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Lecture5 - Lecture 5 Algebra of propositional formulas...

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Definition (by recursion) of substitution function on PROP (def. 1.2.5) Given a propositional formula φ , the function replaces a propositional formula p i in it with ψ . It is denoted as φ [ ψ /p i ]. Base case φ [ ψ /p i ] = φ if φ is atomic and φ p i . = ψ if φ is atomic and φ = p i . Inductive cases ( φ 1 φ 2 )[ ψ /p i ] = φ 1 [ ψ /p i ] φ 2 [ ψ /p i ] ( ¬ φ ) [ ψ /p i ] = ( φ [ ψ /p i ]) Example ((p 1 ˭ p 2 ) p 3 )[ p 4 ˮ p 5 / p 2 ] = ((p 1 ˭ ( p 4 ˮ p 5 )) p 3 ) Lecture 5 Algebra of propositional formulas/ Interdefinability of connectives
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Theorem 1.2.6 If |= φ 1  φ 2 , then |= ψ [ φ 1 /p]  ψ [ φ 2 /p] for any atom p. Proof Observe that the theorem is proved if we can prove the following: for any valuation such that [| φ 1 |] v = [| φ 2 |] v we have [| ψ [ φ 1 /p] |] v = [| ψ [ φ 2 /p] |] v . (Why?) This is easily proved by induction on ψ . In the base case, the substitution results in either the equation [| ψ |] v = [| ψ |] v or the equation [| φ 1 |] v = [| φ 2 |] v . The induction cases are immediate given the recursive definition of the substitution function. By definition: ( ψ 1 ψ 2 )[ φ i /p] = ψ 1 [ φ i /p] ψ 2 [ φ i /p] ( ¬ ψ ) [ φ i /p] = ( ψ [ φ i /p i ]). By the induction hypothesis, the truth values of the expressions on the right under v are the same for i = 1 and i = 2. But this then means that the truth values of the expressions on the left are the same under v for i=1 and i=2. So the claim holds for the induction cases. Substitution theorem
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Implicit in the substitution theorem is a notion of equivalence between propositional formulas. In satisfying the condition |= φ 1  φ 2 the formulas φ 1 and φ 2 of the theorem are in a strong sense equivalent. Every valuation gives the same truth value to both. Let the expression ‘ φ 1 φ 2 denote this kind of equivalence. Example p 1 p 2 ¬ p 1 ˮ p 2 . p 1 p 2 p 1 p 2 p 1 ˮ p 2 1 1 1 1 1 0 0 0 0 1 1 1 0 0 1 1 For any for any valuation v [|p 1 p 2 |] v = [| p 1 ˮ p 2 |] v Equivalence between propositional formulas
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Lemma 1.3.5
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Lecture5 - Lecture 5 Algebra of propositional formulas...

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