Kleene’s (Second) Recursion Theorem
Let Q:N2→N be partial computable. Then exists an index e so that
φe(n)=Q(e,n)
on all n. That is, the program φe is able to “see itself”
We prove this using Rogers’ fixpoint theorem. Take some Q partial computable, and let
f(e)=the index of (n↦Q(e,n))
then by Rogers’, we know that exists some e for which
φe=φf(e)
Which is to say that
φe=n↦Q(e,n)