Notes for Math 601, Fall based on Introduction to Mathematical Logic by Elliott Mendelson Fifth edition, 2010, Chapman & Hall

Notes for Math 601, Fall 2010 based on Introduction to Mathematical Logic by Elliott Mendelson Fifth edition, 2010, Chapman & Hall All first-order languages contain the variables: v 0, v 1, v 2,... the connectives:,, the quantifier:. First-Order Predicate Calculus. A specific first-order language L is completely determined by a nonempty set of relation symbols R,..., a (possibly empty) set of function symbols f,..., a (possibly empty) set of constants a,..., the assignment of a positive integer to each relation and function symbol. Here is a somewhat more formal presentation. A first-order language L consists of several pairwise disjoint sets and operations among them, as follows. There is a nonempty set V, whose elements are called variables, a set C, whose elements are called constants, a nonempty set R, whose elements are called relation symbols, a set F, whose elements are called function symbols. Usually, V = ω, but other choices are studied. C and F may be empty. Every element of F R has a rank, which is a positive integer. (The rank of a constant, if it were defined, would be 0.) Let Sym(L) be the set of symbols of L, namely the set of all constants, relation symbols, and function symbols of L. Note that Sym(L) 1 since L has at least one relation symbol. The set of terms of L is the smallest set that contains the variables and the constants of L (if any), and is closed under the following rule for forming terms: If f is a function symbol of rank n and t 1,..., t n are terms, then the finite sequence f, t 1,..., t n is a term, usually written as ft 1 t n or f(t 1,..., t n ) (or even t 1 ft 2 in case n = 2). Let Tm(L) be the set of terms of L. [This set forms an absolutely free algebra with function symbols as operations.] If R R and R has rank n > 0 and t 1,..., t n are terms, then the finite sequence R, t 1,..., t n is called an atomic formula, and is usually written as Rt 1 t n or R(t 1,..., t n ) (or even t 1 Rt 2 in case n = 2). The set of formulas of L is defined as the smallest set that contains all atomic formulas and is closed under the following rules for forming formulas: If A, B are formulas and v V, then the finite sequences, A, B,, A, and, v, A are formulas, usually written as A B, A, and va (or ( v)a or v A or v(a) or ( v)(a)), respectively. Let Fm(L) be the set of formulas of L. Usually the variables and connectives are regarded as fixed, so that a language is specified by the constants, relation symbols, and function symbols: L = L(C, F, R). 1

2 Some abbreviations: A B := A B A B := (A B) A B := ((A B) (B A)) xa := x A Definition 1 (and theorem!). The function Fv(-) is uniquely determined by the following rules: Fv(x) = {x} for every variable x, Fv(a) = for every constant a of L, Fv(ft 1... t n ) = Fv(t 1 ) Fv(t n ), Fv(Rt 1... t n ) = Fv(t 1 ) Fv(t n ), Fv(A B) = Fv(A) Fv(B), Fv( A) = Fv(A), Fv( xa) = Fv(A) {x}. For every formula A, Fv(A) is the set of free variables of A. Note that, for every term t, Fv(t) is the set of variables that occur in t, and for every atomic formula A, Fv(A) is the set of variables that occur in A. By the last rule, x is not a free variable of xa. Definition 2. A Fm(L) is a sentence if Fv(A) =. Let Sen(L) be the set of sentences of L, that is, Sen(L) = {A : A Fm(L) and Fv(A) = }. Definition 3 (and theorem!). If x is a variable and t is a term, then Sub x t is the map that replaces free occurrences of x with t. It is uniquely determined by the following rules. Sub x t (x) = t Sub x t (y) = y Sub x t (a) = a Sub x t (ft 1... t n ) = f(sub x t (t 1 )... Sub x t (t n )) Sub x t (Rt 1... t n ) = R(Sub x t (t 1 )... Sub x t (t n )) Sub x t (A B) = Sub x t (A) Sub x t (B) Sub x t ( A) = Sub x t (A) Sub x t ( xa) = xa Sub x t ( ya) = ysub x t (A) for every variable y x for every constant a for every variable y x Lemma 4. If x / Fv(t) then Sub x t (t) = t. If x / Fv(A) then Sub x t (A) = A. Definition 5. Let A Fm(L), t Tm(L), and let x be a variable. Then t is free for x in A if A belongs to the smallest set that (1) contains every formula in which x is not free, (2) contains all atomic formulas, (3) is closed under and, (4) is closed under y whenever y / Fv(t). The statement t is free for x in B is short for t is free to be substituted for x in B. Another way to say it is that t is free for x in B if the computation of Sub x t (B) requires the rule Sub x t ( ya) = ysub x t (A) only when y / Fv(t) or x / Fv(A).

3 Theorem 6. Fm(L) = ω + Sym(L). 1. Semantics. An interpretation M of a first-order language L consists of a nonempty set D (called the domain of M), and a mapping of relation symbols to relations on D, function symbols to functions from D to itself, and constants to elements of D, such that each relation symbol R R of rank n is assigned by M to a relation R M D n, each function symbol f F of rank n is assigned by M to a function f M : D n D, each constant a C is assigned by M to an element a M D. Example. Suppose L has one binary relation symbol R, one binary function symbol f, and two constants a and b. Define an interpretation M by setting D = the set of real numbers, R M = { r, s : r, s D and r < s}, f M : D 2 D where f(r, s) = r + s, a M = 0, and b M = 1. We need some special notations for the next definition. For every assignment s : V D, every x V, and every d D, let s(x/d) be the assignment that coincides with s except that s(x/d) maps x to d. Thus { s(y) if x y s(x/d)(y) = d if x = y The satisfaction relation defined next may or may not hold between an interpretation M, a formula A, and an assignment s. It is written symbolically as M = A [s] and is read, s satisfies A in M or A is satisfied by s in M. Definition 7. Let L be a first-order language and let M be an interpretation of L. (1) Define a relation M = - [-] and an evaluation map s M : Tm(L) D according to the following rules. For every x V, every constant a C, every function symbol f F, all terms t 1,..., t n Tm(L), every relation symbol R R, all formulas A, B Fm(L), and all assignments s : V D: s M (x) = s(x), s M (a) = a M, s M (ft 1... t n ) = f M (s M (t 1 ),..., s M (t n )), M = Rt 1... t n iff s M (t 1 ),..., s M (t n ) R M, M = A [s] iff it is not the case that M = A [s], i.e., M = A [s], M = A B [s] iff either M = A [s] or M = B [s], M = xa [s] iff M = A [s(x/d)] for every d D. (2) A formula A is satisfiable iff there are M and s such that M = A [s]. (3) M is a model of A iff M = A iff = M A iff M = A [s] for every s : V D. (4) A theory is any subset of Fm(L). (5) For any theory Γ Fm(L) and any interpretation M, M is a model of Γ iff M = Γ iff M = A for every A Γ. (6) A is logically valid iff = A iff = A iff for every interpretation M, M = A. (7) A theory Γ logically implies A iff Γ = A iff for every interpretation M, if M = Γ, then M = A. Example. Suppose L has R R, f F, a, b C. Define M by setting D = R = the set of real numbers, R M = { p, q : p, q D, p < q}, f M : D 2 D where f(p, q) = p + q, a M = 0, and

4 b M = 1. Suppose s : V D is such that s(x) = 5, s(y) = 3, and s(z) = 4. Determine which of the following statements are true and which are not. M = Rab [s] M = Rba [s] M = Rxy [s] M = Rzfxy [s] M = xrax [s] M = xrxb [s] M = x yrxy [s] Example. Suppose L has a binary relation symbol R and distinct variables x, y, z. Let Γ = { Rxy (Ryz Rxz), Rxx, yrxy }. Let M be any intepretation of L. Prove that if M = Γ, then D is infinite. The next theorem and its corollary justifies the use of the rule MP. Theorem 8 (See (III), p. 53). If M = A and M = A B, then M = B. Corollary 9. If = A and = A B, then = B. The next theorem and its corollary justifies the use of the rule Gen. Theorem 10 (See (VI), p. 53). For every x V, M = A iff M = xa. Corollary 11. For every x V, = A iff = xa. To define tautology as used in the next theorem, let S = {A : A Fm(L) is atomic or A = xb for some B Fm(L) and some variable x}. Then it can be shown that for every function v : S {T, F } there is a unique extension of v to a valuation ˆv : Fm(L) {T, F } such that ˆv(A B) = ˆv(A) ˆv(B) and ˆv( A) = ˆv(A) for all A, B Fm(L). A Fm(L) is a tautology iff ˆv(A) = T for every v : S {T, F }. The next theorem justifies the use of axiom sets (A1), (A2), and (A3). Theorem 12 (See (VII), p. 53). If A is a tautology then = A. Lemma 13. Let M be an interpretation for a language L, with domain D. For every t Tm(L), if s, s : V D agree on Fv(t) (that is, s(x) = s (x) whenever x Fv(t)), then s M (t) = s M (t). Proof. Let P := {t : t Tm(L), and if s, s : V D agree on Fv(t) then s M (t) = s M (t)}. We will show that P contains all variables and constants of L, and is closed under the formation of new terms from terms in P. Consequently, P = Tm(L). Let a be a constant of L (if any). Assume s, s : V D, and s(x) = s (x) whenever x Fv(t). Then s M (a) = a M = s M (a). Thus a P. Consider an arbitrary variable y. Assume s, s : V D and s(x) = s (x) whenever x Fv(y). But Fv(y) = {y}, so this assumption simply says that s(y) = s (y). Then s M (y) = s(y) = s (y) = s M (y). Thus y P. Assume f is a function symbol of L with rank n and t 1,..., t n P. We claim that ft 1... t n P. To prove this, assume s, s : V D and s(x) = s (x) whenever x Fv(ft 1... t n ). Now Fv(ft 1... t n ) = Fv(t 1 ) Fv(t n ), so it follows that s and s agree on Fv(t 1 ), s and s agree

on Fv(t 2 ),..., and s and s agree on Fv(t n ). From these consequences and the assumption that t 1,..., t n P, we may conclude that (1) s M (t 1 ) = s M (t 1 ),..., s M (t n ) = s M (t n ). But then Thus ft 1... t n P. s M (ft 1... t n ) = f M (s M (t 1 ),..., s M (t n )) = f M (s M (t 1 ),..., s M (t n )) by (1) = s M (ft 1... t n ). Theorem 14 (See (VIII), p. 53). If s, s : V D agree on Fv(A), then Proof. Use induction on formulas. Let M = A [s] iff M = A [s ]. P := {A : A Fm(L), if s, s : V D agree on Fv(A), then M = A [s] iff M = A [s ]}. The atomic case. We wish to show Rt 1... t n P. Assume s, s : V D agree on Fv(Rt 1... t n ). Then, for every i {1,..., n}, s, s : V D agree on Fv(t i ), so, by the previous lemma, s M (t i ) = s M (t i ). Then the following statements are equivalent: M = Rt 1... t n [s] s M (t 1 ),..., s M (t n ) R M s M (t 1 ),..., s M (t n ) R M by the previous Lemma M = Rt 1... t n [s ] Assume that A, B P. If s, s : V D agree on Fv( A) = Fv(A), then the following statements are equivalent: M = A [s] M = A [s] M = A [s ] M = A [s ] A P If s, s : V D agree on Fv(A B) = Fv(A) Fv(B), then the following statements are equivalent: M = A B [s] either M = A [s] or M = B [s] either M = A [s ] or M = B [s ] M = A B [s ] since A, B P Assume that s, s : V D agree on Fv( xa). the following statements are equivalent: M = xa [s] M = A [s(x/d)] for every d D M = A [s (x/d)] for every d D M = xa [s ] since A P (see below) Note that, for every d D, s(x/d) and s (x/d) agree on Fv( xa) by hypothesis, but they also agree on {x} since s(x/d)(x) = d = s (x/d)(x), so they agree on Fv( xa) {x}. But Fv(A) Fv( xa) {x}, so s(x/d) and s (x/d) agree on Fv(A). 5

6 Corollary 15. If A is a sentence then, for any interpretation M with domain D and all s, s : V D, M = A [s] iff M = A [s ]. Corollary 16 (See (IX), p. 62). If A is a sentence, then for every interpretation M, either M = A or M = A. For the next theorem, first note that it is not always true that = x(a B) (A xb). For an example, let A = B = Rx, D = {0, 1}, and R M = {1}. Choose s : V D so that s(x) = 1. Then M = x(rx Rx) (Rx xrx)[s] Theorem 17 (See (XI), p. 55). If x / Fv(A), then = x(a B) (A xb). Proof. We wish to show that M = x(a B) (A xb) [s] for every interpretation M and every assignment s : V D, i.e., either M = x(a B) [s] or M = A [s], or else M = xb [s]. Therefore assume (2) (3) M = x(a B) [s], M = A [s]. Let d D. From (2) and the definition of satisfaction we get (4) M = A B [s(x/d)]. From (4) and the definition of satisfaction we conclude that (5) either M = A [s(x/d)] or else M = B [s(x/d)]. By hypothesis, x / Fv(A), so s and s(x/d) agree on Fv(A). It follows by Theorem 14 and (3) that (6) M = A [s(x/d)] By (5) and (6), we get M = B [s(x/d)]. We proved this for every d D, so it follows that M = xb [s], again by the definition of satisfaction. Theorems (X) and (XI) show that all instances of (A4) and (A5) are logically valid. (A4) xa A(x/t) where t is free for x in A. (A5) x(a B) (A xb), where x is not free in A. Without the restrictions, some instances of (A4) and (A5) would not be logically valid. Unrestricted (A4) instance x yrxy yryy fails when D = {0, 1} and R M = { 0, 1, 1, 0 }. Unrestricted (A5) instance x(rx Rx) (Rx xrx) fails when D = {0, 1} and R M = {0}. Lemma 18 (See Lemma 1, p. 55). s M (Sub x t (t )) = s(x/s M (t)) M (t ). Proof. Use induction on terms. First consider the case when t is a variable, say t = y. Then, by the relevant definitions, { s M (Sub x s M (y) = s(y) if y x t (y)) = s M (t) if y = x and s(x/s M (t)) M (y) = In case t is a constant, say t = a, we have { s(x/s M (t))(y) = s(y) s(x/s M (t))(x) = s M (t) if y x if y = x s M (Sub x t (a)) = s M (a) = a M = s(x/s M (t)) M (a).

7 Assume the lemma holds for terms t 1,..., t n and f is a function symbol of rank n. Then s M (Sub x t (ft 1... t n )) = s M (fsub x t (t 1 )... Sub x t (t n )) = f M (s M (Sub x t (t 1 )),..., s M (Sub x t (t n ))) = f M (s(x/s M (t)) M (t 1 ),..., s(x/s M (t)) M (t n )) = s(x/s M (t)) M (ft 1... t n ). Lemma 19 (See Lemma 2, p. 55). Let M be an interpretation for L with domain D, t Tm(L), C Fm(L), and x V. Suppose t is free for x in C. Then for all s : V D, M = C [s(x/s M (t))] iff M = Sub x t (C) [s]. Proof. The proof is by induction. Let P := {A : A Fm(L) and for all s : V D, M = A [s(x/s M (t))] iff M = Sub x t (A) [s]}. We will show that P has the closure properties listed in the definition of is free to be substituted for, which implies that A P, as desired. Claim. If x / Fv(A), then A P. Proof. Let s : V D. Then the following statements are equivalent: M = A [s(x/s M (t))] M = A [s] M = Sub x t (A) [s] see note (a) below see note (b) below Note (a): s(x/s M (t)) and s agree everywhere except possibly at x, but x is not free in A by assumption, so s and s(x/s M (t)) agree on the free variables of A. Apply Theorem 14. Note (b): By Lemma 4, Sub x t (A) = A. Claim. Every atomic formula is in P. Proof. Consider an atomic formula Rt 1... t n where R is a relation symbol of rank n and t 1,..., t n Tm(L). Let s : V D. Then the following statements are equivalent: M = Rt 1... t n [s(x/s M (t))] s(x/s M (t)) M (t 1 ),..., s(x/s M (t)) M (t n ) R M s M (Sub x t (t 1 )),..., s M (Sub x t (t n )) R M by Lemma 18 M = RSub x t (t 1 )... Sub x t (t n ) [s] M = Sub x t (Rt 1... t n ) [s] Claim. If A, B P then A B P. Proof. Let s : V D. Then, assuming A, B P, we have M = A [s(x/s M (t))] iff M = Sub x t (A) [s]. and M = B [s(x/s M (t))] iff M = Sub x t (B) [s].

8 Then the following statements are equivalent: Thus A B P. Claim. If A P then A P. M = Sub x t (A B) [s], M = Sub x t (A) [s] or M = Sub x t (B) [s], M = A [s(x/s M (t))] or M = B [s(x/s M (t))], M = A B [s(x/s M (t))]. Proof. Let s : V D. Then, assuming A P, we have M = A [s(x/s M (t))] iff M = Sub x t (A) [s], so the following statements are equivalent: Thus A P. M = Sub x t ( A) [s], M = Sub x t (A) [s], M = A [s(x/s M (t))], M = A [s(x/s M (t))]. Claim. If A P and y / Fv(t), then ya P. Proof. If y = x then x / Fv( ya), so ya P by the first Claim. Therefore, assume y x. We wish to show that ya P. Let s : V D. Then the following statements are equivalent: M = ya [s(x/s M (t))] M = A [s(x/s M (t))(y/d)] for all d D M = A [s(y/d)(x/s M (t))] for all d D M = A [s(y/d)(x/s(y/d) M (t))] for all d D M = Sub x t (A) [s(y/d)] for all d D M = y Sub x t (A) [s] M = Sub x t ( ya) [s] see (c) below see (d) below see (e) below (c) Note that s(x/s M (t))(y/d) = s(y/d)(x/s M (t)) since x y. (d) Note that, since y / Fv(t), we have s M (t) = s(y/d) M (t) by Lemma 13. (e) We can apply the assumption that A P, with s(y/d) in place of s. Theorem 20 (See (X), p. 55). If t is free for x in A, then = xa Sub x t (A). Proof. Let M be an interpretation for L with domain D, and let s : V D. We wish to show that M = xa Sub x t (A) [s]. For this purpose, assume It follows from this that M = xa [s]. M = A [s(x/s M (t))]. We may apply Lemma 19 since t is free for x in A by assumption, so M = Sub x t (A) [s], as desired.

9 The hypothesis that t is free for x in A is necessary, since there are formulas of the form xa Sub x t (A) that are not logically valid. For an example, let L be a language with exactly one binary relation symbol R. Let D = {0, 1} and R M = { 0, 1, 1, 0 } (so R is the diversity relation on the two-element set D). Then x yrxy Sub x y( yrxy) = x yrxy ysub x y(rxy) = x yrxy yrsub x y(x)sub x y(y) = x yrxy yryy. But x yrxy yryy is not logically valid, since for every s : V D we have M = x yrxy [s] ( for everything there is something different ), but M = yryy [s] ( there is something different from itself ). Definition 21 (Axioms for First-Order Logic). (A1) := { A (B A) : A, B Fm(L) }, (A2) := { (A (B C)) ((A B) (A C)) : A, B, C Fm(L) }, (A3) := { ( B A) (( B A) B) : A, B Fm(L) }, (A4) := { xa Sub x t (A) : t is free for x in A }, (A5) := { x(a B) (A xb) : x / Fv(A) }, Ax(L) := (A1) (A2) (A3) (A4) (A5). Definition 22. Let T Fm(L) and A Fm(L). A proof of A from T is a finite sequence of formulas B 1,..., B n such that B n = A and, for every i {1,..., n}, one of the following holds: Hyp B i T, Ax B i Ax(L), MP there are j, k < i such that B k = B j B i, Gen there is some j < i and some variable x such that B i = xb j. We say that T proves A in L (in symbols, T L A) iff there is a proof of A from T. Furthermore, L A iff L A. Use the notation T A instead of T L A if L is clear from context. Theorem 23 (Properties of Provability). (1) T A iff T A for some finite T T. (2) If T T and T A, then T A. (3) If T B for every B T, and T A, then T A. Theorem 24 (Soundness Theorem). If T A then T = A. Proof. Suppose B 1,..., B n is a proof of A from T. We show by induction that for every i {1,..., n}, T = B i. Assume T = B 1,..., T = B i 1. We wish to show that T = B i. If B i T then T = B i by the definition of =. If B i Ax(L) then = B i by Theorem 12, Theorem 20, and Theorem 17. If B i is obtained by MP from earlier formulas, i.e., there are j, k < i such that B k = B j B i, then T = B j and T = B k by assumption, so T = B i by Theorem 8. If B i is obtained by Gen from an earlier formula, i.e., there is some j < i and some variable x such that B i = xb j, then T = B j by assumption, so T = B i by Theorem 10.

10 It is not necessarily the case that if A B then A B. For example, we have Rx xrx, since Rx, xrx is a proof of xrx from {Rx}. However, it is not the case that Rx xrx. To see this, we need only show that = Rx xrx. The desired conclusion then follows by the Soundness Theorem. Suppose D = {0, 1}, R M = {0}, and s(x) = 0. Then = Rx [s] and = xrx [s], so = Rx xrx [s]. There is a version of the deduction theorem for first-order logic. To prove it we start with a definition and a lemma. Definition 25. Suppose T Fm(L), A Fm(L), and B 1,..., B n is a proof of B n from T {A} and A / T. For each i {1,..., n}, we say B i is independent of A in B 1,..., B n if B i Ax(L) T or B i is a consequence by MP or Gen of some earlier formulas that are independent of A. Lemma 26. Suppose B 1,..., B n is a proof of B n from T {A} and A / T. {1,..., n}, if B i is independent of A in B 1,..., B n then T B i. Proof. The proof is by induction on i {1,..., n}. For each i Let i = 1. Assume B 1 is independent of A in B 1,..., B n. Then B 1 Ax(L) T, since B 1 cannot be a consequence of earlier formulas. But then T B 1. Let 1 < i n and assume the result holds for all B 1,..., B i 1. Assume B i is independent of A. If B i Ax(L) T, then T B i. If there are k, j < i such that B k = B j B i, then, by assumption, we have T B j and T B k, so T B i by MP. If there is some variable x and some j < i such that B i = xb j, then, by assumption, we have T B j, so T B i by Gen. Theorem 27 (Deduction Theorem, I). Assume B 1,..., B n is a proof of B n from T {A} and A / T. Assume, for every i {1,..., n}, that if B i can only be obtained by Gen from previous formulas that are not independent of A, then every such application of Gen involves a variable x that is not free in A. Then T A B i for every i {1,..., n}. Proof. The proof is by induction on i. Assume T A B 1,..., T A B i 1. 1. If B i is independent of A, then, by Lemma 26, T B i, but B i (A B i ), so T A B i by MP. and but 2. If B i = A, then T A B i since A A. 3. Suppose there are j, k < i such that B k = B j B i. By the inductive hypothesis, we have by (A2), so T A B i by MP twice. T A B j T A B k (A B k ) ((A B j ) (A B i )) 4. Assume B i is not independent of A (hence B i / Ax(L) T ), B i A, and B i does not follow from any previous formulas by MP. Consequently, B i can only be obtained by Gen from previous formulas that are not independent of A (since otherwise B i would be independent of A). By hypothesis, the application of Gen involves a variable x that is not free in A. Thus there is some j < i such that B i = xb j and x / Fv(A). By the inductive hypothesis, hence, T A B j T x(a B j )

11 by Gen, but by (A5), since x / Fv(A), so by MP. T x(a B j ) (A xb j ) T A B i Corollary 28 (Deduction Theorem, II). (1) If there is a proof of B from T {A} in which every application of Gen involves a variable not free in A, then T A B. (2) If A is a sentence, T is a theory, and B is any formula, then T, A B iff T A B. Definition 29. T is consistent if there is no formula A Fm(L) such that T A and T A. T is inconsistent if there is some formula A Fm(L) such that T A and T A. Note that consistency is preserved going down, while inconsistency is preserved going up, in the following sense. Assume T T. If T is consistent, so is T. If T is inconsistent, so is T. Theorem 30. (1) T is consistent iff there is some A Fm(L) such that T A. (2) T is inconsistent iff T A for every A Fm(L). Proof. The two parts are equivalent. We prove only part (2). First, if T proves every formula, it surely proves some formula and its negation and hence is inconsistent. If T is inconsistent, then by definition there is some A Fm(L) such that T A and T A. Let B Fm(L). We have A ( A B) since every tautology is provable, so A, A B. Therefore T B by Theorem 23(3). Theorem 31. is consistent. Proof. Let M be the interpretation determined as follows. Let D = {0}, a M = 0 for every constant a of L, R M = D (rank of R) for every relation symbol R of L, and f M (0,..., 0) = 0 for every function symbol f of L. Suppose s(x) = 0 for every x V. If were not consistent, then A and A for some formula A. By the Soundness Theorem, = A and = A, so M = A [s] and M = A [s], a contradiction. Lemma 32. Assume T is consistent and A Sen(L). Then (1) either T {A} is consistent, or else T { A} is consistent, (2) T {A} is inconsistent iff T A, (3) T { A} is inconsistent iff T A. Proof. (1): Assume T {A} and T { A} are both inconsistent. Let B Fm(L). Then T, A B and T, A B. The Deduction Theorem may be applied since A is a sentence, so T A B and T A B. We have (A B) (( A B) B) since every tautology is provable, so T B. But B was arbitrary, so T is inconsistent, contrary to assumption. (2): If T A, then T {A} is inconsistent because T, A A and T, A A. Conversely, if T {A} is inconsistent, then T, A A. But A is a sentence, so T A A by the Deduction Theorem 28(ii). However, (A A) A since every tautology is provable, hence T A. Part (3) follows from part (2) and the logical equivalence of A with A. Definition 33. A theory T Fm(L) is complete if, for every sentence A Sen(L), either T A or T A.

12 Let M be any interpretation of L. Let T h(m) = {A : A Sen(L) and M = A}. T h(m) is called the theory of M. Then T h(m) is complete and consistent. In fact, T h(m) is maximal consistent, in the sense that if T h(m) T Fm(L) and T is consistent, then T h(m) = T. Theorem 34 (Lindenbaum s Lemma). Every consistent theory is contained in a complete consistent theory. Proof. Let T Fm(L) be a consistent theory. Let α = Sen(L) and let A κ : κ < α be an enumeration of the sentences of L. (The assumption α = Sen(L) means that there exists such a one-to-one correspondence between the sentences and the cardinal α.) Set T 0 = T, { T κ {A κ } T κ+1 = T κ if T κ {A κ } is consistent if T κ {A κ } is inconsistent, for every ordinal κ < α, T λ = κ<λ T κ for every limit ordinal λ α. Claim 1. T κ is consistent for every κ α. Proof. The proof is by (possibly transfinite) induction on κ. T 0 is consistent by hypothesis. Assume, as inductive hypothesis, that T κ is consistent. If T κ+1 = T κ {A κ } then, by the definition of T κ+1, T κ {A κ } is consistent, so T κ+1 is consistent. On the other hand, if T κ+1 = T κ then, by the definition of T κ+1, T κ {A κ } is inconsistent, but T κ+1 itself is consistent because it coincides with T κ (which is consistent by assumption). Either way, T κ+1 is consistent. Let λ α be a limit ordinal, and assume T µ is consistent for every µ < λ. If T λ is inconsistent, then there are formulas B 1,..., B n T λ and A Fm(L) such that B 1,..., B n A and B 1,..., B n A. For each i = 1,..., n, B i T λ = κ<λ T κ, so there is some κ i < λ such that B i T κi. Let µ = max(κ 1,..., κ n ). Then µ < λ and B 1,..., B n T µ, since T κi T µ for i = 1,..., n. Hence T µ is inconsistent, a contradiction. By Claim 1, T α is consistent. Claim 2. T α is complete. Proof. Let A be any sentence. We will show that either A T α or A T α (hence either T α A or T α A). Since A κ : κ < α be an enumeration of all the sentences of L, there must be some κ < α such that A κ = A and some λ < α such that A λ = A. If T κ {A κ } is consistent, then the first case in the definition of T κ+1 applies, and we conclude that A T α since A = A κ T κ {A κ } = T κ+1 T α. Therefore we assume T κ {A κ } is inconsistent. Since T κ T α and A = A κ, it follows that T α {A} is also inconsistent. Similarly, if T λ {A λ } is consistent, then A T α since A = A λ T λ {A λ } = T λ+1 T α, so we also assume that T α { A} is inconsistent. At this point we have assumed that T α {A} and T α { A} are both inconsistent. But this case cannot arise, because by Lemma 32, T α is therefore inconsistent, contrary to Claim 1. Therefore either

13 There is a variation on the proof. Alter the definition of T α to T 0 = T, { T κ {A κ } T κ+1 = T κ { A κ } if T κ {A κ } is consistent if T κ {A κ } is inconsistent, T λ = κ<λ T κ for every limit ordinal λ α. for every ordinal κ < α, Then the completeness of T α is easy, while Lemma 32 is used in the inductive proof that T α is consistent. In the proof above, consistency is easy, while Lemma 32 is used instead for completeness. Definition 35. Let C be a set of constants of a language L, and let T Fm(L) be any theory. We say C is a set of witnesses for T if for every formula A with one free variable x, there is some constant c C such that T Sub x c (A) xa. Lemma 36 (Witness Model Lemma). Assume T Fm(L), T is consistent, T is complete, C is a set of constants of L, C = Fm(L), and C is a set of witnesses for T. Define an interpretation M as follows. Let the domain of M be D = {t : t Tm(L) and Fv(t) = }. For every constant a, every function symbol f, and every relation symbol R of L, let (7) (8) (9) a M = a, f M (t 1,..., t n ) = ft 1... t n if t 1,..., t n D and n is the rank of f, R M = { t 1,..., t n : t 1,..., t n D and n = rank of R and T Rt 1... t n }. Then D = Fm(L) and M = T. Proof. First we need Claim 1. For every s : V D and every t D, s M (t) = t. Proof. Let s : V D. We prove s M (t) = t by induction on the complexity of terms. If t is a constant, then s M (t) = t M = t by (7). Suppose t = ft 1... t n and s M (t 1 ) = t 1,..., s M (t n ) = t n. Then Claim 2. Let A Sen(L). Then s M (ft 1... t n ) = f M (s M (t 1 ),..., s M (t n )) = f M (t 1,..., t n ) = ft 1... t n by (8) (10) for every s : V D, M = A [s] iff T A Proof. We prove (10) by induction on the number of connectives and quantifiers in A. Suppose A is atomic. Then there are t 1,..., t n D and some relation symbol R with rank n such that A = Rt 1... t n. Then, for every s : V D, the following statements are equivalent: M = Rt 1... t n [s] s M (t 1 ),..., s M (t n ) R M definition of = t 1,..., t n R M Claim 1 T Rt 1... t n (9)

14 Assume (10) holds for A Sen(L). We prove it for A Sen(L). Let s : V D. Then 1. M = A [s] Hyp. 2. M = A [s] definition of = 3. T A (10) holds for A 4. T A T is complete 1. T A Hyp. 2. T A T is consistent 3. M = A [s] (10) holds for A 4. M = A [s] definition of = Assume (10) holds for A Sen(L) and B Sen(L). We prove it for A B. 1. M = A B [s] Hyp. 2. M = A [s] and M = B [s] definition of = 3. T A and T B (10) holds for A and B 4. T A and T B T is complete 5. T (A B) A ( B (A B)), MP twice 6. T A B T is consistent 1. T A B Hyp. 2. T (A B) T is complete 3. T A and T B (A B) A, and (A B) B 4. T A and T B T is consistent 5. M = A [s] and M = B [s] (10) holds for A and B 6. M = A B [s] definition of = Suppose xa Sen(L), and assume (10) holds for Sub x t (A) whenever t D. (This is the inductive assumption notice that Sub x t (A) has fewer occurrences of than xa.) We prove (10) for xa. Since Fv( xa) =, we have Fv(A) {x}. This yields two cases. Case 1. Suppose Fv(A) =. Then Sub x t (A) = A whenever t D, so (10) holds for A. Let s : V D. Then 1. M = xa [s] Hyp. 2. M = A [s(x/t)] for all t D definition of = 3. M = A [s] with t = s(x) 4. T A (10) holds for A 5. T xa by Gen

15 1. T xa Hyp. 2. T A xa A by (A4), 1, MP 3. M = A [s] (10) holds for A 4. M = A [s(x/t)] for all t D s, s(x/t) agree on Fv(A) =, Theorem 14 5. M = xa [s] definition of = Case 2. Suppose Fv(A) = {x}. Since C is a set of witnesses for T, there is some c C such that (11) T Sub x c (A) xa. Assume s : V D and M = xa [s]. Note that c is free for x in A since c contains no free variables, so xa Sub x c (A) is an instance of axiom scheme (A4). Hence M = xa Sub x c (A) [s], so M = Sub x c (A) [s]. By the inductive hypothesis, this yields so, by (11) and MP, For the converse, assume Then, for some t D, T Sub x c (A) T xa. M = xa [s]. M = A [s(x/t)], By Claim 1, t = s M (t), so M = A [s(x/s M (t))], By Lemma 19, M = Sub x t (A) [s]. By the inductive hypothesis, T Sub x t (A). Since T is complete and Sub x t (A) Sen(L), T Sub x t (A). However, t has no free variables, and hence is free for x in A. So, by (A4), but every tautology is provable, so xa Sub x t (A), ( xa Sub x t (A)) ( Sub x t (A) xa). By MP, Sub x t (A) xa, hence T xa. Finally, since T is consistent and xa Sen(L), Thus (10) holds for xa. T xa.

16 Lemma 37 (Witness Consistency Lemma). Suppose T is a consistent theory of L, c is a constant of L, A Fm(L), {x} = Fv(A), and c does not occur in A or in any formula in T. Then T {Sub x c (A) xa} is consistent. Proof. Assume T {Sub x c (A) xa} is inconsistent. Since Sub x c (A) xa is a sentence, it follows by Lemma 32 that T (Sub x c (A) xa). But every tautology is provable, so hence, by MP, (Sub x c (A) xa) Sub x c (A), (Sub x c (A) xa) xa, (12) T Sub x c (A) and (13) T xa. By (12), there is a proof B 1,..., B n of Sub x c (A) from T. This proof has only finitely many formulas in it, so there are most finitely many variables occurring free or bound in any formula of the proof. Let y be a variable that does not occur in any formula of the proof. This is possible because there are infinitely many variables, but only finitely many of them can occur in the proof. Replace every occurrence of c with y in the proof B 1,..., B n, resulting in the sequence Sub c y (B 1 ),..., Sub c y(b n ). Claim. Sub c y(b 1 ),..., Sub c y(b n ) is a proof from T. Proof. Let 1 i n. We must check that if B i is an axiom, then so is Sub c y(b i ), that if B i is in T, then so is Sub c y(b i ), and if B i follows by MP or Gen, then so does Sub c y(b i ). If B i is an instance of (A1), (A2), or (A3), then so is Sub c y(b i ). For instance, if B i is an instance of (A1), say C (D C), then Sub c y(b i ) is Sub c y(c) (Sub c y(d) Sub c y(c)), which is also an instance of (A1). Suppose B i is an instance of (A4), say zc Sub z t (C), where t is a term that is free for z in C. Then Sub c y(b i ) = zsub c y(c) Sub c y(sub z t (C)) = zsub c y(c) Sub z Sub c y(t) Subc y(c). This last formula will be an instance of (A4) if Sub c y(t) is free for z in Sub c y(c). By the selection of y we know that y does not occur in B i, so y z. Therefore, no new free occurrences of z are introduced in Sub c y(c). Thus Sub z Sub c y(t) will put Subc y(t) into the same locations in Sub c y(c) that Sub z t puts t in C. Note that Sub c y(t) contains y free, so we need to know that no free occurrence of z is in the scope of a quantified y. But this is true because y does not occur in B i. Thus Sub c y(b i ) is an instance of (A4). Suppose B i is an instance of (A5), say z(c D) (C zd), where z / Fv(C). Then Sub c y(b i ) is z(sub c y(c) Sub c y(d)) (Sub c y(c) zsub c y(d)). To see this is an instance of (A5), we need to know that z / Fv(Sub c y(c)). But Fv(Sub c y(c)) Fv(C) {y}, and z y since y does not occur in B i. So far we have shown that if B i is an axiom, so is Sub c y(b i ). If B i T, then Sub c y(b i ) = B i since c does not occur in any formula in T, so Sub c y(b i ) T.

If B i is obtained by MP, say j, k < i and B k = (B j B i ), then Sub c y(b k ) = (Sub c y(b j ) Sub c y(b i )), so Sub c y(b i ) is also obtained by MP. If B i is obtained by Gen, say j < i and B i = zb j, then Sub c y(b i ) = zsub c y(b j ), so Sub c y(b i ) is also obtained by Gen. Thus Sub c y(b 1 ),..., Sub c y(b n ) is a proof from T. Now c does not occur in A by hypothesis, so all the occurrences of c in Sub x c (A) are at places where x occurs free in A. Hence Sub c y(b n ) = Sub c y(sub x c (A)) = Sub x y(a). Therefore By Gen, T Sub x y(a). T ysub x y(a). Now y does not occur in A, so all the occurrences of y in Sub x y(a) are at places where x occurs free in A. Hence Sub y x(sub x y(a)) = A. Therefore, by (A4), Then, by MP, whence, by Gen, contradicting (13). T ysub x y(a) A. T A, T xa, Lemma 38 (Witness Extension Lemma 2.15). For every consistent theory T Fm(L) there is a set C, a language L +, and a theory T + Fm(L + ) such that (14) (15) (16) (17) (18) (19) (20) T T +, T + is consistent in L +, C is a set of witnesses for T +, C = Fm(L) = Fm(L + ), L and L + have the same function symbols and the same relation symbols, every constant of L + is either a constant of L or belongs to C, no constant in C is a constant of L. Proof. Choose a set C that is disjoint from the variables, connectives, quantifier, symbols, terms, and formulas of L, such that C = Fm(L). The elements of C are called new constants. Let L + be the language whose relation and function symbols are the same as those of L (with the same ranks they have in L), and whose constants are all the constants of L, together with all the elements of C. Hence Sym(L + ) = Sym(L) C, so Fm(L + ) = ω + Sym(L + ) = ω + Sym(L) + C = Fm(L) + Fm(L) = Fm(L). L + contains many more axioms than L, namely, all instances of (A1) (A5) that contain new constants. With respect to this larger language with more axioms, T is still consistent any proof of a contradictory formula using formulas from T and axioms from L + can be converted to a proof in L by replacing the finitely many new constants that appear in the proof with variables that do not appear in the proof. (See the proof of the Witness Consistency Lemma for details.) 17

18 Thus (17) (20) hold. Let α = Fm(L) = Fm(L + ). Note that α is also the number of formulas of L + that have exactly one free variable. Let A κ : κ < α be a well-ordering of all the formulas of L + that have exactly one free variable. For every κ < α, let C κ = {c : c C and, for some λ κ, c occurs in A λ }. There are at most finitely many constants from C occurring in each A λ, so C κ < ω whenever κ < ω, and C κ = κ whenever ω κ < α (although no such case arises when α = ω). Since C = α, it follows that C C κ = α whenever κ < α. Choose a well-ordering of C. For each κ < α, let c κ be the least element of C (C κ {c λ : λ < κ}). (Note that C (C κ {c λ : λ < κ}) = α whenever κ < α.) It follows that if κ < α, then (21) for every λ κ, c κ does not occur in A λ, and c κ c λ if λ < κ. For every κ α let and let Clearly (14) and (16) hold. { } T κ = T Sub x c λ (A λ ) xa λ : λ < κ, {x} = Fv(A λ ), T + = T α. Claim. T κ is consistent in L + for every κ α. Proof. Use induction on κ α. First we prove the Claim for κ = 0. Note that T 0 = T. We know T is consistent in L, but we must show T is also consistent in L +. To do this, note that any proof in L + of a formula in Fm(L) from T can be converted to a proof in L by replacing the finitely many constants in C that occur in the proof by variables that do not occur anywhere in the proof (as shown in the proof of the Witness Consistency Lemma). Suppose T is inconsistent in L +. Then T proves all formulas, so choose a contradictory formula B Fm(L). Then T L + B. By replacing constants in C by new variables in a proof of B from T in L +, we get a proof of B from T in L, hence T L B, contradicting the consistency of T in L. It follows that T is consistent in L +. Assuming T κ is consistent in L +, the consistency of T κ+1 in L + follows by the Witness Consistency Lemma. To see this, note that T κ+1 = T κ {Sub x c κ (A κ ) xa κ } where {x} = Fv(A κ ), and, by (21), c κ does not occur in any formula in T κ, and does not occur in A κ. The consistency of T λ for limit ordinals λ α follows immediately from the consistency of T κ for all κ < λ (as in the proof of Claim 1 in the proof of Lindenbaum s Lemma). By the Claim, (15) holds. Theorem 39 (Prop. 2.17, Completeness Theorem, I). has model of cardinality Fm(L). (2) A theory is consistent iff it has a model. (1) Every consistent theory T Fm(L) Proof. Part (1): By the the Witness Extension Lemma, there are L +, T +, and C such that (14) (20) hold. By Lindenbaum s Lemma, there is a complete consistent T T +. Note that C is a set of witnesses for T as well as T +. By the Witness Model Lemma, T has a model M of cardinality Fm(L + ). By (4) of the Witness Extension Lemma, M has cardinality Fm(L), and M = T since T T. Part (2) follows from part (1) and the observation that any theory with a model must be consistent.

Theorem 40 (Cor. 2.19, Gödel s Completeness Theorem). For every A Fm(L), T = A iff T A. Proof. If T A, then T = A by the Soundness Theorem. For the converse, assume T A. Let B be a closure of A, that is, the result of adding universal quantifiers for all the variables that occur free in A. More specifically, if Fv(A) = {x 1,..., x n }, then a closure of A is x 1 x n A. Notice that B is a sentence and T B, since otherwise we would get T A because B A. To see this, make repeated use of (A4) and MP: B {}}{ x 1 x 2 x 3 x n A x 2 x 3 x n A x 2 x 3 x n A x 3 x n A B x 3 x n A (A4) (A4). etc. B A (B C) ((C D) (B D)), MP twice, Since B Sen(L), it follows that T { B} is consistent by Lemma 32. By the first Completeness Theorem, there is a model M of T { B}. Thus M = T and M = B. Since B is a sentence, it follows that M = B by Corollary 16. It follows, by repeated use of Theorem 10, that M = A. Thus M is a model of T that is not a model of A, so T = A. Theorem 41 (See Problem 2.54, Compactness Theorem). If every finite subset of T Fm(L) has a model, then T has a model. Proof. Assume every finite subset of T Fm(L) has a model. Suppose T has no model. Then, by the Completeness Theorem, T is inconsistent. Hence there is a proof of (A A) from T, for any A. Let T be the finite set of formulas of T that appear in this proof. Then T (A A), so T is inconsistent, and therefore has no model, contrary to hypothesis. Here is the proof again every finite subset of T Fm(L) has a model, so every finite subset of T Fm(L) is consistent (by the simple fact that having a model implies consistency), so T is consistent (for if T were inconsistent then there would be a proof of a contradiction from T, but proofs are finite, so there would be a finite inconsistent subset of T, contradiction), so T has a model (by the Completeness Theorem, I). Theorem 42 (Cor. 2.22). Suppose α Fm(L) and T Fm(L) is consistent. Then T has a model of cardinality α. Proof. Let C be a set of constants with cardinality α, and arrange for C to be disjoint from all the symbols, terms, and formulas of L. Let L + be the language obtained from L by adding C to the set of constants. Note that Fm(L + ) = α. By hypothesis, T is consistent, so, by the Completeness Theorem, there is an interpretation M of L with domain D such that D = Fm(L) and M = T. Extend M to an interpretation M + of L +, with the same domain D as M, as follows. Choose d D, and set c M + = d for every c C. No constant of C appears in any formula in T, so M + = T. Thus T is consistent (in L + ). By the Completeness Theorem (for L + ), T has a model of cardinality Fm(L + ) = α. Theorem 43 (Cor. 2.21, Löwenheim-Skolem Theorem). If a theory in L has a model then it has a model of cardinality no larger than the cardinality of L. Skolem s Paradox. Let L be a language that has a single binary relation symbol E. Note that L is countable (has only countably many formulas). Consider a theory of sets in L. The axioms are 19

20 called Extensionality (sets having the same members must be members of the same sets), Union (two sets have a union, any set of sets has a union), Pairing ({x, y} exists for any two sets x and y), Infinity (there is an infinite set), Powerset (every set x has a power set, the set of subsets of x), Comprehension (an axiom scheme: for every first-order formula A and every set x there is a set of elements of x that have satisfy A). This theory could also include the Replacement Scheme, the Axiom of Choice, and the Generalized Continuum Hypothesis. Skolem s Paradox is that if this theory is consistent then it has a countable model.