dtps - My Rendition of Hunter's Strong Inductive Proof of...

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My Rendition of Hunter’s Strong Inductive Proof of The Deduction Theorem for PS Branden Fitelson 02/13/07 Theorem . Let Γ be an arbitrary set of formulas of P, and let A and B be arbitrary formulas of P. If Γ ∪{ A } ‘ PS B , then Γ PS A B . If there is a derivation (in PS) of B from Γ ∪{ A } , then there is a derivation (in PS) of A B from Γ alone. Proof. We will prove this by strong induction on the length of derivations establishing Γ ∪{ A } ‘ PS B . More precisely, we will prove by strong induction that the statement S(n) is true for all n 1, where S(n) is: S(n) : If there is a derivation (in PS) of B from Γ ∪{ A } of length n , then there is a derivation (in PS) of A B from Γ alone. To prove that S(n) is true for all n 1 by strong induction, we will proceed in two steps: I. Basis Step : Prove that S( 1 ) is true. II. Inductive Step : Assume as our inductive hypothesis that S(i) is true for all i such that 1 < i < n . Then, use this inductive hypothesis to show that S(i) is true when i = n ( i.e. , that S(n) is true). Having accomplished both (I) and (II), we will have succeeded in showing that S(n) is true for all n 1. In what follows, I will use D to stand for some derivation (in PS) of B from Γ ∪ { A } , and I will use D 0 to stand for some derivation (in PS) of A B from Γ alone. Since we are concerned about the length of D , I will use D 1 , D 2 ,..., D m to stand for the m lines of a derivation D with length m . Now, the inductive proof.
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dtps - My Rendition of Hunter's Strong Inductive Proof of...

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