Wouldn’t a counterexample itself be proof of the existence of the counterexample?
If counterexamples are a subset of proofs of the existence of a counterexample, then:
(No counterexamples < N AND no simple proof of a counterexample) IMPLIES
(No counterexamples < N AND no counterexample) IMPLIES
(No counterexample)
But the absence of a counterexample is identical with the truth of the conjecture, which is what we were trying to prove. Once you are willing to look at N+1, by induction you are looking at all the positive integers, which takes forever.
Wouldn’t a counterexample itself be proof of the existence of the counterexample?
If counterexamples are a subset of proofs of the existence of a counterexample, then:
(No counterexamples < N AND no simple proof of a counterexample) IMPLIES
(No counterexamples < N AND no counterexample) IMPLIES
(No counterexample)
But the absence of a counterexample is identical with the truth of the conjecture, which is what we were trying to prove. Once you are willing to look at N+1, by induction you are looking at all the positive integers, which takes forever.