1. Propositional Calculus

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1 1. Propositional Calculus Some notes for Math 601, Fall 2010 based on Elliott Mendelson, Introduction to Mathematical Logic, Fifth edition, 2010, Chapman & Hall. 2. Syntax ( grammar ). 1.1, p. 1. Given: a set S = {S 1, S 2, } of statement letters or propositional variables. In the book, it is assumed that S is a countable infinite set (i.e., there are as many elements of S as natural numbers). Everything that follows holds even if S is finite or uncountable. Assume that {,,,, } is a 5-element set disjoint from S: S {,,,, } =. The elements of {,,,, } are called propositional connectives. They will also be used as symbols denoting particular operations on sets. Ordered pairs x, y and ordered triples (see p. xix) are defined thus (pp. 231): For any sets x and y let x, y = {{x}, {x, y}}, x, y, z = x, y, z = {{{{x}, {x, y}}}, {{{x}, {x, y}}, z}}. x y =, x, y, x y =, x, y, x y =, x, y, x y =, x, y, x =, x. Assume S contains no ordered pairs. Let Sent be the closure of S under the operations,,,, and. Then Sent,,,,, is an algebra called the sentence algebra, Sent itself is the called the set of sentences or formulas (in the book, statement forms ) built from S, and Sent is the language of propositional calculus. The essential properties of the sentence algebra are as follows: Theorem 1. For every set S that contains no ordered pairs there exists a set Sent with the following properties: (1) S Sent. (2) Sent is closed under the operations,,,, and defined above. (3) Let A, B, C, D, E, F, G, H, J Sent. Then (a) A B, C D, E F, G H, and J are not in S. (b) A B, C D, E F, G H, and J are all different from each other. (c) If A B = C D then A = C and B = D. (d) If A B = C D then A = C and B = D (e) If A B = C D then A = C and B = D. (f) If A B = C D then A = C and B = D. (g) If A = B then A = B. (4) Sent is the intersection of all sets that contain S and are closed under,,,, and : Sent = {X : S X and if A, B X, then A B, A B, A B, A B, A X}. 1

2 2 Theorem 2 (principle of induction for Sent). If S X and X is closed under,,,, and, then Sent X. Theorem 3 ( unique readability for Sent). If A Sent, then exactly one of the following statements holds: (1) A S, (2) A = B C for some B, C Sent, (3) A = B C for some B, C Sent, (4) A = B C for some B, C Sent, (5) A = B C for some B, C Sent, (6) A = B for some B Sent. The book constructs statements by a method that involves some notational difficulties: see footnotes, p. 1, and read Parentheses in 1.2., pp Semantics ( meanings ). 1.1, pp Let {T, F} be a set of two truth values, where T F. Define five operations on {T, F} as follows. T F T F + T F T F T T T F T T F T T T T T F T F F T T F F F F T F F F T F T Then {T, F},,, +,, is an algebra called the truth-value algebra. Definition 4. A valuation is a homomorphism from the sentence algebra Sent,,,,, into the truth-value algebra. Theorem 5. For every function f : S {T, F} there is a unique valuation v extending f. This means that for all A, B Sent, (1) v(a) = f(a) if A S, i.e., v extends f, (2) v(a B) = v(a) v(b), (3) v(a B) = v(a) v(b), (4) v(a B) = v(a) + v(b), (5) v(a B) = v(a) v(b), (6) v( A) = v(a). Proof. To show that there can be at most one such extension, suppose that v and v are extensions of f satisfying (1) (6). Let X = {A : A Sent and v (A) = v (A)}. We will show that S X and X is closed under,,,, and. Since X Sent, it then follows that X = Sent, hence v = v. To see that S X, assume A S. v (A) = v (A). Thus A X. Assume A, B X. Then A B X because Then v (A) = f(a) and v (A) = f(a) by (1), hence v (A B) = v (A) v (B) by (2) for v = v (A) v (B) since A, B X

3 = v (A B) by (2) for v Similar calculations show that A B X, A B X, A B X, and A X. Thus X is closed under,,,, and. To show the existence of v we use Zorn s Lemma [Prop. 4.42(e), p. 279]. Say that a function w : dom(w) {T, F} is f-acceptable if (a) S dom(w) Sent, (b) w(a) = f(a) for every A S, i.e., w extends f, (c) if A B dom(w) then A, B dom(w) and w(a B) = w(a) w(b), (d) if A B dom(w) then A, B dom(w) and w(a B) = w(a) w(b), (e) if A B dom(w) then A, B dom(w) and w(a B) = w(a) + w(b), (f) if A B dom(w) then A, B dom(w) and w(a B) = w(a) w(b), (g) if A dom(w) then A dom(w) and w( A) = w(a). Let be the set of f-acceptable functions. is partially ordered by the inclusion relation. A chain is a subset C such that if v, v C then either v v or v v. We show below that the union of a chain of f-acceptable functions is f-acceptable. The union of a chain is trivially an upper bound with respect to inclusion, so this will show that every chain in has an upper bound in. Claim 1. f is f-acceptable, i.e. f. Proof. The domain of f is S, so (a) holds. Clearly (b) holds. The hypotheses of (c) (g) are always false by Theorem 1(3)(a), so (c) (g) hold as well. Claim 2. The union of every chain C is in. Proof. Let w = C. We will show w is f-acceptable. First we show w is a function. Let A, t 0, A, t 1 w. We must show t 0 = t 1. From A, t 0 w = C we get A, t 0 v for some v C, and, similarly, A, t 1 v for some v C. Since C is a chain, either v v or v v. In the first case, v v, we have A, t 0, A, t 1 v, hence t 0 = t 1 since v is a function. In the second case, v v, we have A, t 0, A, t 1 v, hence t 0 = t 1 since v is a function. Since w is the union of functions with values in {T, F}, its values must also be in {T, F}. Thus we know w : dom(w) {T, F}. Since w is the union of functions defined on all of S, w is also defined on all of S. Since w is the union of functions whose domains are sets of formulas, the domain of w is also a set of formulas. Thus (a) holds for w. Next we prove (c) for w. Note that dom(w) = u C dom(u). Assume A B dom(w). Then A B dom(u) for some u C. But u is f-acceptable, so (c) holds for u, hence A, B dom(u) and u(a B) = u(a) u(b). Since w is a function extending u, we have u(a B) = w(a B), u(a) = w(a), and u(b) = w(b). Combining these equations yields w(a B) = w(a) w(b). The proofs of (d) (g) for w are similar. Claim 3. has a maximal element. Call it v. Proof. By Claim 2 and Zorn s Lemma [Prop. 4.42(e), p. 279]. Claim 4. The domain of v is Sent. 3

4 4 Proof. By Claim 3, v. Hence dom(v) Sent by (a). We prove the opposite inclusion by induction. So we wish to prove S dom(v) and dom(v) is closed under the operations on formulas. We have S dom(v) by (a), since v. Claim 5. dom(v) is closed under. Proof. Assume A, B dom(v). We wish to show A B dom(v). Assume, to the contrary, that A B / dom(v). Let w = v { A B, v(a) v(b) }. We will prove that w. But this will contradict the maximality of v, since v w but v w, thus proving that the assumption was wrong, hence A B dom(v). First note that w is a function because the single additional pair that is in w but not in v begins with a formula that is not in the domain of v. From v and (a) we get S dom(v) Sent, and we have A B Sent. Consequently S dom(v) {A B} = dom(w) Sent, so (a) holds for w. If C S, then v(c) = f(c) by (b) for v. But w is a function extending v, so w(c) = f(c) as well. Thus (b) holds for w. Now we prove (c) for w. Assume C D dom(w). Then either C D dom(v) or else C D = A B. First case: C D dom(v). By (c) for v we get C, D dom(v) and v(c D) = v(c) v(d). Since w extends v, we have C, D, C D dom(w), v(c D) = w(c D), v(c) = w(c), and v(d) = w(d). Therefore w(c D) = w(c) w(d), as desired. So (c) holds in the first case. Second case: C D = A B. By property (3)(iii) of Sent, we get C = A and D = B. But we assumed A, B dom(v), so C, D dom(v) dom(w) and w(c D) = w(a B) = v(a) v(b). So (c) holds in the second case. Now we prove (d) for w. Assume C D dom(w). Then, by the definition of w, either C D dom(v) or else C D = A B. But the second possibility cannot occur, according to property (3)(ii) of Sent. Hence C D dom(v). By (d) for v we get C, D dom(v) w and v(c D) = v(c) v(d), hence, since w extends v, w(c D) = w(c) w(d). Similarly, (e), (f), and (g) hold for w. This completes the proof of Claim 5 that dom(v) is closed under. Similar proofs show that dom(v) is also closed under,, and. Since dom(v) contains S and is closed under,,, and, we conclude that Sent dom(v). But dom(v) Sent by (a) since v. Therefore dom(v) = Sent. P(S) is the set of all subsets of S, and is called the powerset of S. The next theorem proves the existence of a function st such that st(a) is the set of statement letters occurring in A. Theorem 6. There exists a unique map st : Sent P(S) such that (1) st(a) = {A} if A S, (2) st(a B) = st(a B) = st(a B) = st(a B) = st(a) st(b), (3) st( A) = st(a).

5 Proof. There was nothing in the proof of Theorem 5 that depended on the exact nature of the set {T, F}, so {T, F} can be replaced with P(S), and other parts of the proof can be appropriately changed. The next two theorems have similar inductive proofs. Theorem 7. For every A Sent, st(a) is finite, i.e., st(a) < ω. Theorem 8. If B Sent, and v, w are valuations with w(a) = v(a) for every A st(b), then w(b) = v(b). By the next theorem, there is a function sf such that sf(a) is the set of subformulas of A. Theorem 9. There exists a unique map sf : Sent P(S) such that (1) sf(a) = {A} if A S, (2) sf(a B) = sf(a) sf(b) {A B}, (3) sf(a B) = sf(a) sf(b) {A B}, (4) sf(a B) = sf(a) sf(b) {A B}, (5) sf(a B) = sf(a) sf(b) {A B}, (6) sf( A) = sf(a) { A}. Definition 10. Let A, B Sent and Γ Sent. 4. Tautologies (1) A is a tautology if v(a) = T for every valuation v. (2) A is a contradiction if v(a) = F for every valuation v. (3) A is logically equivalent to B, in symbols, A B, if v(a) = v(b) for every valuation v. (4) Γ logically implies A, and A is a logical consequence of Γ, in symbols, Γ = A, if v(a) = T for every valuation v with the property that v(g) = T for every G Γ. Write B = A instead of {B} = A, B, C = A instead of {B, C} = A, = A instead of = A. Note that = A iff A is a tautology. Theorem 11. There is an algorithm for determining whether a sentence A is a tautology. Proof. The truth-table method is such an algorithm. The previous theorems show that the only values of a valuation v that contribute to the value of v(a) are the values of v on the variables in st(a), and st(a) is finite, so there are only 2 st(a) cases to check to see whether v(a) = T for all valuations v. Theorem 12 (Prop. 1.1). Let A, B Sent. (1) A = B iff A B is a tautology. (2) A B iff A B is a tautology. Proof. Note first that v(a B) = T iff v(a) = v(b). The result follows from this observation and the relevant definitions. For a proof of the second part is suffices to observe that the following statements are equivalent: A B, 5

6 6 for all v, v(a) = v(b), for all v, v(a) v(b) = T, for all v, v(a B) = T. Prop. 1.2: modus ponens is sound. Prop. 1.3: Substitution into a tautology produces a tautology. Prop. 1.4: Substitution into equivalent statements yields equivalent statements. 5. Adequate Sets of Connectives Prop. 1.5:,, generate all truth functions. Prop. 1.6: Adequate sets are {, }, {, }, and {, }. 6. Axioms and proofs, 1.4. Now we assume that Sent is built up from the set S using only the operations and. We no longer have the fundamental operations,, and, which are introduced instead by the following definitions. Definition 13. A B := A B, A B := (A B), A B := ((A B) (B A)). (A1) := {A (B A) : A, B Sent}, (A2) := {(A (B C)) ((A B) (A C)) : A, B, C Sent}, (A3) := {( B A) (( B A) B) : A, B Sent}, (A1)(A2) := (A1) (A2), (A1)(A2)(A3) := (A1) (A2) (A3). Theorem 14. If A (A1)(A2)(A3) then A is a tautology. Definition 15. Let Γ, Sent, and A Sent. Then Γ A if there is a finite sequence B 1,..., B n of sentences of length n ω such that (1) B n = A, (2) for every i {1,..., n}, either (a) B i or (b) B i Γ or (c) there are j, k {1,..., i 1} such that B k = B j B i. The sequence B 1,..., B n is called a proof of A from hypotheses Γ using axioms. A less formal definition of proof is that it a sequence of formulas in which every formula is either a hypothesis or an axiom or is obtained from two previous formulas by modus ponens (MP), where the axioms and rule MP are (A1) A (B A)

7 (A2) (A (B A)) ((A B) (A C)) (A3) ( B A) (( B A) B) A, A B MP B A set of formulas Ω Sent is MP-closed iff C Ω whenever B, C Sent and B, B C Ω. The principle of induction on provability is Theorem 16. Γ A iff A {Ω : Γ Ω and Ω is MP-closed}. Definition 17. Let Γ Sent and A Sent. (1) Γ A iff Γ (A1)(A2)(A3) A, (2) Γ d A iff Γ (A1)(A2) A. Lemma 18. [See Lemma 1.8] d A A for every A Sent. 7 Proof. Let B 1 = A ((A A) A), B 2 = (A ((A A) A)) ((A (A A)) (A A)), B 3 = (A (A A)) (A A), B 4 = A (A A), B 5 = A A. Then B 1, B 2, B 3, B 4, B 5 is a proof of A A from using (A1)(A2), since B i i = 1,..., 5, B 1 (A1), B 2 (A2), B 2 = B 1 B 3, B 4 (A1), and B 3 = B 4 B 5. Sent for 7. Soundness, 1.4. Theorem 19. If Γ A and B is a tautology for every B, then Γ = A. Proof. Assume Γ A. Then there is a proof B 1,..., B n of A from Γ using. Claim. Γ = B i whenever 1 i n. The claim is proved by induction on i. Suppose i = 1. Then there are no previous formulas in the sequence, so B 1 must be in Γ. If B 1 Γ then we get Γ = B 1 immediately from the definition of =. If B 1 then B 1 is a tautology by hypothesis, hence = B 1, hence Γ = B 1. Thus the Claim holds when i = 1. Assume Γ = B 1,..., Γ = B i 1. We wish to show that Γ = B i. If B i Γ, then either B i, B i is a tautology by hypothesis so Γ = B i, or else B i Γ and Γ = B i. Suppose B i is obtained from previous formulas by modus ponens. Then there are j, k < i such that B k = (B j B i ). By the inductive hypothesis, Γ = B j and Γ = B j B i. Consider an arbitrary valuation v such that v(c) = T for every C Γ. Then v(b j ) = T since Γ = B j, and v(b j B i ) = v(b j ) v(b i ) = T since Γ = B j B i. But the equations v(b j ) = T and v(b j ) v(b i ) = T imply v(b i ) = T by the definition of. This shows that Γ = B i. Corollary 20. If Γ A and B is a tautology for every B Γ, then A is also a tautology. Corollary 21 (Prop. 1.12, Weak Soundness). If A then A is a tautology.

8 8 8. Properties of provability, 1.4. The next six theorems are the fundamental properties of provability (see p. 26). Theorem 22 (weakening). If Γ Γ and Γ A then Γ A. Theorem 23 (weakening). If and Γ A then Γ A. Theorem 24 (reiteration). If A Γ, then Γ A. Theorem 25 (MP-lemma). If Γ A and Γ A B then Γ B. Theorem 26 (compactness). If Γ A then Γ A for some finite subsets Γ Γ and. Theorem 27 (transitivity). If Γ B for every B Γ and Γ A, then Γ A. 9. The Deduction Theorem, 1.4. Theorem 28 (Deduction Theorem; see Prop. 1.9). Let A, B Sent and Γ Sent. Then Γ, A d B iff Γ d A B. Proof. Assume Γ d A B. Then Γ, A d A B by weakening (Th 22), and Γ, A d A by reiteration (Th 24). Hence Γ, A d B by the MP-lemma (Th 25). For the converse, assume Γ, A d B. Then there is a proof C 1,..., C n of B from Γ {A} using (A1)(A2). Claim. Γ d A C i whenever 1 i n. Proof. The claim is proved by induction on i. Case: i = 1. We need to show Γ d A C 1. Subcase: C 1 Γ (A1)(A2). Then C 1, C 1 (A C 1 ), A C 1 is a proof of A C 1 from Γ using (A1), so Γ d A C 1. Subcase: C 1 arises via MP. This case can t occur because there are no previous formulas. Subcase: C 1 = A. Then Γ d A C 1 by Lemma 18. Case: 1 < i n. Assume Γ d A C j for j {1,..., i 1}. We want to show Γ d A C i. Subcase: C i Γ (A1)(A2). Then C i, C i (A C i ), A C i is a proof of A C i from Γ using (A1)(A2), so Γ d A C i. Subcase: C i = A. Then d A C i by Lemma 18. Subcase: there are j, k {1,..., i 1} such that C k = C j C i. By assumption we have Γ d A C j and Γ d A C k. The latter fact, restated, says that Γ d A (C j C i ). Hence there is a proof B 1,..., B r of A C j from Γ using (A1)(A2), and there is a proof C 1,..., C s of A (C j C i )) from Γ using (A1)(A2). Then the following sequence of formulas is a proof of A C i from Γ using (A1)(A2). B 1. B r 1 A C j proof of A C j from Γ

9 D 1. D s 1 A (C j C i )) proof of A (C j C i )) from Γ (A (C j C i )) ((A C j ) (A C i )) (A2) (A C j ) (A C i ) MP A C i MP 9 The next theorem follows from the properties of provability and the Deduction Theorem. Theorem 29 ([see Cor. 1.10]). Let A, B, C Sent. Then (1) d A (B A), (2) d A ((A B) B), (3) d (A (B C)) (B (A C)), (4) d (A B) ((A (B C)) (A C)), (5) d A ((A B) ((A (B C)) C)), (6) d (A B) ((B C) (A C)), (7) A B, B C d A C, (8) B, A (B C) d A C. 10. The implicational fragment (no negation) References for this section are Church [1] and Tarski [2]. A formula is an arrow-tautology if it is a tautology with no occurrences of in it. Consideration of arrow-tautologies is simplified by assuming that negation is absent and Sent is simply the closure of S under alone. Problem. Which 2-letter, 3-arrow formulas are tautologies? Determine which formulas with at most 2 statement letters and exactly 3 arrows are tautologies. In other words, if A, B S and {C 1, C 2, C 3, C 4 } {A, B}, then which of the following formulas are tautologies? (see Problem 1.5(a)(b)(c)(f), p. 7). (C 1 C 2 ) (C 3 C 4 ), ((C 1 C 2 ) C 3 ) C 4, C 1 (C 2 (C 3 C 4 )), (C 1 (C 2 C 3 )) C 4, C 1 ((C 2 C 3 ) C 4 ). Also determine which of these formulas are provable by the Deduction Theorem, that is, are probable from using (A1)(A2). By symmetry, it suffices to consider only the formulas in which C 1 = A, and among those, only the ones that are tautologies. Solution: There are 40 such tautologies. Formula (37), ((A B) A) A, is known as Peirce s tautology. All but (37) can be proved using only (A1)(A2). Straighforward application of the Deduction Theorem is enough see this very easily in all cases except (35) and (36), which

10 10 require a little more work. (1) (A A) (A A) (11) A ((A A) A) (2) (A A) (B B) (12) A ((A B) A) (3) (A B) (A A) (13) A ((A B) B) (4) (A B) (A B) (14) A ((B A) A) (5) (A B) (B B) (15) A ((B B) A) (6) (B A) (A A) (16) B ((A A) B) (7) (B A) (B A) (17) B ((A B) B) (8) (B A) (B B) (18) B ((B A) A) (9) (B B) (A A) (19) B ((B A) B) (10) (B B) (B B) (20) B ((B B) B) (21) A (A (A A)) (35) ((A A) A) A (22) A (A (B A)) (36) ((A A) B) B (23) A (A (B B)) (37) ((A B) A) A (24) A (B (A A)) (38) ((B A) B) B (25) A (B (A B)) (39) ((B B) A) A (26) A (B (B A)) (40) ((B B) B) B (27) A (B (B B)) (28) B (A (A A)) (29) B (A (A B)) (30) B (A (B A)) (31) B (A (B B)) (32) B (B (A A)) (33) B (B (A B)) (34) B (B (B B)) It turns out that Peirce s tautology cannot be proved by the Deduction Theorem. To prove this, use an alternative truth table T 3 for using three truth values T, M, F: T 3 T M F T T M F M T T F F T T T A formula A is a T 3 -tautology if it always gets value T under any valuation. It can be checked that (A1) and (A2) are T 3 -tautologies, that being a T 3 -tautology is preserved by modus ponens, and that Peirce s formula is not a T 3 -tautology since ((M F) M) M = M. There is, up to isomorphism, there is just one function on T, M, F which has these properties. Theorem 30. d ((A B) A) A, i.e., (A1)(A2) (Peirce) Axiomatizing arrow-tautologies. Since Peirce s tautology cannot be derived using only (A1)(A2), one can ask what axiom could be added to (A1)(A2) to obtain all arrow-tautologies? Is Peirce s tautology itself enough? It turns that the answer is yes, but to prove this we invoke some results of Tarski and Bernays. Consider the following axiom sets of arrow-tautologies. (A1) A (B A) (A2) (trans) (A (B C)) ((A B) (A C)) (A B) ((B C) (A C))

11 11 (Peirce) ((A B) A) A (Tarski) ((A B) C) ((A C) C) Theorem 31 (Tarski [2, p. 52 3]). A is an arrow-tautology iff (A1)(trans)(Tarski) A. Bernays pointed out that (Tarski) may be replaced by (Peirce) in Tarski s theorem, thus giving Theorem 32 (Tarski-Bernays [2, Th. 29]). A is an arrow-tautology iff (A1)(trans)(Peirce) A. From these results we can now obtain Corollary 33. A is an arrow-tautology iff (A1)(A2)(Peirce) A. Proof. Only arrow-tautologies can be deduced from (A1)(A2) and (Peirce) by MP. Conversely, if A is an arrow-tautology then (A1)(trans)(Peirce) A by the Tarski-Bernays Theorem, but (A1)(A2) (trans) by Th. 29, so (A1)(A2)(Peirce) A by transitivity of provability. Independence was proved for the axiom systems in the theorems of Tarski and Tarski-Bernays by Lukasiewicz [2, p. 52 3]. For the independence of (A1), use the 4-element table T 5 with 1 as distinguished value. Then (A2), (trans), (Peirce), and (Tarski) are T 5 -tautologies but (A1) is not, since 2 (3 2) = 2 4 = 4 1. T For the independence of (A2) and (trans) from (A1), (Peirce), and (Tarski), use the 4-element table T 6, with 1 as distinguished element. Then (A1), (Peirce) and (Tarski) are T 6 -tautologies but (A2) and (trans) are not. T (Tarski) fails in T 3 because ((M F) M) ((M M) M) = (F M) (T M) = T M = M. (A1) (trans) (Tarski) (A1) (trans) (Peirce) (A1) (A2) (Peirce) T 3 true true false T 5 false true true T 6 true false true Theorem 34 ( Lukasiewicz 1948, see [2, footnote to Th. 30]). The shortest single axiom scheme that axiomatizies the implicational propositional calculus is (A (B C)) ((C A) (D A)). For some hints, see Church [1, 18.4]. Church [1, 18.3] also has suggestions for proving Theorems 31 and 32.

12 12 Theorem 35. (1) (A1), (trans) (A2). (2) (A1)(trans) A A. Proof. Use either table below. T M F T T M M M T M T F T M T T M F T T M M M T M T F T T T 11. Completeness Theorem for Propositional Calculus of and Definition 36. For every valuation v and every A Sent, let { A v A if v(a) = T = A if v(a) = F Lemma 37. Assume v is a valuation and A is a sentence built up from sentence letters S 1,..., S n. Let (v) = {S v 1,..., S v n}. Then (v) A v, i.e., ( ) If v(a) = T then (v) A and if v(a) = F then (v) A. Proof. The proof is by induction on the complexity of A. The boxed formulas are lemmas that need to be proved. (1) Suppose A is a statement letter, say A = S i. If v(a) = T then v(s i ) = T, hence A = S i = S v i (v), hence (v) A. If v(a) = F then v(s i ) = F, hence A = S i = Si v (v), hence (v) A. (2) Suppose A = B, and ( ) holds for B. If v(a) = T = v( B) then v(b) = F, so by ( ), (v) B, hence (v) A. If v(a) = F = v( B) then v(b) = T, so by ( ), (v) B, hence (v) B since B B, hence (v) A. (3) Suppose A = (B C), and ( ) holds for B and C. If v(a) = T = v(b C) then either v(c) = T, so by ( ), (v) C, hence (v) A since C B C, or else v(b) = F, so by ( ), (v) B, hence (v) A since B B C. If v(a) = F = v(b C) then v(b) = T and v(c) = F, so by ( ), (v) B and (v) C, hence (v) A since B, C (B C).

13 13 Theorem 38 (Weak Completeness). If = A then A. Proof. Assume = A. Note that A v = A for every v, since we are assuming that A is a tautology. Claim. If i {1,..., n + 1}, then S v i, S v i+1,..., S v n A for every valuation v. Proof. By induction. Case i = 1. We have S v 1,..., S v n A for every v by Lemma 37. Case 1 < i n + 1. Assume the Claim for i 1. For a given valuation v, we have S v i 1, S v i,..., S v n A by the Claim. Let w be the same valuation as v except that v(s i 1 ) w(s i 1 ). By the inductive hypothesis S v i 1, S v i,..., S v n A S w i 1, S w i,..., S w n A By the Deduction Theorem, plus S v i = S w i,..., S v n = S w n, (1) (2) Note that {S v i 1, S w i 1} = {S i 1, S i 1 }. But (3) S v i,... S v n S v i 1 A S v i,... S v n S w i 1 A S i 1 A, S i 1 A A by B C, B C C, so Si v,..., Sn v A by (1), (2), (3), and transitivity of. Now apply the Claim with i = n + 1 to get A. 12. The boxed lemmas Items (1) (30) below are theorems that hold for all A, B Sent. (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15) (16) (17) (18) ( B A) (( B A) B) (A3) B A, B A B (1), DT A B A (A1), DT A B A (3) with (B = B) ( ) A = A A B A (3) with B = B A, A B (2), (4), (5) B A, A B (2), (5) B A, A B (2), (4) A A B (6), DT B A A B (7), DT B A A B (8), DT B B, B B (7) with (A = B) B B Lemma 18 B B (12), (13) B B (14), DT

14 14 (19) (20) (21) (22) (23) (24) (25) (26) (27) (28) (29) (30) (31) (32) (33) B B, B B B B B B (16), (17) B B (15) with (B = B) ( ) A = B (8) with B = B (18), DT A, A B B (14) with (B = A), Th. 4, (18) A B A B A B B A (20), DT (11) with A B B A (21), (22) A B B A (10) with A B, A B B (23), (24), (2) ( ) A = B B = A ( ) A = B B = A A, A B B A, A B, B A (A B) B (26), DT (A B) B B (A B) (23) with (A = A B) A B (A B) (27), (28) A, B (A B) (29), DT The boxed lemmas have now been proved, since C B C is (3), B B C is (9), B B is (18), B, C (B C)) is (30), and B C, B C C is (25). References 1. Alonzo Church, Introduction to Mathematical Logic, Princeton University Press, Princeton, NJ, 1996, Reprint of the second (1956) edition, Princeton Paperbacks. MR 98g: Alfred Tarski, Logic, Semantics, Metamathematics, second ed., Hackett Publishing Co., Indianapolis, IN, 1983, Papers from 1923 to 1938, Translated by J. H. Woodger, Edited and with an introduction by John Corcoran. MR 85e:01065