Proof
Let f: N→N be the function defined by:
- f(#(θ)) = #(θ(#(θ))
for each T-formula θ in one free variable, and f(n) = 0 otherwise. The function f is computable, so there is a formula δ representing f in T. Thus for each formula θ, T proves
- (∀y) ,
which is to say
- (∀y) .
Now define the formula β(z) as:
- β(z) = (∀y) ,
then
- β(#(θ)) ⇔ (∀y) ,
which is to say
- β(#(θ)) ⇔ ψ(#(θ(#(θ))))
Let φ be the sentence β(#(β)). Then we can prove in T that:
- (*) φ ⇔ (∀y) ⇔ (∀y) .
Working in T, analyze two cases:
1. Assuming φ holds, substitute #(β(#(β)) for y in the rightmost formula in (*), and obtain:
- (#(β(#(β)) = #(β(#(β))) → ψ(#(β(#(β))),
Since φ = β(#(β)), it follows that ψ(#(φ)) holds.
2. Conversely, assume that ψ(#(β(#(β)))) holds. Then the final formula in (*) must be true, and φ is also true.
Thus φ ↔ ψ(#(φ)) is provable in T, as desired.
Read more about this topic: Diagonal Lemma
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