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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 + +*** END OF THIS PROJECT GUTENBERG EBOOK THE ALGEBRA OF LOGIC *** + +***** This file should be named 10836-t.tex or 10836-t.zip ***** +This and all associated files of various formats will be found in: + https://www.gutenberg.org/1/0/8/3/10836/ + +Produced by David Starner, Arno Peters, Susan Skinner +and the Online Distributed Proofreading Team. + +Updated editions will replace the previous one--the old editions +will be renamed. + +Creating the works from public domain print editions means that no +one owns a United States copyright in these works, so the Foundation +(and you!) can copy and distribute it in the United States without +permission and without paying copyright royalties. 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Anyone seeking to utilize +this eBook outside of the United States should confirm copyright +status under the laws that apply to them. 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) |
