Metatheory - University of Cambridge

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6. Completeness 59 if Γ are infinite, SimpleSearch will not terminate. Indeed, we will not even get past Step 1 of SimpleSearch, since we will never finish writing all of Γ and ¬C as assumptions! A similar problem arises for the more abstract proof of Lemma 6.2 offered in §6.3. In the course of that proof, I assumed that only finitely many atomic sentences occur as subsentences in the sentences among ∆, namely, A 1 , A 2 , . . . , A m , and I used this assumption to select my jointly polarising sentences, Ω m . But if ∆ are infinite, it might be that infinitely many atomic sentences occur as subsentences among ∆, and my proof will break down. Fortunately, this last problem can be repaired. In making our repair, the crucial insight is that every TFL-proof is only ever finitely long. This insight will allow us to prove Lemma 6.2 in full generality. Proof of Lemma 6.2, in the fully general case. Suppose ∆ ⊬ ⊥; let A 1 , A 2 , . . ., be all the atomic sentences that occur as subsentences in any of the sentences among ∆. We define: Base clause. If ∆, A 1 ⊬ ⊥, then let Ω 1 be A 1 . Otherwise, let Ω 1 be ¬A 1 Recursion clause. For each number n > 1: • If ∆, Ω n , A n+1 ⊬ ⊥, then let Ω n+1 be all of Ω n and A n+1 • Otherwise, let Ω n+1 be all of Ω n and ¬A n+1 Just as before, we can prove by induction that: Claim 1. ∆, Ω n ⊬ ⊥, for all n. Now let Ω be all and only the sentences that are to be found among some (any) Ω n . I claim: Claim 2. ∆, Ω ⊬ ⊥ To establish this claim, suppose that ∆, Ω ⊢ ⊥. Since every TFL-proof is only finitely long, there must be finitely many sentences, ±A i1 , ±A i2 , . . . , ±A ik , among Ω such that: ∆, ±A i1 , ±A i2 , . . . , ±A ik ⊢ ⊥ But then all of these sentences, ±A i1 , ±A i2 , . . . , ±A ik are among some Ω n . (Specifically, they are in Ω n , when n is the largest number among i 1 , . . . , i k .) So ∆, Ω n ⊢ ⊥, which contradicts Claim 1. It follows from Claim 2 that Ω are jointly polarising. The remainder of the proof now proceeds exactly as before. ■ In short, pursuing our more abstract proof-strategy allows us to offer a ‘completely general’ proof of the Completeness Theorem. What is more, we can obtain an important result as an immediate consequence. Theorem 6.12. Compactness Theorem. Let ∆ be any sentences. Suppose that, whenever you take only finitely many sentences of ∆, those finitely many sentences are jointly consistent. Then ∆ themselves are jointly consistent. Proof. We prove the contrapositive. Suppose that ∆ are jointly inconsistent. By the (completely general) Completeness Theorem, ∆ ⊢ ⊥. Since every TFLproof is only finitely long, there are finitely many sentences, ∆ 0 , among ∆ such that ∆ 0 ⊢ ⊥. By the Soundness Theorem, ∆ 0 are jointly inconsistent. ■
6. Completeness 60 Practice exercises A. Justify the following principles, which are appealed to in the proof of Lemma 6.10 and in the more abstract proof of Lemma 6.2. 1. If Γ ⊢ C then Γ ⊢ ¬¬C 2. If Γ ⊬ C ∧ D, then either Γ ⊬ C or Γ ⊬ D 3. If Γ ⊢ ¬C or Γ ⊢ ¬D, then Γ ⊢ ¬(C ∧ D) 4. If Γ ⊢ ¬C and Γ ⊢ ¬D, then Γ ⊢ ¬(C ∨ D). 5. If Γ ⊬ ⊥, then either Γ, A ⊬ ⊥ or Γ, ¬A ⊬ ⊥. 6. If Γ ⊢ A and Γ ⊬ ⊥ then Γ, A ⊬ ⊥. 7. If Γ, A ⊬ ⊥, then Γ ⊬ ¬A 8. If Γ, ¬A ⊢ ⊥, then Γ ⊢ A B. Work through Cases 5 and 6 of the proof of Lemma 6.10.
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Metatheory for truth-functional log
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Contents 1 Induction 3 1.1 Stating
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Contents 2 The corresponding notion
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1. Induction 4 knocking the first o
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1. Induction 6 So the induction pro
Page 11 and 12: 1. Induction 8 sentences. But thing
Page 13 and 14: 1. Induction 10 Case 1: A is atomic
Page 15 and 16: Substitution 2 In this chapter, we
Page 17 and 18: 2. Substitution 14 2.2 Interpolatio
Page 19 and 20: 2. Substitution 16 Step 3: Repeat s
Page 21 and 22: 2. Substitution 18 We shall use the
Page 23 and 24: Normal forms 3 In this chapter, I p
Page 25 and 26: 3. Normal forms 22 Case 2: S is tru
Page 27 and 28: 3. Normal forms 24 until no (bi)con
Page 29 and 30: 3. Normal forms 26 4a: B ∗ contai
Page 31 and 32: 3. Normal forms 28 (cnf2) Every occ
Page 33 and 34: Expressive adequacy 4 In chapter 3,
Page 35 and 36: 4. Expressive adequacy 32 4.2 Indiv
Page 37 and 38: 4. Expressive adequacy 34 In fact,
Page 39 and 40: 4. Expressive adequacy 36 negations
Page 41 and 42: 5. Soundness 38 Again, it is easy t
Page 43 and 44: 5. Soundness 40 m i j k l A ∨ B A
Page 45 and 46: 5. Soundness 42 assumptions. So, ap
Page 47 and 48: 6. Completeness 44 6.2 An algorithm
Page 49 and 50: 6. Completeness 46 been expanded. N
Page 51 and 52: 6. Completeness 48 We have managed
Page 53 and 54: 6. Completeness 50 1 (A ∨ (¬B
Page 55 and 56: 6. Completeness 52 Step 1. Start a
Page 57 and 58: 6. Completeness 54 Now let D be any
Page 59 and 60: 6. Completeness 56 Now, when we are
Page 61: 6. Completeness 58 Finally: by Lemm
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