{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Lecture5

# Lecture5 - Lecture 5 Algebra of propositional formulas...

This preview shows pages 1–5. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Lemma 1.3.5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 14

Lecture5 - Lecture 5 Algebra of propositional formulas...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online