Diagonal Lemma - Proof

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.