University:
Massachusetts Institute of Technology
Course:
24.242 | Logic II
Academic year:

2020
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111

Pages:

15
Author:

kopiena3

Peano Arithmetic Peano Arithmetic1or PA is the system we get from Robinson's Arithmetic by adding the induction axiom schema: ((R(0) A (Yx)(R(x) - R(sx))) - (Yx)R(x)). What this means is that any sentence of the language of arithmetic that you get from the schema by replacing the schematic letter "R" with a formula, then prefixing universal quantifiers to bind all the free variables is an axiom of PA. Thus PA consist of the axioms (Ql) through (Q1 I), together with infinitely many induction axioms. The induction axiom schema formalizes a familiar method of reasoning about the natural numbers. To show that every natural number has the property expressed by the formula we substitute for "R," we begin by showing that 0 has the property; this is the base case. Next we derive, by conditional proof, the conditional Wx)(R(x) - R(sx)); we assume R(x) as inductive hypothesis, then derive R(sx). The rule of mathematical induction permits us to infer (Vx)R(x). Virtually all of our ordinary mathematical reasoning about the natural numbers can be formalized in PA. Indeed, after some initial awkwardness, in which we produce proofs of facts of elementary arithmetic that we've taken for granted since childhood, reasoning in PA is nearly indistinguishable from ordinary arithmetical thinking. 1'11 do a couple of these early proofs informally here, just to get an idea of what's going on. 1 The so-called Peano axioms were first formulated by Richard Dedekind. Peano said as much in a footnote, but somehow "Peano Arithmetic" was the name that stuck. Peano Arithmetic, p. 2 Proposition 1. PA t(Vx)(x = 0 V (3y)x = sy). Proof: Use the following induction axiom: [[(0 = 0 V (3y)O = sy) A (Vx)((x = 0 V (3y)x = sy) - ('V'x)(x =0 - (sx = 0 V (3y)sx = sy))] v (3y)x = sy)] The antecedent is a theorem of pure 1ogic.m Proposition 2. PA ~(VX)(O + x) = x. Proof: Use the following induction axiom: [[(0 + 0) = 0 A (Vx)((O + x) = x The base clause, "(0 + (0 + sx) = sx)] - (Vx)(O + x) = x] + 0) = 0," follows from (43). To get the induction step, assume, as inductive hypothesis (M) that (0 + x) = x. We have (0 + sx) = s(0 + x) [by (44)] [by IH].q = sx Proposition 3. PA t(Vx)(Vy)(sx + y) = s(x + y). Proof: We use the following induction axiom: (Vx)[[(sx + 0) = s(x + 0) A (Vy)((sx + y) = s(x + y) - ('V'Y)(SX + Y)= s(x + Y)I. The base clause is easy. Two applications of (43) yield (sx + 0) = sx = s(x + 0) To get the induction step, assume, as IH, (SX+ y) = s(x + y). We have: - (sx + sy) = s(x + sy))] Peano Arithmetic, p. 3 (SX+ SY)= s(sx + Y) [by (4411 = ss(x + y) [by IH] + sy) = S(X [by (44) again].H Proposition 4 (Commutative law of addition). PA t(b'x)(Vy)(x + y) = (y + x). Proof: We use this induction axiom: (b'x)[[(x + 0) = (0 + x) - WY)((X+ Y)= (Y+ x) + (x + SY)= (SY+ x))l (b'y)(x + Y)= (Y+ x)l. (43) gives us "(x + 0) = x," and Proposition 2 gives us "(0 + x) = x"; these together yield the base clause, "(x + 0) = (0 + x)." To get the induction step, assume as IH: (x + Y)= (Y+ XIWe have: (X+ SY)= s(x + Y) [by (4411 = s(y + x) [by IH] = (sy + x) [by Proposition 31H Proposition 5 (Associative law of addition). PA t(b'x)(Vy)(b'z)((x = (x + y) + z) + (y + z)). Proof: Two applications of (43) give us the basis clause, "((x the induction step, assume as IH: ((x + Y)+ z) = (x + (Y+ z)). We have: ((x + y) + sz) = s((x + y) + z) = s(x + (y + z)) [by (4411 [by IHI + y) + 0) = (x + (y + O))." To get Peano Arithmetic, p. 4 = (x + s(y + z)) [by (4411 + (y + sz)) [by (4411-• = (X Proposition 6. PA t('dx)(o.x) Proof: The base clause, "(0.0) (0.x) = 0," = 0. comes from (Q5). To get the induction, assume as IH: = 0. We have: (0.s~) = ((0.x) + 0) [by (4611 = (0.x) [by (4411 =O [by IH].q Proposition 7. PA t(~x)(b'~)(sx.~) = ((xmy) + y). Proof: We derive the base clause as follows: (sx.0) = o [by (Q5) = (x.0) [by (Q5) again1 = ((x0y) + 0) [by (Q3)I. Assuming, as IH, (sx0y) = ((x0y) + Y), we compute: (sxosy) = ((sxoy) + sx) [by (4611 = (((X~Y) + Y)+ sx) [by IHI =((x.y) + (y + sx)) [by Proposition 51 = ((X~Y) + S(Y+ x)) [by (4411 =((x.y) + s(x + y)) [by Proposition 41 Peano Arithmetic, p. 5 = ( ( x * ~+ ) (x + SY)) = (((xmy) + x) +sy) = ((x*sY)+ SY) [by (4411 [by Proposition 51 [by (4611-rn Proposition 8 (Commutative Law of Multiplication). PA t(x*y) = (y*x). Proof: The base clause, "(x*O) = (O*x)," uses (Q5) and Proposition 6. As inductive hypothesis, assume: (x*Y)= (Y*x). We compute: (X*SY)= ( ( x * ~+) x) [by 4611 = ((y*x) + x) [by IHI = (sy*x) [by Proposition 71. rn Proposition 9 (Distributive law). PA t(b'x)(Vy)(Vz)(x*(y + z)) = ((x*y) + (xmz)). Proof: We prove this equivalent formula: (b'y)(~z)Wx)(x*(y+ 4 ) = ((x*y) + (xaz)), by using this induction axiom: (b'Y)(W [[(Om(Y+ 4 ) = ((O*Y)+(0.~1) A Wx)((x*(Y+ 4 ) = ((x*y) + (x*z)) (sx*(y + z)) = ((sx*y) + (sxez)))l - (b'x)(x*(y + z)) = ((x*y) + (x*z))l To get the base clause, we compute: (O*(y + z)) = 0 = (0 [by Proposition 61 + 0) = ((0.y) + (0.z)) [by (4311 [by Proposition 6 again]. In proving the induction step, we assume the IH: - Peano Arithmetic, p. 6 (x*(Y+ z)) = ((x*Y)+ (xaz)). Now we calculate: [by Proposition 71 (SX*(Y + z)) = ((x*(Y+ z)) + (Y+ z)) = (((x0y) + (xaz)) + (Y+ 4 ) [by IHI = ((x*y) + ((x*z) + (y + z))) [by Proposition 51 = ((x*y) + (((xoz) + y) + z)) [by Proposition 51 = ((x*y) + ((y + (x*z)) + z)) [by Proposition 41 = ((x*y) + (y + ((x*z) + z))) [by Proposition 51 = (((x*y) + y) + ((x*z) + z)) [by Proposition 51 = ((sxoy) + (sx*z)) [by Proposition 71. H Proposition 10 (Associative law of multiplication). PA t(~x)(b'~)(Vz)((x~~)*z) = (x*(y*z)). Proof: The induction axiom we intend to employ is this: (V~)(VY)[[((~*Y = )(x0(y*O)) ~~) A (Vz)(((x0y).z) - (x*(yasz)))l (Vz)((xay)*z)= (xa(y*z))l. We get the base clause thus: ((x*y ) * ~=) o [by (4611 = (x*O) [by (4611 = (x*(yaO)) [by (4611. To get the induction step, we assume this IH: ((xay)*z) = (xa(y*z)). We compute: = (x*(y0z)) - ((xay)*sz) = Peano Arithmetic, p. 7 ((xay)*sz) = (((xay)*z) + ( x * ~ ) ) [by (4611 = ((xa(y*z)) + (x0y)) [by IHI = (x*((yaz) + Y)) [by Proposition 91 = (x*(yosz)) [by (4611-rn We could keep going like this for a very long time. The induction axiom schema we have been using is sometimes called the "weak induction schema," to distinguish it form the following strong induction schema: - - (WX)(WY< X)SY Sx) 0'~)s~)- In applying this schema, we assume as inductive hypothesis that every number less than x has the property represented by Sx, then try to show that x has the property. If we succeed, we conclude that every number has the property. We don't need to assume the instances of the strong induction schema as additional axioms, because we can derive them using the regular induction schema. Specifically, the induction axiom we use is this: [[(VY< 0)SY A (Vx)((Vy < x)Sy - (VY< sx)Sy)l - Wx)(Vy < x)Syl. The inductive hypothesis, "(Vy < O)Sy," is a consequence of (Q9). (Q10) tells us that the induction clause, "(Vx)((Vy < x)Sy - - (Vy < sx)Sy)," is equivalent to this: WX)((VY< X)SY (VY)((Y< x v Y = x) which, in turn is equivalent to this: - ((VY< x)Sy A Sx)), - Sx). (Vx)((Vy < x)Sy which is equivalent to (Vx)((Vy is in the collection. (D2) If is in the collection, so is . That is, f is the intersection of all subcollections of IUI x 1231 that satisfl (Dl) and (D2). f is a function from 1 3 1to 1231. To see this, note, first, that f pairs 0' with one and only one element of 1231: E f by (Dl). If y implies, since f is smallest, that f + oB, f - {) satisfies (Dl) and (D2), which - {) =f and C f. Next, assume that f pairs x with one and only one element y of 1931. Because f satisfies (D2), the pair E g. If sU(a)+ sU(x),will be in g because it's in f. If sU(a)= sU(x),then, because U satisfies (Q2), a = x. Because f pairs x with only one element of 131, b must be equal to y, and so sB(b) + Z; hence, again, is in g. Thus g satisfies (Dl) and (D2). Because f is the smallest class that satisfies (Dl) and (D2), g must be equal to f, which means that isn't ) , with nothing else. in f. Consequently, f pairs sU(x)with s ~ ( ~and f(x 0%y) = f(x) 0% f(y). f(x E' y) = f(x) E~ f(y). x 3 y iff f(x) <' f(y). If o is a variable assignment for U, then, for any formula 4 , o satisfies 4 in U iff foo satisfies 4 in 23. (foo is defined by setting foo(v) equal to f(o(v)).) It follows that the same sentences are true in U and in 23. Peano Arithmetic, p. 12 Let C be the set of elements of IUI that are paired by f with exactly one element of 131.We see that 0%is in C and that, whenever x is in C, sU(x)is in C. Because U satisfies the principle of mathematical induction, C must be equal to IUI, which means that f is a function fiom IUI to 1231. A similar argument, this time using the fact that 23 satisfies the principle of mathematical induction, shows that f is a bijection. To complete the proof that f is an isomorphism, we have to show several things. We have to show that f(oU)= oB;this follows immediately fiom the way f was defined. For each of the function signs of the language, we have to show that f respects the operation of the function sign; for example, we have to show that f(x +' y) = f(x) +' f(y). Finally, we have to show that f preserves the "<" relation, that is, that x 3 y iff f(x) <' f(y). Of these, we'll only write out the proofs for "s" and "+" here. (Dl) tells us that, if E f, Consequently, for x E IUI, since E f, E f, that is, f(s3(x)) = sB(f(x)). To get the clause for "+," pick x E 1 3 '11.Let E = {y E IUI: f(x +' y) = f(x) +' f(y)). We want to show that 0' is in E, and also to show that, if y is in E, so is s'(~). Because U satisfies the principle of mathematical induction, this will suffice to show that every member of IUI is in E. Because U satisfies (Q3), x +' 0' = X. Because 23 satisfies (Q3), f(x) +' 0%= f(x). Consequently, f(x +' 0%)= f(x) = f(x) +%oB= f(x) +B f(oQL), and 0' is in E. Suppose that y is in E. We compute f(x +UsyY))= f(sU(x y)) = sB(f(x y)) = sB(f(x)+%f(y)) [because U satisfies (Q4)] [because f respects "s"] [because y E El Peano Arithmetic, p. 13 23 23 = f(x) + S (f(~)) = f(x) +23f(syY)) [because U satisfies (Q4)] [because f respects "s"]. Therefore, s'(~) is in E.H Now we have a puzzle. Dedekind's theorem tells us that any model of Q that satisfies the principle of mathematical induction is isomorphic to the standard model. In particular, since true arithmetic includes Q and it also includes all the instances of the induction axiom schema, all models of true arithmetic ought to be isomorphic to the standard model. But they aren't. The Compactness Theorem tells us that there are nonstandard models of true arithmetic, that is, models of true arithmetic that aren't isomorphic to the standard model. The solution to this puzzle is to realize that the induction axiom schema doesn't hlly succeed in expressing the content of the principle of mathematical induction. What the induction axiom schema tells us is that the principle of mathematical induction is satisfied by every There's no way the schema could collection that is named by some predicate of the lang~age.~ tell us about collections that aren't named by predicates of the language. The collections that appear in the proof of Dedekind's theorem - the domain of the function f, and so on - aren't named by predicates of the language. 3 To put the matter a little more precisely, let U be a model of the language of arithmetic. Extend the language of arithemetic by adding a new constant to serve as a standard name of each element of the universe of U. If U satisfies all the induction axioms, we are assured that the priniciple of mathematical induction holds for every subcollection of IUI t hat is the extension of some predicate of the extended language. The slogan is that the priniciple holds for collections that are named by some "predicate with parameters" in U. Peano Arithmetic, p. 14 To realize the full strength of the principle of mathematical induction, we have to go beyond the familiar language of arithmetic to the language of second-order arithmetic. In addition to the familiar symbols of the language of arithmetic, this new language includes the second-order variables, "X,," "X,," "X,," "X,," and so on.4The definition of "formula" is changed in two ways: For any term T, KTis an atomic formula. Also, if 4 is a formula, so are (3XJ4 and @ & 'a.) The distinction of "free" and "bound" occurrences of second-order variables, and the distinction between sentences and other formulas, works exactly the way it did for the first-order language. The definition of "model" is unchanged, but there are small changes in the semantics. A variable assignment for a model U assigns an element of IUI to each ordinary variable (or each individual variable, as they're called in this context), and it assigns a subset of IUI to each second-order variable. a satisfies &T of o(X).,, iff the individual T denotes with respect to a is an element An &-variant of a variable assignment a agrees with a except perhaps in what it 4), ,X 3,( assigns to K. a satisfies in U iff some &-variant of a satisfies 4 in U. a satisfies (VX,,,)$ in U iff every &-variant of a satisfies 4 in U. 4 We are allowing second-order variables to take the place of unary predicates. We could also, if we wanted, allow second-order variables that take the place of predicates of more than one arguments. As far as what what we're doing here goes, this wouldn't make any differencebecause we can use the h c t i o n Pair to translate things we want to say about binary relations on the natural numbers into statements about properties of natural numbers. Peano Arithmetic, p. 15 Second-order PA consists of axioms (Q 1) through (Q 1I), together with the following second-order induction axiom: (vxo)((xoo A WY)(XOY - Xosy)) - (~Y)XOY). Thus we can write down second-order PA as a single sentence of the second-order language of arithmetic. Because the second-order variables range over all subcollections of the universe of discourse, not just those subcollections that happen to be named by some formula or other, the second-order induction axiom expresses the full strength of the principle of mathematical induction. Dedekind's theorem amounts to the following: Corollary. Second-order PA is categorical; that is, any two models of the theory are isomorphic.

Peano Arithmetic

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