Is there some fixed number of cases that is sufficient to establish conclusively an arbitrary proposition? Is there some fixed N that is large enough for any P?
I don’t think so. There are only O(k^N) strings of length N. It would be very strange if there were only finitely many true claims about the integers.
Here is a proposed proof sketch, by the way.
Suppose for contradiction that there were some N such that all true claims about natural numbers had proofs of length less than N. Necessarily, there is a least such N, call it N’, for which that’s true. Take P as some proposition with a proof of length N’. Take P2 as some other proposition. The proof of (P and P’) is going to have length longer than N’, since it has to include P as a subproof and at least one additional line for P and P’.
This violates our assumption, QED.
P and P’ doesn’t seem like a very interesting proposition, but I don’t know a way to formalize “interesting”. I think there are probably are only finitely many claims that a finite mathematical mind can possibly find interesting.
Pretty sure this can’t happen. We stipulated that we had the shortest proof for P. How did P and P’ get proved without proving P first? Or alternatively, without there being a shorter proof available for P without P’?
What if P = “The prime numbers are infinite or there are exactly 1337 of them” and P’ = “There are not exactly 1337 prime numbers.”
The shortest way to prove P involves proving (P and P’), which is the statement “The prime numbers are infinite.” It would take a roundabout argument indeed to exclude all finite cardinalities besides 1337, without also excluding 1337 and accidentally proving (P and P’).
I don’t have an example to hand. I just have a feeling that there might be some case where the shortest proof of (P and P’) is a component of the shortest proof of P, rather than the other way around.
Could you elaborate on what question that is your answer to? Is that an answer to one of my questions? I might just be confused.
Is there some fixed number of cases that is sufficient to establish conclusively an arbitrary proposition? Is there some fixed N that is large enough for any P?
I don’t think so. There are only O(k^N) strings of length N. It would be very strange if there were only finitely many true claims about the integers.
Here is a proposed proof sketch, by the way. Suppose for contradiction that there were some N such that all true claims about natural numbers had proofs of length less than N. Necessarily, there is a least such N, call it N’, for which that’s true. Take P as some proposition with a proof of length N’. Take P2 as some other proposition. The proof of (P and P’) is going to have length longer than N’, since it has to include P as a subproof and at least one additional line for P and P’.
This violates our assumption, QED.
P and P’ doesn’t seem like a very interesting proposition, but I don’t know a way to formalize “interesting”. I think there are probably are only finitely many claims that a finite mathematical mind can possibly find interesting.
I don’t think this necessarily follows. Suppose the last few lines of the proof of P read something like:
Pretty sure this can’t happen. We stipulated that we had the shortest proof for P. How did P and P’ get proved without proving P first? Or alternatively, without there being a shorter proof available for P without P’?
What if P = “The prime numbers are infinite or there are exactly 1337 of them” and P’ = “There are not exactly 1337 prime numbers.”
The shortest way to prove P involves proving (P and P’), which is the statement “The prime numbers are infinite.” It would take a roundabout argument indeed to exclude all finite cardinalities besides 1337, without also excluding 1337 and accidentally proving (P and P’).
I don’t have an example to hand. I just have a feeling that there might be some case where the shortest proof of (P and P’) is a component of the shortest proof of P, rather than the other way around.