and this
+was evident \emph{a priori}.\index{Boole!Problem of|)}
+
+\section{The Method of Poretsky}\label{ch:42}
+The method of \author{}{Boole} and \author{}{Schröder} which we
+have heretofore discussed is clearly inspired by the example of
+ordinary algebra, and it is summed up in two processes analogous
+to those of algebra, namely the solution of equations with
+reference to unknown quantities and elimination of the unknowns.
+Of these processes the second is much the more important from a
+logical point of view, and \author{}{Boole} was even on the point
+of considering deduction as essentially consisting in the
+\emph{elimination of middle terms}. This notion, which is too
+restricted, was suggested by the example of the syllogism, in
+which the conclusion results from the elimination of the middle
+term, and which for a long time was wrongly considered as the only
+type of mediate deduction.\footnote{In fact, the fundamental
+formula of elimination
+\begin{displaymath}
+ (ax + bx' = 0) < (ab = 0)
+\end{displaymath}
+is, as we have seen, only another form and a consequence of the
+principle of the syllogism
+\begin{displaymath}
+ (b < x < a') < (b < a').
+\end{displaymath}}
+
+However this may be, \author{}{Boole} and \author{}{Schröder} have
+exaggerated the analogy between the algebra of logic and ordinary
+algebra. In logic, the distinction of known and unknown terms is
+artificial and almost useless. All the terms are---in principle at
+least---known, and it is simply a question, certain relations
+between them being given, of deducing new relations (unknown or
+not explicitly known) from these known relations. This is the
+purpose of \author{}{Poretsky's} method which we shall now
+expound. It may be summed up in three
+laws, the \emph{law of forms}, the \emph{law of consequences}%
+\index{Consequences!Law of}\index{Law of Consequences} and the
+\emph{law of causes}.
+
+\section{The Law of Forms}\label{ch:43}
+This law answers the following
+problem: An equality being given, to find for any term
+(simple or complex) a determination equivalent to this equality.
+In other words, the question is to find all the \emph{forms}
+equivalent to this equality, any term at all being given as
+its first member.
+
+We know that any equality can be reduced to a form in which the
+second member is~0 or~1; \emph{i.e.}, to one of the two equivalent
+forms
+\begin{displaymath}
+ N = 0, \qquad N' = 1.
+\end{displaymath}
+
+The function~$N$ is what \author{}{Poretsky} calls the \emph{logical zero}
+of the given equality; $N'$~is its logical \emph{whole.}%
+\footnote{They are called ``logical'' to distinguish them from the
+identical \emph{zero} and \emph{whole}, \emph{i.e.}, to indicate
+that these two terms are not equal to 0 and 1 respectively except
+by virtue of the data of the problem.}
+
+Let~$U$ be any term; then the determination of~$U$:
+\begin{displaymath}
+ U = N'U + NU'
+\end{displaymath}
+is equivalent to the proposed equality; for we know it is
+equivalent to the equality
+\begin{displaymath}
+ (NU + NU' = 0) = (N = 0).
+\end{displaymath}
+
+Let us recall the signification of the determination
+\begin{displaymath}
+ U = N'U + NU'.
+\end{displaymath}
+
+It denotes that the term~$U$ is contained in~$N'$ and contains
+$N$. This is easily understood, since, by hypothesis,
+$N$~is equal to~0 and $N'$ to~1. Therefore we can formulate
+the \emph{law of forms} in the following way:
+
+\emph{To obtain all the forms equivalent to a given equality, it
+is sufficient to express that any term contains the logical zero
+of this equality and is contained in its logical whole.}
+
+The number of forms of a given equality is unlimited; for
+any term gives rise to a form, and to a form different from
+the others, since it has a different first member. But if we
+are limited to the universe of discourse determined by~$n$
+simple terms, the number of forms becomes finite and determinate.
+For, in this limited universe, there are $2^n$ constituents.
+Now, all the terms in this universe that can be
+conceived and defined are sums of some of these constituents.
+Their number is, therefore, equal to the number
+of combinations that can be made with $2^n$ constituents,
+namely $2^{2^n}$ (including~0, the combination of~0 constituent,
+and~1, the combination of all the constituents). This will
+also be the number of different forms of any equality in the
+universe in question.
+
+\section{The Law of Consequences}\label{ch:44}
+
+We shall now pass to the law of consequences.%
+\index{Consequences!Law of}\index{Law of Consequences} Generalizing
+the conception of \author{}{Boole,} who made deduction%
+\index{Deduction} consist in the elimination of middle terms,
+\author{}{Poretsky} makes it consist in the elimination of known terms
+(\emph{connaissances}\index{connaissances@\emph{Connaissances}}). This
+conception is explained and justified as follows.
+
+All problems in which the data are expressed by logical
+equalities or inclusions can be reduced to a single logical
+equality by means of the formula%
+\footnote{We employ capitals to denote complex terms (logical functions) in
+contrast to simple terms denoted by small letters ($a, b, c, \ldots$)}
+\begin{displaymath}
+ (A = 0)(B = 0) (C = 0) \ldots = (A + B + C \ldots = 0).
+\end{displaymath}
+
+In this logical equality, which sums up all the data of the
+problem, we develop the first member with respect to all
+the simple terms which appear in it (and not with respect
+to the unknown quantities). Let $n$ be the number of simple
+terms; then the number of the constituents of the development
+of~1 is $2^n$. Let $m$ ($\leq 2^n$) be the number of those
+constituents appearing in the first member of the equality.
+All possible consequences of this equality (in the universe
+of the $n$ terms in question) may be obtained by forming all
+the additive combinations of these $m$~constituents, and equating
+them to~0; and this is done in virtue of the formula
+\begin{displaymath}
+ (A + B = 0) < (A = 0).
+\end{displaymath}
+
+We see that we pass from the equality to any one of its
+consequences by suppressing some of the constituents in its first
+member, which correspond to as many elementary equalities
+(having~0 for second member), \emph{i.e.}, as many as there are
+data in the problem. This is what is meant by ``eliminating the
+known terms''.
+
+The number of consequences that can be derived from an equality
+(in the universe of~$n$ terms with respect to which it is
+developed) is equal to the number of additive combinations that
+may be formed with its $m$ constituents; \emph{i.e.}, $2^m$. This
+number includes the combination of 0~constituents, which gives
+rise to the identity $0 = 0$, and the combination of the
+$m$~constituents, which reproduces the given equality.
+
+Let us apply this method to the equation with one unknown
+quantity
+\begin{displaymath}
+ ax + bx' = 0.
+\end{displaymath}
+Developing it with respect to the \emph{three} terms $a, b, x$:
+\begin{align*}
+ (abx &+ ab'x + abx' + a'bx' = 0) \\
+ &= [ab(x + x') + ab'x + a'bx' = 0] \\
+ &= (ab = 0) (ab'x = 0) (a'bx'=0).
+\end{align*}
+
+Thus we find, on the one hand, the resultant $ab = 0$,
+and, on the other hand, two equalities which may be transformed
+into the inclusions
+\begin{displaymath}
+ x < a' + b, \qquad a'b < x.
+\end{displaymath}
+
+But by the resultant which is equivalent to $b < a'$, we have
+\begin{displaymath}
+ a' + b' = a', \qquad a'b = b.
+\end{displaymath}
+
+This consequence may therefore be reduced to the double
+inclusion
+\begin{displaymath}
+ x < a', \qquad b < x,
+\end{displaymath}
+that is, to the known solution.
+
+Let us apply the same method to the premises of the
+syllogism
+\begin{displaymath}
+ (a < b) (b < c).
+\end{displaymath}
+
+Reduce them to a single equality
+\begin{displaymath}
+ (a < b) = (ab' = 0), \quad (b < c) = (bc' = 0), \quad (ab' + bc' = 0),
+\end{displaymath}
+and seek all of its consequences.
+
+Developing with respect to the three terms $a, b, c$:
+\begin{displaymath}
+ abc' + ab'c + ab'c' + a'bc' = 0.
+\end{displaymath}
+
+The consequences\index{Consequences!Sixteen} of this equality, which
+contains four constituents, are~16 ($2^4$) in number as follows:
+\begin{gather*}
+ \tag*{1.} (abc' = 0) = (ab < c);\\
+ \tag*{2.} (ab'c = 0) = (ac < b);\\
+ \tag*{3.} (ab'c'= 0) = (a < b + c);\\
+ \tag*{4.} (a'bc' = 0) = (b < a + c);\\
+ \tag*{5.} (abc' + ab'c = 0) = (a < bc + b'c');\\
+ \tag*{6.} (abc' + ab'c'= 0) = (ac' = 0) = (a < c).
+\end{gather*}
+
+This is the traditional conclusion of the syllogism.%
+\footnote{It will be observed that this is the only consequence (except the
+two extreme consequences [see the text below]) independent of b; therefore
+it is the resultant of the elimination of that middle term.}
+
+\begin{displaymath}
+ \tag*{7.} (abc' + a'bc' = 0) = (bc' = 0) = (b < c).
+\end{displaymath}
+
+This is the second premise.
+
+\begin{displaymath}
+ \tag*{8.} (ab'c + ab'c' = 0) = (ab' = 0) = (a < b).
+\end{displaymath}
+
+This is the first premise.
+
+\begin{gather*}
+ \tag*{9.} (ab'c + a'bc' = 0) = (ac < b < a + c);\\
+ \tag*{10.} (ab'c'+ a'bc' = 0) = (ab' + a'b < c);\\
+ \tag*{11.} (abc' + ab'c + ab'c' = 0) = (ab' + ac' = 0) = (a < bc);\\
+ \tag*{12.} (abc' + ab'c + a'bc' = 0) = (ab'c + bc' = 0) = (ac < b < c);\\
+ \tag*{13.} (abc' + ab'c' + a'bc' = 0) = (ac' + bc' = 0) = (a + b < c);\\
+ \tag*{14.} (ab'c + ab'c' + a'bc' = 0) = (ab' + a'bc' = 0) = (a < b < a + c).
+\end{gather*}
+
+The last two consequences (15 and 16) are those obtained by combining
+0~constituent and by combining all; the first is the identity
+\begin{displaymath}
+ \tag*{15.} 0 = 0,
+\end{displaymath}
+which confirms the paradoxical proposition that the true
+(identity) is implied by any proposition (is a consequence
+of it); the second is the given equality itself
+\begin{displaymath}
+ \tag*{16.} ab' + bc' = 0,
+\end{displaymath}
+which is, in fact, its own consequence by virtue of the
+principle of identity. These two consequences may be called
+the ``extreme consequences'' of the proposed equality. If
+we wish to exclude them, we must say that the number of
+the consequences properly so called of an equality of~$m$
+constituents is $2^m - 2$.
+
+\section{The Law of Causes}\label{ch:45}\index{Causes!Law of|(}
+The method of finding the consequences of a given equality
+suggests directly the method of finding its \emph{causes}, namely,
+the propositions of which it is the consequence. Since we pass
+from the cause to the consequence by eliminating known terms,
+\emph{i.e.}, by suppressing constituents, we will pass conversely
+from the consequence to the cause by adjoining known terms,
+\emph{i.e.}, by adding constituents to the given equality. Now,
+the number of constituents that may be added to it, \emph{i.e.},
+that do not already appear in it, is $2^n-m$. We will obtain all
+the possible causes (in the universe of the $n$ terms under
+consideration) by forming all the additive combinations of these
+constituents, and adding them to the first member of the equality
+in virtue of the general formula
+\begin{displaymath}
+ (A + B = 0) < (A = 0),
+\end{displaymath}
+which means that the equality $(A = 0)$ has as its cause the
+equality $(A + B = 0)$, in which $B$ is any term. The number
+of causes thus obtained will be equal to the number of the
+aforesaid combinations, or $2^{2n}-m$.
+
+This method may be applied to the investigation of the
+causes of the premises of the syllogism
+\begin{displaymath}
+ (a < b) (b < c)
+\end{displaymath}
+which, as we have seen, is equivalent to the developed
+equality
+\begin{displaymath}
+ abc' + ab'c + ab'c' + a'bc' = 0.
+\end{displaymath}
+
+This equality contains four of the eight $(2^3)$ constituents
+of the universe of three terms, the four others being
+\begin{displaymath}
+ abc, a'bc, a'b'c, a'b'c'.
+\end{displaymath}
+
+The number of their combinations is 16 $(2^4)$, this is also
+the number of the causes sought, which are:
+\begin{gather}
+ \begin{split}
+ (abc &+ abc' + ab'c + ab'c' + a'bc' = 0)\\
+ &= (a + bc' = 0) = (a = 0) (b < c);
+ \end{split}\\
+ \begin{split}
+ (abc' &+ ab'c + ab'c' + a'bc + a'bc' = 0)\\
+ &= (abc'+ ab' + a'b = 0) = (ab < c) (a = b);
+ \end{split}\\
+ \begin{split}
+ (abc' &+ ab'c + ab'c' + a'bc' + a'b'c = 0)\\
+ &= (bc' + b'c + ab'c' = 0) = (b = c) (a < b + c);
+ \end{split}\\
+ \begin{split}
+ (abc' &+ ab'c + ab'c' + a'bc' + a'b'c' = 0)\\
+ &= (c' + ab' = 0) = (c=1) (a < b);
+ \end{split}\\
+ \begin{split}
+ (abc &+ abc' + ab'c + ab'c' + a'bc + a'bc' = 0)\\
+ &= (a + b = 0) = (a = 0) (b = 0);
+ \end{split}\\
+ \begin{split}
+ (abc &+ abc' + ab'c + ab'c' + a'bc' + a'b'c = 0)\\
+ &= (a + bc' + b'c = 0) = (a = 0) (b = c);
+ \end{split}\\
+ \begin{split}
+ (abc &+ abc' + ab'c + ab'c' + a'bc' + a'b'c' = 0)\\
+ &= (a + c' = 0) = (a = 0) (c = 1)\footnote{
+ It will be observed that this cause is the only one which is independent
+ of $b$; and indeed, in this case, whatever $b$ is, it will always
+ contain $a$ and will always be contained in $c$. Compare Cause 5, which
+ is independent of $c$, and Cause 10, which is independent of $a$.};\\
+ \end{split}\\
+ \begin{split}
+ (abc' &+ ab'c + ab'c' + a'bc + a'bc' + a'b'c = 0)\\
+ &= (ac' + a'c + ab'c + a'bc' = 0)\\
+ &= (a = c)(ac < b < a + c) = (a = b = c);
+ \end{split}\\
+ \begin{split}
+ (abc' &+ ab'c + ab'c' + a'bc + a'bc' + a'b'c = 0)\\
+ &= (c' + ab' + a'b = 0) = (c = 1)(a = b);
+ \end{split}\\
+ \begin{split}
+ (abc' &+ ab'c + ab'c' + a'bc' +a'b'c + a'b'c' = 0)\\
+ &= (b' + c' = 0) = (b = c = 1).
+ \end{split}
+\end{gather}
+
+Before going any further, it may be observed that when
+the sum of certain constituents is equal to~0, the sum of
+the rest is equal to~1. Consequently, instead of examining
+the sum of seven constituents obtained by ignoring one of
+the four missing constituents, we can examine the equalities
+obtained by equating each of these constituents to~1:
+\begin{alignat}{2}
+ (a'b'c' = 1) &= (a + b + c = 0) &&= (a = b = c = 0);\\
+ (a'b'c = 1) &= (a + b + c' = 0) &&= (a = b = 0) (c = 1);\\
+ (a'bc = 1) &= (a + b' + c' = 0) &&= (a = 0) (b = c = 1);\\
+ (abc = 1) & &&= (a = b = c = 1).
+\end{alignat}
+
+Note that the last four causes are based on the inclusion
+\begin{displaymath}
+ 0 < 1.
+\end{displaymath}
+
+The last two causes (\ref{eq:absurdity} and \ref{eq:equality}) are obtained either by
+adding \emph{all} the missing constituents or by not adding any.
+In the first case, the sum of all the constituents being equal
+to~1, we find
+\begin{equation}\label{eq:absurdity}
+ 1 = 0,
+\end{equation}
+that is, absurdity, and this confirms the paradoxical proposition
+that the false (the absurd) implies any proposition
+(is its cause). In the second case, we obtain simply the
+given equality, which thus appears as one of its own causes
+(by the principle of identity):
+\begin{equation}\label{eq:equality}
+ ab' + bc' = 0.
+\end{equation}
+
+If we disregard these two extreme causes, the number of
+causes properly so called will be
+\begin{displaymath}
+ 2^{2^n - m} - 2.
+\end{displaymath}
+\index{Causes!Law of|)}
+
+\section{Forms of Consequences and Causes}\label{ch:46}
+\index{Causes!Forms of}\index{Forms of Causes}
+\index{Consequences!Forms of}\index{Forms of Consequences}
+We can
+apply the law of forms to the consequences and causes of a
+given equality so as to obtain all the forms possible to each
+of them. Since any equality is equivalent to one of the two forms
+\begin{displaymath}
+ N=0, \qquad N'=1,
+\end{displaymath}
+each of its consequences has the form%
+\footnote{In \S\ref{ch:44} we said that a consequence is obtained by taking a part
+of the constituents of the first member $N$, and not by multiplying it by
+a term~$X$; but it is easily seen that this amounts to the same thing.
+For, suppose that $X$ (like $N$) be developed with respect to the $n$ terms
+of discourse. It will be composed of a certain number of constituents.
+To perform the multiplication of $N$ by $X$, it is sufficient to multiply
+all their constituents each by each. Now, the product of two identical
+constituents is equal to each of them, and the product of two different
+constituents is 0. Hence the product of $N$ by $X$ becomes reduced to
+the sum of the constituents common to $N$ and $X$, which is, of course,
+contained in $N$. So, to multiply $N$ by an arbitrary term is tantamount
+to taking a part of its constituents (or all, or none).}
+\begin{displaymath}
+ NX = 0, \qquad \text{or } N' + X' = 1,
+\end{displaymath}
+and each of its causes has the form
+\begin{displaymath}
+ N + X = 0, \qquad \text{or } N' X' = 1.
+\end{displaymath}
+
+In fact, we have the following formal implications:
+\begin{gather*}
+ (N + X = 0) < (N = 0) < (N X = 0),\\
+ (N' X' = 1) < (N' = 1) = (N' + X' = 1).
+\end{gather*}
+
+Applying the law of forms, the formula of the consequences
+becomes
+\begin{displaymath}
+ U = (N' + X') U + N X U',
+\end{displaymath}
+and the formula of the causes
+\begin{displaymath}
+ U = N' X' U + (N + X) U';
+\end{displaymath}
+or, more generally, since~$X$ and~$X'$ are indeterminate terms,
+and consequently are not necessarily the negatives of each
+other, the formula of the consequences will be
+\begin{displaymath}
+ U = (N' + X) U + N Y U',
+\end{displaymath}
+and the formula of the causes
+\begin{displaymath}
+ U = N' X U + (N + Y) U'.
+\end{displaymath}
+
+The first denotes that $U$ is contained in $(N' + X)$ and
+contains $N Y$; which indeed results, \emph{a fortiori}, from the hypothesis
+that $U$ is contained in $N'$ and contains $N$.
+
+The second formula denotes that $U$ is contained in $N' X$
+and contains $N' + Y$ whence results, \emph{a fortiori}, that $U$ is
+contained in $N'$ and contains $N$.
+
+We can express this rule verbally if we agree to call
+every class contained in another a \emph{sub-class}, and every
+class that contains another a \emph{super-class}. We then say:
+To obtain all the consequences of an equality (put in the
+form $U = N' U + N U'$), it is sufficient to substitute for its
+logical whole $N'$ all its super-classes, and, for its logical
+zero N, all its sub-classes. Conversely, to obtain all the
+causes of the same equality, it is sufficient to substitute for
+its logical whole all its sub-classes, and for its logical zero,
+all its super-classes.
+
+\section{Example: Venn's Problem}\label{ch:47}
+\index{Council!Members of}
+\emph{The members of the administrative council of a financial society
+ are either bondholders or shareholders, but not both. Now, all the
+ bondholders
+form a part of the council. What conclusion must we draw?}
+
+Let~$a$ be the class of the members of the council; let~$b$
+be the class of the bondholders and~$c$ that of the shareholders.
+The data of the problem may be expressed as
+follows:
+\begin{displaymath}
+ a < bc' + b'c, \qquad b < a.
+\end{displaymath}
+
+Reducing to a single developed equality,
+\begin{gather}
+ \notag a(b c = b' c') = 0, \qquad a' b = 0,\\
+ \label{eq:47.1} a b c + a b' c' + a' b c + a' b c' = 0.
+\end{gather}
+
+This equality, which contains 4 of the constituents, is
+equivalent to the following, which contains the four others,
+\begin{equation}\label{eq:47.2}
+ a b c' + a b'c + a' b' c + a' b' c' = 1.
+\end{equation}
+
+This equality may be expressed in as many different forms
+as there are classes in the universe of the three terms
+$a, b, c$.
+
+\begin{gather*}
+ \tag*{Ex. 1.} a = a b c' + a b' c + a' b c + a' b c',\\
+ \intertext{that is,}
+ b < a < b c' + b' c, \\
+ \tag*{Ex. 2.} b = a b c' + a b' c = a c';\\
+ \tag*{Ex. 3.} c = a b' c + a' b' c + a b' c' + a' b c'\\
+ \intertext{that is,}
+ a b' + a' b < c < b'.
+\end{gather*}
+
+These are the solutions obtained by solving equation (\ref{eq:47.1})
+with respect to~$a$, $b$, and~$c$.
+
+From equality (\ref{eq:47.1}) we can derive 16 consequences%
+\index{Consequences!Sixteen} as follows :
+\begin{align*}
+\tag*{1.} a b c &= 0;\\
+\tag*{2.} (a b' c'= 0) &= (a < b + c);\\
+\tag*{3.} (a' b c = 0) &= (b c < a);\\
+\tag*{4.} (a' b c = 0) &= (b < a + c);\\
+\tag*{5.} (a b c + a b' c' = 0) &= (a < b c' + b' c) \text{ [1$^{\text{st}}$ premise]};\\
+\tag*{6.} (a b c + a' b c = 0) &= (b c = 0);\\
+\tag*{7.} (a b c + a' b c' = 0) &= (b < a c' + a' c);\\
+\tag*{8.} (a b' c' + a' b c = 0) &= (b c < a < b + c);\\
+\tag*{9.} (a b' c' + a' b c' = 0) &= (a b' + a' b < c);\\
+\tag*{10.} (a' b c + a' b c' = 0) &= (a' b = 0) \text{ [2$^{\text{d}}$ premise]};\\
+\tag*{11.} (a b c + a b' c' + a' b c = 0) &= (b c + a b' c' = 0);\\
+\tag*{12.} a b c + a b' c + a' b c' &= 0;\\
+\tag*{13.} (a b c + a' b c + a' b c' = 0) &= (b c + a' b c') = 0;\\
+\tag*{14.} a b' c' + a' b c + a' b c' &= 0.
+\end{align*}
+
+The last two consequences, as we know, are the identity
+$(0 = 0)$ and the equality (\ref{eq:47.1}) itself. Among the preceding
+consequences will be especially noted the 6\th{} ($b c = 0$), the
+resultant of the elimination of $a$, and the 10\th{} ($a' b = 0$),
+the resultant of the elimination of $c$. When $b$ is eliminated
+the resultant is the identity
+\begin{displaymath}
+ [(a' + c) a c' = 0] = (0=0).
+\end{displaymath}
+
+Finally, we can deduce from the equality (\ref{eq:47.1}) or its equivalent
+(\ref{eq:47.2}) the following 16 causes:\index{Causes!Sixteen}
+\begin{align*}
+\tag*{1.} (a b c' = 1) &= (a=1)(b=1)(c=0);\\
+\tag*{2.} (a b' c = 1) &= (a=1)(b=0)(c=1);\\
+\tag*{3.} (a' b' c = 1) &= (a=0)(b=0)(c=1);\\
+\tag*{4.} (a' b' c' = 1) &= (a=0)(b=0)(c=0);\\
+\tag*{5.} (a b c' + a b' c = 1) &= (a=1)(b'=c);\\
+\tag*{6.} (a b c' + a' b' c = 1) &= (a=b=c');\\
+\tag*{7.} (a b c' + a' b' c' = 1) &= (c=0)(a=b);\\
+\tag*{8.} (a b' c + a' b' c = 1) &= (b=0)(c=1);\\
+\tag*{9.} (a b' c + a' b' c' = 1) &= (b=0)(a=c);\\
+\tag*{10.} (a' b' c + a' b' c' = 1) &= (a=0)(b=0);\\
+\tag*{11.} (a b c' + a b' c + a' b' c = 1) &= (b = c')(c' < a);\\
+\tag*{12.} (a b c' + a b' c + a' b' c' = 1) &= (b c = 0)(a = b + c);\\
+\tag*{13.} (a b c' + a' b' c + a' b' c' = 1) &= (a c = 0)(a = b);\\
+\tag*{14.} (a b' c + a' b' c + a' b' c' = 1) &= (b = o)(a < c).
+\end{align*}
+
+The last two causes, as we know, are the equality (\ref{eq:47.1})
+itself and the absurdity ($1 = 0$). It is evident that the
+cause independent of $a$ is the 8\th{} $(b = 0)(c = 1)$, and the
+cause independent of $c$ is the 10\th{} $(a = 0)(b = 0)$. There
+is no cause, properly speaking, independent of $b$. The most
+``natural'' cause, the one which may be at once divined
+simply by the exercise of common sense, is the 12\th{}:
+\begin{displaymath}
+ (b c = 0)(a = b + c).
+\end{displaymath}
+
+But other causes are just as possible; for instance the 9\th{}
+$(b = 0) (a = c)$, the 7\th{} $(c = 0) (a = b)$, or the 13\th{}
+$(a c = 0) (a = b)$.
+
+We see that this method furnishes the complete enumeration
+of all possible cases. In particular, it comprises, among
+the \emph{forms} of an equality, the solutions deducible therefrom
+with respect to such and such an ``unknown quantity'', and,
+among the \emph{consequences} of an equality, the resultants of the
+elimination of such and such a term.
+
+\section{The Geometrical Diagrams of Venn}\label{ch:48}
+\author{}{Poretsky's}
+method may be looked upon as the perfection of the methods of
+\author{}{Stanley Jevons} and \author{}{Venn.}
+
+Conversely, it finds in them a geometrical and mechanical
+illustration, for \author{}{Venn's} method is translated in
+geometrical diagrams which represent all the constituents, so
+that, in order to obtain the result, we need only strike out (by
+shading) those which are made to vanish by the data of the
+problem. For instance, the universe of three terms $a, b, c$,
+represented by the unbounded plane, is divided by three simple
+closed contours into eight regions which represent the eight
+constituents (Fig.~\ref{fig:1}).
+
+\begin{figure}[htbp]
+ \centering
+ \includegraphics[scale=0.3]{venn.1}
+\caption{}
+\label{fig:1}
+\end{figure}
+
+To represent geometrically the data of \author{}{Venn's} problem
+we must strike out the regions $a b c$, $a b' c'$, $a' b c$ and
+$a' b c'$; there will then remain the regions $a b c'$, $a b' c$,
+$a' b' c$, and $a' b' c'$ which will constitute the universe
+\emph{relative to the problem}, being what \author{}{Poretsky}
+calls his \emph{logical whole} (Fig.~\ref{fig:2}). Then every
+class will be contained in this universe, which will give for each
+class the expression resulting from the data of the problem. Thus,
+simply by inspecting the diagram, we see that the region $b c$
+does not exist (being struck out); that the region $b$ is reduced
+to $a b' c'$ (hence to $a b$); that all $a$ is $b$ or $c$, and so
+on.
+
+\begin{figure}[htbp]
+ \centering
+ \includegraphics[scale=0.25]{venn.2}\\
+ \caption{}
+ \label{fig:2}
+\end{figure}
+
+This diagrammatic method has, however, serious inconveniences as a
+method for solving logical problems. It does not show how the data
+are exhibited by canceling certain constituents, nor does it show
+how to combine the remaining constituents so as to obtain the
+consequences sought. In short, it serves only to exhibit one
+single step in the argument, namely the equation of the problem;
+it dispenses neither with the previous steps, \emph{i.e.},
+``throwing of the problem into an equation'' and the
+transformation of the premises, nor with the subsequent steps,
+\emph{i.e.}, the combinations that lead to the various
+consequences. Hence it is of very little use, inasmuch as the
+constituents can be represented by algebraic symbols quite as well
+as by plane regions, and are much easier to deal with in this
+form.
+
+\section{The Logical Machine of Jevons}\label{ch:49}
+In order to make his diagrams more tractable, \author{}{Venn}
+proposed a mechanical device by which the plane regions to be
+struck out could be lowered and caused to disappear. But
+\author{}{Jevons} invented a more complete mechanism, a sort of
+\emph{logical piano}. The keyboard of this instrument was composed
+of keys indicating the various simple terms $(a, b, c, d)$, their
+negatives, and the signs $+$ and $=$. Another part of the
+instrument consisted of a panel with movable tablets on which were
+written all the combinations of simple terms and their negatives;
+that is, all the constituents of the universe of discourse.
+Instead of writing out the equalities which represent the
+premises, they are ``played'' on a keyboard like that of a
+typewriter. The result is that the constituents which vanish
+because of the premises disappear from the panel. When all the
+premises have been ``played'', the panel shows only those
+constituents whose sum is equal to~1, that is, forms the universe
+with respect to the problem, its logical whole. This mechanical
+method has the advantage over \author{}{Venn's} geometrical method
+of performing automatically the ``throwing into an equation'',
+although the premises must first be expressed in the form of
+equalities; but it throws no more light than the geometrical
+method on the operations to be performed in order to draw the
+consequences from the data displayed on the panel.
+
+\section{Table of Consequences}
+\label{ch:50}\index{Consequences!Table of|(} But
+\author{}{Poretsky's} method can be illustrated, better than by
+geometrical and mechanical devices, by the construction of a table
+which will exhibit directly all the consequences and all the
+causes of a given equality. (This table is relative to this
+equality and each equality requires a different table). Each table
+comprises the $2^n$ classes that can be defined and distinguished
+in the universe of discourse of $n$ terms. We know that an
+equality consists in the annulment of a certain number of these
+classes, viz., of those which have for constituents some of the
+constituents of its \emph{logical zero} $N$. Let $m$ be the number
+of these latter constituents, then the number of the subclasses of
+$N$ is $2^m$ which, therefore, is the number of classes of the
+universe which vanish in consequence of the equality considered.
+Arrange them in a column commencing with 0 and ending with $N$
+(the two extremes). On the other hand, given any class at all, any
+preceding class may be added to it without altering its value,
+since by hypothesis they are null (in the problem under
+consideration). Consequently, by the data of the problem, each
+class is equal to $2^m$ classes (including itself). Thus, the
+assemblage of the $2^n$ classes of discourse is divided into
+$2^{n-m}$ series of $2^m$ classes, each series being constituted
+by the sums of a certain class and of the $2^m$ classes of the
+first column (sub-classes of $N$). Hence we can arrange these
+$2^m$ sums in the following columns by making them correspond
+horizontally to the classes of the first column which gave rise to
+them. Let us take, for instance, the very simple equality $a = b$,
+which is equivalent to
+\begin{displaymath}
+ ab' + a'b = 0.
+\end{displaymath}
+
+The logical zero ($N$) in this case is $a b' + a' b$. It comprises
+two constituents and consequently four sub-classes: $0$, $a b'$,
+$a' b$, and $a b' + a' b$. These will compose the first column.
+The other classes of discourse are $a b$, $a' b'$, $a b + a' b'$,
+and those obtained by adding to each of them the four classes of
+the first column. In this way, the following table is obtained:
+\begin{displaymath}
+ \begin{matrix}
+ 0 & a b & a' b' & a b + a' b' \\
+ a b' & a & b' & a + b' \\
+ a' b & b & a' & a' + b \\
+ a b' + a'b & a + b & a' + b' & 1
+ \end{matrix}
+\end{displaymath}
+
+By construction, each class of this table is the sum of
+those at the head of its row and of its column, and, by the
+data of the problem, it is equal to each of those in the
+same column. Thus we have 64 different consequences for
+any equality in the universe of discourse of 2 letters. They
+comprise 16 identities (obtained by equating each class to
+itself) and 16 forms of the given equality, obtained by
+equating the classes which correspond in each row to the
+classes which are known to be equal to them, namely
+
+\begin{displaymath}
+ \begin{matrix}
+ 0 = a b' + a' b, & a b = a + b, & a' b' = a' + b' & a b + a' b' = 1 \\
+ a = b, & b' = a', & a b' = a' b, & a + b' = a' + b.
+ \end{matrix}
+\end{displaymath}
+
+Each of these 8 equalities counts for two, according as it
+is considered as a determination of one or the other of its
+members.
+\index{Consequences!Table of|)}
+
+\section{Table of Causes}\label{ch:51}\index{Causes!Table of|(}
+The same table may serve to
+represent all the causes of the same equality in accordance
+with the following theorem:
+
+When the consequences of an equality $N = 0$ are expressed
+in the form of determinations of any class $U$, the
+causes of this equality are deduced from the consequences
+of the \emph{opposite} equality, $N = 1$, put in the same form,
+by changing $U$ to $U'$ in one of the two members.
+
+For we know that the consequences of the equality $N = 0$
+have the form
+\begin{displaymath}
+ U = (N' + X) U + N Y U',
+\end{displaymath}
+and that the causes of the same equality have the form
+\begin{displaymath}
+ U = N' X U + (N + Y) U'.
+\end{displaymath}
+
+Now, if we change $U$ into $U'$ in one of the members of
+this last formula, it becomes
+\begin{displaymath}
+ U = (N + X') U + N' Y' U',
+\end{displaymath}
+and the accents of $X$ and $Y$ can be suppressed since these
+letters represent indeterminate classes. But then we have
+the formula of the consequences of the equality $N' = 0$ or
+$N = 1$.
+
+This theorem being established, let us construct, for instance,
+the table of causes of the equality $a = b$. This will
+be the table of the consequences of the opposite equality
+$a = b'$, for the first is equivalent to
+\begin{displaymath}
+ a b' + a' b = 0,
+\end{displaymath}
+and the second to
+\begin{align*}
+ (a b + a' b' = 0) = (a b' + a' b = 1).\\
+ \begin{matrix}
+ 0 & a b' & a' b & a b' + a' b \\
+ a b & a & b & a + b \\
+ a' b' & b' & a' & a' + b' \\
+ a b + a' b' & a + b' & a' + b & 1
+ \end{matrix}
+\end{align*}
+
+To derive the causes of the equality $a = b$ from this table
+instead of the consequences of the opposite equality $a = b'$,
+it is sufficient to equate the negative of each class to each
+of the classes in the same column. Examples are:
+\begin{displaymath}
+ \begin{matrix}
+ a' + b' = 0, & a' + b' = a' b', & a' + b' = a b + a' b', \\
+ a' + b = a, & a' + b = b', & a' + b = a + b';\ldots.
+ \end{matrix}
+\end{displaymath}
+
+Among the 64 causes of the equality under consideration
+there are 16 absurdities (consisting in equating each class of
+the table to its negative); and 16 forms of the equality (the
+same, of course, as in the table of consequences, for two
+equivalent equalities are at the same time both cause and
+consequence of each other).
+
+It will be noted that the table of causes differs from the table
+of consequences only in the fact that it is symmetrical to the
+other table with respect to the principal diagonal $(0, 1)$; hence
+they can be made identical by substituting the word ``row'' for
+the word ``column'' in the foregoing statement. And, indeed, since
+the rule of the consequences concerns only classes of the same
+column, we are at liberty so to arrange the classes in each column
+on the rows that the rule of the causes will be verified by the
+classes in the same row.
+
+It will be noted, moreover, that, by the method of construction
+adopted for this table, the classes which are the
+negatives of each other occupy positions symmetrical with
+respect to the center of the table. For this result, the subclasses
+of the class $N'$ (the logical whole of the given
+equality or the logical zero of the opposite equality) must
+be placed in the first row in their natural order from 0 to $N'$;
+then, in each division, must be placed the sum of the classes
+at the head of its row and column.
+
+With this precaution, we may sum up the two rules in the
+following practical statement:
+
+To obtain every consequence of the given equality (to
+which the table relates) it is sufficient to equate each class
+to every class in the same column; and, to obtain every
+cause, it is sufficient to equate each class to every class in
+the row occupied by its symmetrical class.
+
+It is clear that the table relating to the equality $N = 0$
+can also serve for the opposite equality $N = 1$, on condition
+that the words ``row'' and ``column'' in the foregoing statement
+be interchanged.
+
+Of course the construction of the table relating to a given
+equality is useful and profitable only when we wish to
+enumerate all the consequences or the causes of this equality.
+If we desire only one particular consequence or cause
+relating to this or that class of the discourse, we make use
+of one of the formulas given above.
+\index{Causes!Table of|)}
+
+\section{The Number of Possible Assertions}\label{ch:52}
+If we regard logical functions and equations as developed with
+respect to \emph{all} the letters, we can calculate the number of
+assertions or different problems that may be formulated about $n$
+simple terms. For all the functions thus developed can contain
+only those constituents which have the coefficient 1 or the
+coefficient 0 (and in the latter case, they do not contain them).
+Hence they are additive combinations of these constituents; and,
+since the number of the constituents is $2^n$, the number of
+possible functions is $2^{2^n}$. From this must be deducted the
+function in which all constituents are absent, which is
+identically 0, leaving $2^{2^n}-1$ possible equations (255 when $n
+= 3$). But these equations, in their turn, may be combined by
+logical addition, \emph{i.e.}, by alternation; hence the number of
+their combinations is $2^{2^{2^n}-1}-1$, excepting always the
+null combination. This is the number of possible assertions%
+\index{Assertions!Number of possible}
+affecting $n$ terms. When $n = 2$, this number is as high as
+32767.%
+\footnote{\author{}{G. Peano,} \emph{Calcolo geometrico} (1888)
+p.~x; \author{}{Schröder,} \emph{Algebra der Logik}, Vol. II,
+p.~144--148.} We must observe that only universal premises are
+admitted in this calculus, as will be explained in the following
+section.
+
+\section{Particular Propositions}\label{ch:53}
+Hitherto we have only considered propositions with an
+\emph{affirmative} copula (\emph{i.e.}, inclusions or equalities)
+corresponding to the \emph{universal} propositions
+of classical logic.%
+\footnote{The \emph{universal affirmative}, ``All $a$'s are $b$'s'', may be expressed by
+the formulas
+\begin{displaymath}
+ (a < b) = (a = a b) = (a b' = 0) = (a' + b = 1),
+\end{displaymath}
+and the \emph{universal negative}, ``No $a$'s are $b$'s'', by the formulas
+\begin{displaymath}
+ (a < b') = (a = a b') = (a b = 0) = (a' + b' = 1).
+\end{displaymath}} It remains for us to study propositions
+with a \emph{negative} copula (non inclusions or inequalities),
+which translate \emph{particular} propositions%
+\footnote{For the \emph{particular affirmative}, ``Some $a$'s are $b$'s'', being the negation
+of the universal negative, is expressed by the formulas
+\begin{displaymath}
+ (a \nless b') = (a \neq a b') = (a b \neq 0) = (a'+ b' \neq 1),
+\end{displaymath}
+and the \emph{particular negative}, ``Some $a$'s are not $b$'s'', being the negation
+of the universal affirmative, is expressed by the formulas
+\begin{displaymath}
+ (a \nless b) = (a \neq a b) = (a b' \neq 0) = (a'+ b \neq 1).
+\end{displaymath}}; but the calculus of
+propositions having a negative copula results from laws already
+known, especially from the formulas of \author{}{De Morgan}%
+\index{De Morgan!Formulas of} and the law of contraposition. We shall
+enumerate the chief formulas derived from it.
+
+The principle of composition gives rise to the following
+formulas:
+\begin{align*}
+ (c \nless ab) &= (c \nless a) + (c \nless b), \\
+ (a + b \nless c) &= (a \nless c) + (b \nless c),
+\end{align*}
+whence come the particular instances
+\begin{align*}
+ (a b \neq 1) &= (a \neq 1) + (b \neq 1), \\
+ (a + b \neq 0) &= (a \neq 0) + (c \neq 0).
+\end{align*}
+
+From these may be deduced the following important implications:
+\begin{align*}
+ (a \neq 0) &< (a + b \neq 0),\\
+ (a \neq 1) &< (a b \neq 1).
+\end{align*}
+
+From the principle of the syllogism, we deduce, by the
+law of transposition,
+\begin{align*}
+ (a < b) (a \neq 0) < (b \neq 0),\\
+ (a < b) (b \neq 1) < (a \neq 1).
+\end{align*}
+
+The formulas for transforming inclusions and equalities give
+corresponding formulas for the transformation of non-inclusions
+and inequalities,
+\begin{align*}
+ (a \nless b) = (a b' \neq 0) &= (a' + b \neq 1),\\
+ (a \neq b) = (a b' + a' b \neq 0) &= (ab + a'b' + 1).
+\end{align*}
+
+\section{Solution of an Inequation with One Unknown}\label{ch:54}
+If
+we consider the conditional inequality (\emph{inequation}) with
+one unknown
+\begin{displaymath}
+ a x + b x \neq 0,
+\end{displaymath}
+we know that its first member is contained in the sum of
+its coefficients
+\begin{displaymath}
+ a x + b x' < a + b.
+\end{displaymath}
+
+From this we conclude that, if this inequation is verified,
+we have the inequality
+\begin{displaymath}
+ a + b \neq 0.
+\end{displaymath}
+
+This is the necessary condition of the solvability of the
+inequation, and the resultant of the elimination of the
+unknown $x$. For, since we have the equivalence
+\begin{displaymath}
+ \prod_x (ax + bx' = 0) = (a + b = 0),
+\end{displaymath}
+we have also by contraposition the equivalence
+\begin{displaymath}
+ \sum_x (ax + bx' \neq 0) = (a + b \neq 0).
+\end{displaymath}
+
+Likewise, from the equivalence
+\begin{displaymath}
+ \sum_x (ax + bx' = 0) = (ab = 0)
+\end{displaymath}
+we can deduce the equivalence
+\begin{displaymath}
+ \prod_x (ax + bx'\neq 0) = (ab \neq 0),
+\end{displaymath}
+which signifies that the necessary and sufficient condition%
+\index{Condition!Necessary and sufficient} for the inequation to be always
+true is
+\begin{displaymath}
+ (ab \neq 0);
+\end{displaymath}
+and, indeed, we know that in this case the equation
+\begin{displaymath}
+ (ax + bx' = 0)
+\end{displaymath}
+is impossible (never true).
+
+Since, moreover, we have the equivalence
+\begin{displaymath}
+ (ax + bx' = 0) = (x = a'x + bx'),
+\end{displaymath}
+we have also the equivalence
+\begin{displaymath}
+ (ax + bx' \neq 0)=(x \neq a'x + bx').
+\end{displaymath}
+
+Notice the significance of this solution:
+\begin{displaymath}
+ (ax + bx' \neq 0) = (ax \neq 0) + (bx' \neq 0) = (x \nless a') + (b \nless x).
+\end{displaymath}
+
+``Either $x$ is not contained in $a'$, or it does not contain $b$''.
+This is the negative of the double inclusion
+\begin{displaymath}
+ b< x< a.
+\end{displaymath}
+
+Just as the product of several equalities is reduced to one
+single equality, the sum (the alternative) of several inequalities
+may be reduced to a single inequality. But neither several
+alternative equalities nor several simultaneous inequalities can
+be reduced to one.
+
+\section{System of an Equation and an Inequation}\label{ch:55}
+We
+shall limit our study to the case of a simultaneous equality
+and inequality. For instance, let the two premises be
+\begin{displaymath}
+ (ax + bx' = 0) \quad (cx + dx' \neq 0).
+\end{displaymath}
+
+To satisfy the former (the equation) its resultant $ab = 0$
+must be verified. The solution of this equation is
+\begin{displaymath}
+ x=a'x+bx'.
+\end{displaymath}
+
+Substituting this expression (which is equivalent to the
+equation) in the inequation, the latter becomes
+\begin{displaymath}
+ (a'c + ad) x + (bc + b'd) x' \neq 0.
+\end{displaymath}
+
+Its resultant (the condition of its solvability) is
+\begin{displaymath}
+ (a'c + ad + bc + b'd \neq 0) = [(a' + b) c + (a + b') d \neq 0],
+\end{displaymath}
+which, taking into account the resultant of the equality,
+\begin{displaymath}
+ (ab = 0) = (a' + b = a') = (a + b' = b')
+\end{displaymath}
+may be reduced to
+\begin{displaymath}
+ a'c + b'd \neq 0.
+\end{displaymath}
+
+The same result may be reached by observing that the
+equality is equivalent to the two inclusions
+\begin{displaymath}
+ (x < a') (x' < b'),
+\end{displaymath}
+and by multiplying both members of each by the same term
+\begin{align*}
+ (cx < a'c)(dx' < b'd) &< (cx+dx' < a'c + b'd) \\
+ (cx + dx' \neq 0) &< (a'c + b'd \neq 0).
+\end{align*}
+
+This resultant implies the resultant of the inequality taken alone
+\begin{displaymath}
+ c + d \neq 0,
+\end{displaymath}
+so that we do not need to take the latter into account. It is
+therefore sufficient to add to it the resultant of the equality to
+have the complete resultant of the proposed system
+\begin{displaymath}
+ (ab = 0) (a'c + b'd \neq 0).
+\end{displaymath}
+
+The solution of the transformed inequality (which consequently
+involves the solution of the equality) is
+\begin{displaymath}
+ x \neq (a'c' + ad')x + (bc + b'd)x'.
+\end{displaymath}
+
+\section{Formulas Peculiar to the Calculus of Propositions.}\label{ch:56}
+
+All the formulas which we have hitherto noted are valid
+alike for propositions and for concepts. We shall now
+establish a series of formulas which are valid only for propositions,
+because all of them are derived from an axiom
+peculiar to the calculus of propositions, which may be called
+the \emph{principle of assertion}.%
+\index{Assertion!Principle of}
+
+This axiom is as follows:
+\begin{axiom}\label{axiom:X}\index{Axioms}
+ \begin{displaymath}
+ (a = 1) = a.
+ \end{displaymath}
+\end{axiom}
+
+P.~I.: To say that a proposition a is true is to state the
+proposition itself. In other words, to state a proposition is
+to affirm the truth of that proposition.%
+\footnote{We can see at once that this formula is not susceptible of a conceptual
+interpretation (C.~I.); for, if $a$ is a concept, $(a = 1)$ is a proposition,
+and we would then have a logical equality (identity) between
+a concept and a proposition, which is absurd.}
+
+\emph{Corollary}:
+\begin{displaymath}
+ a' = (a' = 1) = (a = 0).
+\end{displaymath}
+
+P.~I.: The negative of a proposition $a$ is equivalent to the
+affirmation that this proposition is false.
+
+By Ax.~\ref{axiom:IX} (\S\ref{ch:20}), we already have
+\begin{displaymath}
+ (a = 1) (a = 0) = 0,
+\end{displaymath}
+
+``A proposition cannot be both true and false at the same
+time'', for
+
+\begin{displaymath}
+ \tag{Syll.} (a = 1) (a = 0) < (1=0) = 0.
+\end{displaymath}
+
+But now, according to Ax.~\ref{axiom:X}, we have
+\begin{displaymath}
+ (a = 1) + (a = 0) = a + a' = 1.
+\end{displaymath}
+
+``A proposition is either true or false''. From these two
+formulas combined we deduce directly that the propositions
+$(a = 1)$ and $(a = 0)$ are contradictory, \emph{i.e.},
+\begin{displaymath}
+ (a \neq 1) = (a = 0), \qquad (a \neq 0) = (a = 1).
+\end{displaymath}
+
+From the point of view of calculation Ax.~\ref{axiom:X} makes it possible
+to reduce to its first member every equality whose second member is 1, and
+to transform inequalities into equalities. Of course these equalities and
+inequalities must have propositions as their members. Nevertheless all the
+formulas of this section are also valid for classes in the particular case
+where the universe of discourse contains only one element, for then there
+are no classes but 0 and 1. In short, the special calculus of propositions
+is equivalent to the calculus of classes%
+\index{Classes!Calculus of}\index{Calculus!of classes} when the classes can
+possess only the two values $0$ and $1$.
+
+\section{Equivalence of an Implication and an Alternative}\label{ch:57}
+
+The fundamental equivalence%
+\index{Alternative!Equivalence of an implication and an}
+\begin{displaymath}
+ (a < b) =( a' + b = 1)
+\end{displaymath}
+gives rise, by Ax.~\ref{axiom:X}, to the equivalence
+\begin{displaymath}
+ (a < b) = (a' + b)
+\end{displaymath}
+which is no less fundamental in the calculus of propositions.
+To say that $a$ implies $b$ is the same as affirming ``not-$a$ or
+$b$'', \emph{i.e.}, ``either $a$ is false or $b$ is true.'' This equivalence
+is often employed in every day conversation.
+
+\emph{Corollary}.---For any equality, we have the equivalence
+\begin{displaymath}
+ (a = b) = ab + a'b'.
+\end{displaymath}
+
+\emph{Demonstration:}
+\begin{displaymath}
+ (a = b) = (a < b) (b < a) = (a' + b) (b' + a) = ab + a'b'
+\end{displaymath}
+
+``To affirm that two propositions are equal (equivalent)
+is the same as stating that either both are true or both are
+false''.
+
+The fundamental equivalence established above has important
+consequences which we shall enumerate.
+
+In the first place, it makes it possible to reduce secondary,
+tertiary, etc., propositions to primary propositions, or even
+to sums (alternatives) of elementary propositions. For it
+makes it possible to suppress the copula of any proposition,
+and consequently to lower its order of complexity. An implication
+($A < B$), in which $A$ and $B$ represent propositions
+more or less complex, is reduced to the sum $A' + B$, in
+which only copulas within $A$ and $B$ appear, that is, propositions
+of an inferior order. Likewise an equality ($A = B$)
+is reduced to the sum ($AB + A'B'$) which is of a lower
+order.
+
+We know that the principle of composition%
+\index{Composition!Principle of} makes it possible to combine several
+\emph{simultaneous} inclusions or equalities, but we cannot combine
+alternative inclusions or equalities, or at least the result is not
+equivalent to their alternative but is only a consequence of it. In short,
+we have only the \emph{implications}
+\begin{align*}
+ (a < c) + (b < c) &< (ab < c), \\
+ (c < a) + (c < b) &< (c < a + b),
+\end{align*}
+which, in the special cases where $c = 0$ and $c = 1$, become
+\begin{align*}
+ (a = 0) + (b = 0) &< (ab = 0), \\
+ (a = 1) + (b = 1) &< (a + b = 1).
+\end{align*}
+
+In the calculus of classes,%
+\index{Classes!Calculus of}\index{Calculus!of classes} the converse
+implications are not valid, for, from the statement that the class $ab$ is
+null, we cannot conclude that one of the classes $a$ or $b$ is null (they
+can be not-null and still not have any element in common); and from the
+statement that the sum $(a + b)$ is equal to 1 we cannot conclude that
+either $a$ or $b$ is equal to 1 (these classes can \emph{together} comprise
+all the elements of the universe without any of them \emph{alone}
+comprising all). But these converse implications are true in the calculus
+of propositions
+\begin{align*}
+ (ab < c) &< (a < c) + (b < c), \\
+ (c < a + b) &< (c < a) + (c < b);
+\end{align*}
+for they are deduced from the equivalence established above, or
+rather we may deduce from it the corresponding equalities which
+imply them,
+\begin{align*}
+ \tag{1} (ab < c) &= (a < c) + (b < c), \\
+ \tag{2} (c < a + b) &= (c < a) + (c < b).
+\end{align*}
+
+\emph{Demonstration:}
+\begin{gather*}
+ \tag{1} (ab < c) = a' + b' + c, \\
+ (a < c) + (b < c) = (a' + c) + (b' + c) = a' + b' + c; \\
+ \tag{2} (c < a + b) = c' + a + b, \\
+ (c < a) + (c < b) = (c' + a) + (c' + b) = c' + a + b.
+\end{gather*}
+
+In the special cases where $c = 0$ and $c = 1$ respectively,
+we find
+\begin{gather*}
+ \tag{3} (ab = 0) = (a = 0) + (b = 0), \\
+ \tag{4} (a + b = 1) = (a = 1) + (b = 1).
+\end{gather*}
+
+P.~I.: (1) To say that two propositions united imply a
+third is to say that one of them implies this third proposition.
+
+(2) To say that a proposition implies the alternative of
+two others is to say that it implies one of them.
+
+(3) To say that two propositions combined are false is to
+say that one of them is false.
+
+(4) To say that the alternative of two propositions is true
+is to say that one of them is true.
+
+The paradoxical character of the first three of these statements
+will be noted in contrast to the self-evident character of the
+fourth. These paradoxes are explained, on the one hand, by the
+special axiom which states that a proposition is either true or
+false; and, on the other hand, by the fact that the false implies
+the true and that \emph{only} the false is not implied by the
+true. For instance, if both premises in the first statement are
+true, each of them implies the consequence, and if one of them is
+false, it implies the consequence (true or false). In the second,
+if the alternative is true, one of its terms must be true, and
+consequently will, like the alternative, be implied by the premise
+(true or false). Finally, in the third, the product of two
+propositions cannot be false unless one of them is false, for, if
+both were true, their product would be true (equal to 1).
+
+\section{Law of Importation and Exportation}\label{ch:58}
+The fundamental
+equivalence $(a < b) = a' + b$ has many other interesting
+consequences. One of the most important of these
+is \emph{the law of importation and exportation}, which is expressed
+by the following formula:
+\begin{displaymath}
+ [a < (b < c)] = (ab < c).
+\end{displaymath}
+
+``To say that if $a$ is true $b$ implies $c$, is to say that $a$
+and $b$ imply $c$''.
+
+This equality involves two converse implications: If we
+infer the second member from the first, we \emph{import} into the
+implication $(b < c)$ the hypothesis or condition $a$; if we infer
+the first member from the second, we, on the contrary,
+\emph{export} from the implication $(ab < c)$ the hypothesis $a$.
+
+\emph{Demonstration:}
+\begin{gather*}
+ [a < (b < c)] = a' + (b < c) = a' + b' + c, \\
+ (ab < c) = (ab)' + c = a' + b' + c.
+\end{gather*}
+
+\emph{Cor.} 1.---Obviously we have the equivalence
+\begin{displaymath}
+ [a < (b < c)] = [b < (a < c)],
+\end{displaymath}
+since both members are equal to $(ab < c)$, by the commutative
+law of multiplication.
+
+\emph{Cor.} 2.---We have also
+\begin{displaymath}
+ [a < (a < b)] = (a < b),
+\end{displaymath}
+for, by the law of importation and exportation,
+\begin{displaymath}
+ [a < (a < b)] = (aa < b) = (a < b).
+\end{displaymath}
+
+If we apply the law of importation to the two following
+formulas, of which the first results from the principle of
+identity and the second expresses the principle of contraposition,%
+\index{Contraposition!Principle of}
+\begin{displaymath}
+ (a < b) < (a < b), \qquad (a < b) < (b' < a'),
+\end{displaymath}
+we obtain the two formulas
+\begin{displaymath}
+ (a < b)a < b), \qquad (a < b)b' < a',
+\end{displaymath}
+which are the two types of \emph{hypothetical reasoning}: ``If $a$
+implies $b$, and if $a$ is true, $b$ is true'' (\emph{modus ponens}); ``If $a$
+implies $b$, and if $b$ is false, $a$ is false'' (\emph{modus tollens}).
+
+\emph{Remark}. These two formulas could be directly deduced
+by the principle of assertion, from the following
+\begin{align*}
+ (a < b) (a = 1) &< (b = 1), \\
+ (a < b) (b = 0) &< (a = 0),
+\end{align*}
+which are not dependent on the law of importation and
+which result from the principle of the syllogism.
+
+From the same fundamental equivalence, we can deduce
+several paradoxical formulas:
+\begin{displaymath}
+\tag*{1.} a < (b < a), \qquad a' < (a < b).
+\end{displaymath}
+
+``If $a$ is true, $a$ is implied by any proposition $b$; if $a$ is
+false, $a$ implies any proposition $b$''. This agrees with the
+known properties of~0 and~1.
+
+\begin{displaymath}
+ \tag*{2.} a < [(a < b) < b], \qquad a' < [(b < a) < b'].
+\end{displaymath}
+
+``If $a$ is true, then '$a$ implies $b$' implies $b$; if $a$ is false,
+then '$b$ implies $a$' implies not-$b$.''
+
+These two formulas are other forms of hypothetical reasoning
+(\emph{modus ponens} and \emph{modus tollens}).
+
+\begin{displaymath}
+ \tag*{3.} [(a < b) < a] = a,%
+ \footnote{This formula is \author{}{Bertrand Russell's} ``principle of reduction''.
+ See \emph{The Principles of Mathematics}, Vol. I, p.~17 (Cambridge, 1903).}
+ \qquad [(b < a) < a'] = a',
+\end{displaymath}
+
+``To say that, if $a$ implies $b$, $a$ is true, is the same as affirming
+$a$; to say that, if $b$ implies $a$, $a$ is false, is the same as
+denying $a$''.
+
+\emph{Demonstration:}
+\begin{align*}
+ [(a < b) < a] &= (a' + b < a) = ab' + a = a, \\
+ [(b < a) < a'] &= (b' + a < a') = a'b + a' = a'.
+\end{align*}
+
+In formulas (1) and (3), in which $b$ is any term at all,
+we might introduce the sign $\prod$ with respect to $b$. In the
+following formula, it becomes necessary to make use of this
+sign.
+
+\begin{displaymath}
+\tag*{4.} \prod_{x} \left\{[a < (b < x)] < x \right\} = ab.
+\end{displaymath}
+
+\emph{Demonstration:}
+\begin{align*}
+ \left\{[a < (b < x)] < x \right\} &= \left\{[a' + (b < x)] < x \right\} \\
+ &= [(a' + b' + x) < x] = abx' + x = ab + x.
+\end{align*}
+
+We must now form the product $\prod_{x}(ab + x)$, where $x$
+can assume every value, including 0 and 1. Now, it is
+clear that the part common to all the terms of the form
+$(ab + x)$ can only be $ab$. For, (1) $ab$ is contained in each
+of the sums $(ab + x)$ and therefore in the part common to
+all; (2) the part common to all the sums $(ab + x)$ must be
+contained in $(ab + 0)$, that is, in $ab$. Hence this common
+part is equal to $ab$,%
+\footnote{This argument is general and from it we can deduce the formula
+\begin{displaymath}
+ \prod_{x}(a + x) = a,
+\end{displaymath}
+whence may be derived the correlative formula
+\begin{displaymath}
+ \sum_{x} ax = a.
+\end{displaymath}
+} which proved the theorem.
+
+\section{Reduction of Inequalities to Equalities}\label{ch:59}
+As we
+have said, the principle of assertion enables us to reduce
+inequalities to equalities by means of the following formulas:
+\begin{gather*}
+ (a \neq 0) = (a = 1), \qquad (a \neq 1) = (a = 0), \\
+ (a \neq b) = (a = b').
+\end{gather*}
+For,
+\begin{displaymath}
+ (a \neq b) = (ab' + a'b + 0) = (ab' + ab' = 1) = (a = b').
+\end{displaymath}
+Consequently, we have the paradoxical formula
+\begin{displaymath}
+ (a \neq b) = (a = b').
+\end{displaymath}
+
+This is easily understood, for, whatever the proposition~$b$,
+either it is true and its negative is false, or it is false and
+its negative is true. Now, whatever the proposition~$a$ may
+be, it is true or false; hence it is necessarily equal either to
+$b$ or to $b'$. Thus to deny an equality (between propositions)
+is to affirm the \emph{opposite} equality.
+
+Thence it results that, in the calculus of propositions, we
+do not need to take inequalities into consideration---a fact
+which greatly simplifies both theory and practice. Moreover,
+just as we can combine alternative equalities, we can
+also combine simultaneous inequalities, since they are reducible
+to equalities.
+
+For, from the formulas previously established (\S\ref{ch:57})
+\begin{align*}
+ (ab = 0) &= (a = 0) + (b = 0),\\
+ (a + b = 1) &= (a = 1) + (b = 1),
+\end{align*}
+we deduce by contraposition
+\begin{align*}
+ (a \neq 0) (b \neq 0) &= (ab \neq 0),\\
+ (a \neq 1) (b \neq 1) &= (a + b \neq 1).
+\end{align*}
+
+These two formulas, moreover, according to what we have
+just said, are equivalent to the known formulas
+\begin{align*}
+ (a = 1) (b = 1) &= (ab = 1),\\
+ (a = 0) (b = 0) &= (a + b = 0).
+\end{align*}
+
+Therefore, in the calculus of propositions, we can solve
+all simultaneous systems of equalities or inequalities and all
+alternative systems of equalities or inequalities, which is not
+possible in the calculus of classes.%
+\index{Classes!Calculus of}\index{Calculus!of classes} To this end, it is
+necessary only to apply the following rule:
+
+First reduce the inclusions to equalities and the non-inclusions
+to inequalities; then reduce the equalities so that their second
+members will be 1, and the inequalities so that their second
+members will be 0, and transform the latter into equalities having
+1 for a second member; finally, suppress the second members 1 and
+the signs of equality, \emph{i.e.}, form the product of the first
+members of the simultaneous equalities and the sum of the first
+members of the alternative equalities, retaining the parentheses.
+
+\section{Conclusion}\label{ch:60}
+The foregoing exposition is far from being exhaustive; it does not
+pretend to be a complete treatise on the algebra of logic, but
+only undertakes to make known the elementary principles and
+theories of that science. The algebra of logic is an algorithm
+\index{Algebra!of logic an algorithm} \index{Algorithm!Algebra of
+logic an}with laws peculiar to itself. In some phases it is very
+analogous to ordinary algebra, and in others it is very widely
+different. For instance, it does not recognize the distinction of
+\emph{degrees}; the
+laws of tautology and absorption%
+\index{Absorption!Law of} introduce into it great
+simplifications by excluding from it numerical coefficients.
+It is a formal calculus which can give rise to all sorts of
+theories and problems, and is susceptible of an almost infinite
+development.
+
+But at the same time it is a restricted system, and it is
+important to bear in mind that it is far from embracing all
+of logic. Properly speaking, it is only the algebra of
+classical logic. Like this logic, it remains confined to the
+domain circumscribed by Aristotle,\index{Aristotle} namely, the domain of
+the relations of inclusion between concepts and the relations
+of implication between propositions. It is true that classical
+logic (even when shorn of its errors and superfluities) was
+much more narrow than the algebra of logic. It is almost
+entirely contained within the bounds of the theory of the
+syllogism whose limits to-day appear very restricted and
+artificial. Nevertheless, the algebra of logic simply treats,
+with much more breadth and universality, problems of the
+same order; it is at bottom nothing else than the theory
+of classes or aggregates considered in their relations of inclusion
+or identity. Now logic ought to study many other
+kinds of concepts than generic concepts (concepts of classes)
+and many other relations than the relation of inclusion (of
+subsumption) between such concepts. It ought, in short, to
+develop into a logic of relations, which \author{}{Leibniz} foresaw,
+which \author{}{Peirce} and \author{}{Schröder} founded, and which \author{}{Peano} and
+\author{}{Russell} seem to have established on definite foundations.
+
+While classical logic and the algebra of logic are of hardly any
+use to mathematics, mathematics, on the other hand, finds in the
+logic of relations its concepts and fundamental principles; the
+true logic of mathematics is the logic of relations. The algebra
+of logic itself arises out of pure logic considered as a
+particular mathematical theory, for it rests on principles which
+have been implicitly postulated and which are not susceptible of
+algebraic or symbolic expression because they are the foundation
+of all symbolism and of all
+the logical calculus.\footnote{The principle of deduction%
+ \index{Deduction!Principle of} and the principle of substitution%
+ \index{Substitution!Principle of}. See the author's%
+ \index{Couturat} \emph{Manuel de Logistique}, Chapter 1, \S\S~2
+ and~3 [not published], and \emph{Les Principes des Mathématiques},
+ Chapter~1, A.} Accordingly, we can say that the algebra of logic is
+a \emph{mathematical} logic by its form and by its method, but it must
+not be mistaken for the logic \emph{of mathematics}.
+
+\cleardoublepage
+\begin{theindex}
+\item Absorption, Law of \item Absurdity, Type of \item Addition,
+and multiplication, Logical \subitem and multiplication, Modulus
+of \subitem and multiplication, Theorems on \subitem Logical, not
+disjunctive \item Affirmative propositions \item Algebra, of logic
+an algorithm \subitem of logic compared to mathematical algebra
+\subitem of thought \item Algorithm, Algebra of logic an \item
+Alphabet of human thought \item Alternative \subitem affirmation
+\subitem Equivalence of an implication and an \item Antecedent
+\item Aristotle \item Assertion, Principle of \item Assertions,
+Number of possible \item Axioms \indexspace \item Baldwin \item
+Boole \subitem Problem of \item Bryan, William Jennings
+\indexspace \item Calculus, Infinitesimal \subitem Logical
+\subitem \emph{ratiocinator} \item Cantor, Georg \item Categorical
+syllogism \item Cause \item Causes, Forms of \subitem Law of
+\subitem Sixteen \subitem Table of \item Characters \item Classes,
+Calculus of \item Classification of dichotomy \item Commutativity
+\item Composition, Principle of \item Concepts, Calculus of \item
+Condition \subitem Necessary and sufficient \subitem Necessary but
+not sufficient \subitem of impossibility and indetermination \item
+\emph{Connaissances} \item Consequence \item Consequences, Forms
+of \subitem Law of \subitem of the syllogism \subitem Sixteen
+\subitem Table of \item Consequent \item Constituents \subitem
+Properties of \item Contradiction, Principle of \item
+Contradictory propositions \subitem terms \item Contraposition,
+Law of \subitem Principle of \item Council, Members of \item
+Couturat, v \indexspace \item Dedekind \item Deduction \subitem
+Principle of \item Definition, Theory of \item De Morgan \subitem
+Formulas of \item Descartes \item Development \subitem Law of
+\subitem of logical functions \subitem of mathematics \subitem of
+symbolic logic \item Diagrams of Venn, Geometrical \item
+Dichotomy, Classification of \item Disjunctive, Logical addition
+not \subitem sums \item Distributive law \item Double inclusion
+\subitem expressed by an indeterminate \subitem Negative of the
+\item Double negation \item Duality, Law of \indexspace \item
+Economy of mental effort \item Elimination of known terms \subitem
+of middle terms \subitem of unknowns \subitem Resultant of
+\subitem Rule for resultant of \item Equalities, Formulas for
+transforming inclusions into \subitem Reduction of inequalities to
+\item Equality a primitive idea \subitem Definition of \subitem
+Notion of \item Equation, and an inequation \subitem Throwing into
+an \item Equations, Solution of \item Excluded middle, Principle
+of \item Exclusion, Principle of \item Exclusive, Mutually \item
+Existence, Postulate of \item Exhaustion, Principle of \item
+Exhaustive, Collectively \indexspace \item Forms, Law of \subitem
+of consequences and causes \item Frege \subitem Symbolism of \item
+Functions \subitem Development of logical \subitem Integral
+\subitem Limits of \subitem Logical \subitem of variables \subitem
+Properties of developed \subitem Propositional \subitem Sums and
+products of \subitem Values of \indexspace \item Hôpital, Marquis
+de l' \item Huntington, E. V \item Hypothesis \item Hypothetical
+arguments \subitem reasoning \subitem syllogism \indexspace \item
+Ideas, Simple and complex \item Identity \subitem Principle of
+\subitem Type of \item Ideography \item Implication \subitem and
+an alternative, Equivalence of an \subitem Relations of \item
+Importation and exportation, Law of \item Impossibility, Condition
+of \item Inclusion \subitem a primitive idea \subitem Double
+\subitem expressed by an indeterminate \subitem Negative of the
+double \subitem Relation of \item Inclusions into equalities,
+Formulas for transforming \item Indeterminate \subitem Inclusion
+expressed by an \item Indetermination \subitem Condition of \item
+Inequalities, to equalities, Reduction of \subitem Transformation
+of non-inclusions and \item Inequation, Equation and an \subitem
+Solution of an \item Infinitesimal calculus \item Integral
+function \item Interpretations of the calculus \indexspace \item
+Jevons \subitem Logical piano of \item Johnson, W. E \indexspace
+\item Known terms (\emph{connaissances}) \indexspace \item
+Ladd-Franklin, Mrs \item Lambert \item Leibniz \item Limits of a
+function \indexspace \item MacColl \item MacFarlane, Alexander
+\item Mathematical function \subitem logic \item Mathematics,
+Philosophy a universal \item Maxima of discourse \item Middle,
+Principle of excluded \subitem terms, Elimination of \item Minima
+of discourse \item Mitchell, O \item Modulus of addition and
+multiplication \item \emph{Modus ponens} \item \emph{Modus
+tollens} \item Müller, Eugen \item Multiplication. See \emph{s.
+v.} ``Addition.'' \indexspace \item Negation \subitem defined
+\subitem Double \subitem Duality not derived from \item Negative
+\subitem of the double inclusion \subitem propositions \item
+Non-inclusions and inequalities, Transformation of \item Notation
+\item Null-class \item Number of possible assertions \indexspace
+\item One, Definition of, \indexspace \item Particular
+propositions, \item Peano, \item Peirce, C. S., \item Philosophy a
+universal mathematics, \item Piano of Jevons, Logical, \item
+Poretsky, \subitem Formula of, \subitem Method of, \item
+Predicate, \item Premise, \item Primary proposition, \item
+Primitive idea, \subitem Equality a, \subitem Inclusion a, \item
+Product, Logical, \item Propositions, \subitem Calculus of,
+\subitem Contradictory, \subitem Formulas peculiar to the calculus
+of, \subitem Implication between, \subitem reduced to lower
+orders, \subitem Universal and particular, \item Reciprocal, \item
+\emph{Reductio ad absurdum,} \item Reduction, Principle of, \item
+Relations, Logic of, \item Relatives, Logic of, \item Resultant of
+elimination, \subitem Rule for, \item Russell, B., \indexspace
+\item Schröder, \subitem Theorem of, \item Secondary proposition,
+\item Simplification, Principle of, \item Simultaneous
+affirmation, \item Solution of equations, \subitem of inequations,
+\item Subject, \item Substitution, Principle of, \item
+Subsumption, \item Summand, \item Sums, \subitem and products of
+functions, \subitem Disjunctive, \subitem Logical, \item
+Syllogism, Principle of the, \subitem Theory of the, \item
+Symbolic logic, \subitem Development of, \item Symbolism in
+mathematics, \item Symbols, Origin of, \item Symmetry, \item
+Tautology, Law of \item Term, \item Theorem, \item Thesis, \item
+Thought, \subitem Algebra of, \subitem Alphabet of human, \subitem
+Economy of, \item Transformation \subitem of inclusions into
+equalities, \subitem of inequalities into equalities, \subitem of
+non-inclusions and inequalities, \indexspace \item Universal
+\subitem characteristic of Leibniz, \subitem propositions,
+\indexspace \item Universe of discourse, \item Unknowns,
+Elimination of, \indexspace \item Variables, Functions of, \item
+Venn, John, \subitem metrical diagrams of, \subitem Mechanical
+device of, \subitem Problem of, \item Viète, \item Voigt,
+\indexspace \item Whitehead, A. N., \item Whole, Logical,
+\indexspace \item Zero, \subitem Definition of, \subitem Logical,
+\end{theindex}
+
+\newpage
+\chapter{PROJECT GUTENBERG "SMALL PRINT"}
+\small
+\pagenumbering{gobble}
+\begin{verbatim}
+End of the Project Gutenberg EBook of The Algebra of Logic, by Louis Couturat
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diff --git a/10836-t/venn.2 b/10836-t/venn.2
new file mode 100644
index 0000000..048aa20
--- /dev/null
+++ b/10836-t/venn.2
@@ -0,0 +1,1752 @@
+%!PS-Adobe-3.0
+%%Creator: GIMP PostScript file plugin V 1.12 by Peter Kirchgessner
+%%Title: /usb/images/venn.2
+%%CreationDate: Sun Jan 25 16:14:15 2004
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+{currentfile /ASCII85Decode filter /RunLengthDecode filter gstr readstring pop}
+{currentfile /ASCII85Decode filter /RunLengthDecode filter bstr readstring pop}
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+%%Trailer
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+%%EOF
diff --git a/LICENSE.txt b/LICENSE.txt
new file mode 100644
index 0000000..6312041
--- /dev/null
+++ b/LICENSE.txt
@@ -0,0 +1,11 @@
+This eBook, including all associated images, markup, improvements,
+metadata, and any other content or labor, has been confirmed to be
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+
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diff --git a/README.md b/README.md
new file mode 100644
index 0000000..b48ba69
--- /dev/null
+++ b/README.md
@@ -0,0 +1,2 @@
+Project Gutenberg (https://www.gutenberg.org) public repository for
+eBook #10836 (https://www.gutenberg.org/ebooks/10836)