Writings of Charles S. Peirce: A Chronological Edition, Volume 6. Charles S. Peirce

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СКАЧАТЬ that is, one of those two, a and c, is false, so that a + ac is false whenever a is false and only then.

      And on the same principle a + ab is equal to a, and thus the second member of the last conclusion reduces to a + bc, and the whole conclusion is (a + b)(a + c) equals a + bc, which is the very conclusion we had to prove. The two principles that (a + b)c = ac + bc, and a + bc equals (a + b)(a + c) are commonly referred to by saying that multiplication is distributive with reference to addition, and that addition is distributive with reference to multiplication. I shall now introduce two statements which have special symbols in this system; the first is $, and means any fact necessarily true. All facts that are necessarily true are equal, because we agreed that we should say that two statements are equal provided they are true and false together in all conceivable states of things. The other special symbol for a statement is 0. This signifies any statement that is false. All false statements are equal for the same reason that all true statements are equal. There are a number of rules facilitating the use of these symbols $ and 0. The first is a + 0 = a. This means that to say that either a is true, or else a false statement is true, is the same as to say at once that a is true. In like manner $a = a; for this means that to say a is true and also that any undesignated true statement is true, is no more than to say that a is true. Second, $ + # = $, for this means that to say that either a is true or something true is true, is no more than to say that something true is true, which is not saying anything at all. And in like manner 0a = 0; for this means that to say that a is true and that something false is true, is to say something false. Third, since every statement is either true or false, if we replace any letter, say a, by $ throughout any formula and find the formula is then necessarily true, and if, on afterwards replacing the same letter by 0, we find that the formula so resulting is true also, then the original formula must be true any way. This affords quite a valuable means of proving any doubtful formula. For instance, let us apply it to proving the formula demonstrated above, a + bc = (a + b)(a + c). First, replace a by $ and the formula becomes, $ + bc = ($ + b)($ + c). Now, $ added to anything gives $; so that $ + bc = $, $ + b = $, and $ + c = $. The whole formula thus reduces to $ = $$ which is true. Now replace a by 0 and the formula becomes, 0 + bc = (0 + b)(0 + c). Now 0 added to anything does not alter it, so that we may drop these added Os, and the formula reduces to bc = bc, which is true. Thus, it has been shown that the formula is true when a equals $ or a equals 0; and as a must equal one or the other, it is true any way. We must now introduce a new sign, a = $ is the same as a written alone; it means that the statement a is true. But we have as yet no simple expression for a = 0, meaning that the statement a is false. Let us denote this by making a line over the a; thus, a; this we call the negative or denial of a. There are several rules facilitating the use of denials. First, aā equals 0, or nothing can be true and false at the same time; this is called the principle of contradiction. Second, a + ā equals $, or everything is either true or false; this is called the principle of excluded middle.

      We will now proceed to show this system of signs is to be used for the purpose of drawing conclusions from premises. The simplest possible kind of reasoning is the immediate application of a rule. Thus, a little girl says that whatever mamma forbids is wrong, but mamma forbids this, therefore, this is wrong. Let a mean that anything is forbidden by mamma, b that it is wrong. Then, to say that anything is forbidden by mamma is wrong is the same as to say that either it is not forbidden by mamma, or else it is wrong. This proposition is therefore written ā + b. The other proposition is a. These propositions are asserted to be both true and therefore, they must be multiplied together, and we have, a(ā + b). On performing this multiplication, that is, on applying the distributive principle, we get aā + ab, but aā is 0 by the principle of contradiction, and may therefore be dropped. We therefore have ab. ab is therefore asserted of both the propositions a and b. It therefore asserts b, and therefore the act in question is wrong. Now it would of course be perfectly ridiculous to use this cumbrous system of signs for the purpose of bringing out the conclusion of such a simple mode of argument as this, but it will be found that the system is well adapted to complicated cases but this very feature makes it cumbrous for simple ones.

      It will be observed in the above example that after we get the conclusion ab, we drop the factor a, leaving only b. We obviously have the right to do this at any time. We are always entitled to drop a factor from any additive term, and we are also at liberty to add a term to any factor. In consequence of this, whenever we have given an expression in the form a(b + c) we are at liberty to drop the parenthesis and write ab + c. For the distributive principle gives us ab + ac, and on dropping the factor a from the last term, we get ab + c.

      In my different publications I have used a sign like Y turned over on its side,

, to signify the relation between the antecedent and consequent of an hypothetical proposition. It is a very convenient sign, but as I have no such sign on this typewriter, I shall use a colon for the same purpose, according to the practice of Mr. Hugh McColl. Then, we may write a:b = ā + b. But although the use of this sign does simplify some cases, and I have been one of its principal advocates, and it certainly is useful for a learner, yet it makes a good deal of difficulty in complicated problems. The best way for a beginner to do is to use this sign first, to write down the relations of antecedent and consequent, and afterwards to replace it by +, at the same time negativing the antecedent.

      Let us now take a slightly more complicated kind of reasoning, the direct syllogism. If you tell one lie you will tell a hundred, and if you tell a hundred lies you will corrupt your integrity. Let a mean that you tell a lie, b that you tell a hundred, c that you corrupt your integrity. Then, the premises are a:b and b:c. Multiplying them together we have (a:b)(b:c). This is equivalent to (ā + b)

. Breaking down the first parenthesis, according to the rule just given, we have . Now breaking down the second parenthesis, according to the same rule, we have . But bb is 0 and may be dropped. Thus, we reach the conclusion ā + c or a:c, if you tell a lie you will corrupt your integrity.

      I will now show how to treat a little more complicated arguments called indirect syllogisms. Take these premises: if Enoch and Elijah are mortal the Bible errs; but all men are mortal. Let a mean that any given person is Enoch or Elijah, b that he is a man, c that he is mortal, and d that the Bible errs. Then a:c means that if any person is Enoch or Elijah he is mortal, or what is the same thing, that Enoch and Elijah are mortal. Then, (a:c):d means that if Enoch and Elijah are mortal the Bible errs, which is the first premise. The second premise is b:c, or if any person is a man he is mortal. Multiplying the two premises together, we have [(a:c):d](b:c). Now a:c = ā + c and (ā + c):d is to be converted into the regular form by negativing the antecedent and putting a + instead of the colon. We have, therefore, to find the negative of ā + c. The rule for finding the negative of any expression is this: put a line over every letter that has no line over it, and take a line off every letter that has a line over it, and everywhere substitute multiplication for addition and addition for multiplication. Applying this rule, the negative of ā + c is

. That is to say, to deny that anything is either not Enoch or Elijah, or else is mortal, is equivalent to asserting that something is Enoch or Elijah and at the same time is not mortal. To prove СКАЧАТЬ