Unformatted text preview: ppear in F0(xi0).
(ii) For n > 0, cn ∉ {c0, ..., cn−1} and cn doesn't appear in F0(xi0), ..., Fn(xin).
Let Gk be the wf ∼(∀xik)Fk(xik) → ∼Fk(ck).
Construct the sequence S0, S1, ... as follows:
(i) Let S0 = S+ .
(ii) For each n ≥ 1, let Sn be the extension of Sn−1 obtained by adding Gn−1 as a new axiom. Lemma 2: Each of S0, S1, ..., is consistent.
Proof: By (weak) induction on sequence number n.
Base Step: n = 0. S0 = S+ is consistent (Lemma 1).
Induction Step: For n > 0, suppose Sn is consistent. Now show Sn+1 is consistent.
Suppose Sn+1 is not consistent.
Then: There's a wf A of L+ such that Sn+1 A and Sn+1 (∼A).
Note: Sn+1 (A → (∼A → ∼B)). (This is a tautology of L, and hence of L. By Prop. 4.3, it is a theorem
of K, and hence of the extension Sn+1 of K.)
Thus: Sn+1 (∼B), for any wf B. In particular, Sn+1 (∼Gn). (Even though Gn is an axiom of Sn+1! This is
a consequence of assuming Sn+1 is not consistent.)
So:
{Gn} Sn (∼Gn). (Sn+1 is the same as {Gn} Sn.)
Thus: Sn (Gn → ∼Gn). (By the Deduction Theorem for K. Gn is closed so no ap...
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 Spring '06
 JonathanBain
 Logic, Metalogic, Sn, inductive hypothesis

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