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3. Normal forms 23 allows us to take a sentence, and replace its subsentences with tautologically equivalent subsentences, thereby obtaining a sentence that is tautologically equivalent to what we started with. By choosing cunning substitutions, repeated applications of Lemma 2.2 will allow us to reach a DNF equivalent. I say that we shall choose the substitutions cunningly. In fact, we shall proceed by three lemmas. Each lemma is proved by induction. Each lemma corresponds to one of the three clauses (dnf1)–(dnf3) of the formal definition of a DNF sentence (see §3.1). Each lemma brings us closer to DNF. Indeed – most excitingly of all – each lemma yields an algorithm for pushing a sentence towards DNF (for a reminder about what an algorithm is, see §2.2). I shall start with condition (dnf1). This tells us that any DNF sentence must contain only negations, conjunctions and disjunctions. So our very first task is to show how to generate an equivalent sentence of this form. Lemma 3.2. For any sentence, there is a tautologically equivalent sentence satisfying (dnf1). Proof. Let A be any sentence, and suppose, for induction on length, that for every shorter sentence, B, there is a sentence B ∗ which satisfies (dnf1), and such that B ⊨ B ∗ . There are now six cases to consider: ⊨ Case 1: A is atomic. Then A already satisfies (dnf1), and A ⊨ A, so there is nothing to prove. Case 2: A is ¬B, for some sentence B. Then A ⊨ ¬B ∗ , by Lemma 2.2, and ¬B ∗ satisfies (dnf1). Case 3: A is (B ∨ C ), for some sentences B and C . Then A ⊨ (B ∗ ∨ C ∗ ), by Lemma 2.2, and this satisfies (dnf1). Case 4: A is (B ∧ C ), for some sentences B and C . Then A ⊨ (B ∗ ∧ C ∗ ), by Lemma 2.2, and this satisfies (dnf1). Case 5: A is (B → C ), for some sentence B and C . In this case: A ⊨ ⊨ ⊨ ⊨ (B → C ) given ⊨ (B ∗ → C ∗) by Lemma 2.2 ⊨ (¬B ∗ ∨ C ∗ ) ⊨ by considering truth tables And this last sentence satisfies (dnf1). Case 6: A is (B ↔ C ), for some sentence B and C . Then, as in Case 5, A ⊨ ((B ∗ ∧ C ∗ ) ∨ (¬B ∗ ∧ ¬C ∗ )), which satisfies (dnf1). ⊨ This completes the induction. It is worth noting that this proof doesn’t simply tell us that an equivalent sentence exists; it implicitly provides us with an algorithm for eliminating occurrences of conditionals and biconditionals from a sentence, one at a time, whilst preserving tautological equivalence. (Compare the discussion of the proof of the Interpolation Theorem; see §2.2.) In particular, the algorithm just tells us to rewrite our sentence repeatedly, relying on Lemma 2.2 and the following general equivalences: (A → B) (A ↔ B) ⊨ ⊨ ⊨ (¬A ∨ B) ⊨ ((A ∧ B) ∨ (¬A ∧ ¬B)) ⊨ ⊨ ⊨ ■
3. Normal forms 24 until no (bi)conditionals remain. For example, if we start with the sentence: then we first get: and then: ((¬A ∨ (¬B ∧ C)) → ¬(D ↔ E)) (¬(¬A ∨ (¬B ∧ C)) ∨ ¬(D ↔ E)) (¬(¬A ∨ (¬B ∧ C)) ∨ ¬((D ∧ E) ∨ (¬D ∧ ¬E))) which is a bit closer to DNF. But there is still some way to go. In particular, condition (dnf2) tells us that in any DNF sentence, every occurrence of negation has minimal scope. So our next task is to prove: Lemma 3.3. For any sentence satisfying (dnf1), there is a tautologically equivalent sentence satisfying both (dnf1) and (dnf2) Proof. Let A be any sentence satisfying (dnf1) and suppose, for induction on length, that for every shorter sentence B, if B satisfies (dnf1) then there is a sentence B ∗ satisfying (dnf1) and (dnf2) such that B ⊨ B ∗ . There are now three cases to consider: Case 1: A is atomic. Then there is nothing to prove, since A contains no negations. Case 2: A is either (B ⊛ C ), for some two-place connective ⊛. In this case, A ⊨ (B ∗ ⊛ C ∗ ), which satisfies (dnf1) and (dnf2). Case 3: A is ¬B, for some sentence B. Then there are four subcases: ⊨ 3a: A is ¬B, for some atomic sentence B. Then A itself satisfies (dnf1) and (dnf2). 3b: A is ¬¬C , for some sentence C . Then A ⊨ C ∗ , which satisfies both (dnf1) and (dnf2). 3c: A is ¬(C ∨ D), for some sentences C and D. Then clearly A ⊨ (¬C ∧ ¬D). Moreover, since ¬C is shorter than A, by the induction hypothesis there is some E satisfying (dnf1) and (dnf2) such that E ⊨ ¬C ; similarly, there is some F satisfying (dnf1) and (dnf2) such that F ⊨ ¬D. Hence A ⊨ (E ∧ F ), which satisfies (dnf1) and (dnf2). 3d: A is ¬(C ∧ D), for some sentences C and D. Exactly like Subcase 3c, except that we start by noting that A ⊨ (¬C ∨ ¬D). ⊨ This completes the induction on length. ⊨ As with the previous result, this Lemma implicitly provides us with an algorithm. This time, it allows us to drive negations deep into a sentence, so that they have minimal scope. In particular, we repeatedly apply these equivalences: ¬¬A ¬(A ∨ B) ¬(A ∧ B) ⊨ ⊨ ⊨ ⊨ A ⊨ ⊨ (¬A ∧ ¬B) ⊨ (¬A ∨ ¬B) In more detail, here’s an algorithm based on these rules: ⊨ ⊨ ⊨ ⊨ ■
Page 1 and 2: Metatheory for truth-functional log
Page 3 and 4: Contents 1 Induction 3 1.1 Stating
Page 5 and 6: Contents 2 The corresponding notion
Page 7 and 8: 1. Induction 4 knocking the first o
Page 9 and 10: 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: 3. Normal forms 22 Case 2: S is tru
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 and 62: 6. Completeness 58 Finally: by Lemm
Page 63 and 64: 6. Completeness 60 Practice exercis

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