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Unformatted text preview: t;. Suppose I A.
Now: T (∀xim)Fm(xim) → Fm(cm). (K5, cm is free for xim in Fm(xim), since cm doesn't occur in
Fm(xim).)
So:
I (∀xim)Fm(xim) → Fm(cm). (Prop. 4.4.  axioms are logically valid.)
Hence: I Fm(cm). (Prop. 3.26.)
Thus: T Fm(cm). (Inductive Hypothesis.)
Now: Suppose T A.
Then: T ∼A. (T is complete.) Or T ∼(∀xim)Fm(xim).
But:
T ∼(∀xim)Fm(xim) → ∼Fm(cm). (Gm is an axiom of T.)
So:
T ∼Fm(cm). But T is consistent. Thus it must be that T A.
2. "⇒". Suppose T A. Now suppose I A.
Then: There's a valuation in I that doesn't satisfy A.
So:
There's a valuation v that doesn't satisfy Fm(xim).
Now: v (x im) = d, for some closed term d in DI.
And:
v(d) = d. (Valuations map constants to constants; hence closed terms to closed terms.)
So:
v (x im) = v(d).
Now: We have the following:
1. Fm(xim) is a wf with xim free.
2. d is a (closed) term free for xim in Fm(xim).
3. v (x im) = v(d).
4. v is iequivalent to itself.
Thus: v satisfies Fm(d) iff v satisfies Fm(xim). (Prop. 3.23.)
Hence: v does not satisfy Fm(d).
Thus: I Fm(d) .
Now: T (∀xim)Fm(xim). (assumption T A.)
So:
T Fm(d). (K5, d is free for xim in Fm(xim), and MP.)
Hence: I Fm(d) . (Inductive Hypothesis.) So it must be that I A. Part III.
Lemma 5: For any (open or closed) wf B of L, if S B, then I B.
Proof: Suppose S B, for some wf B of L.
If B is closed, then T B, hence I B. (Lemma 4: If B is a closed wf of L, it is also a closed wf of L+ .)
Suppose B is open.
Then: S B'. (Prop. 4.19, B' is the universal closure of B.)
Hence: T B '.
Thus: I B'. (Lemma 4.)
Hence: I B. (Cor. 3.28.)...
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 Spring '06
 JonathanBain

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