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Dec 3, 2010 at 1:40 comment added user5810 Reeaallyy? Then what is the n in "problem(2n) := 'There is a proof of length n that the original statement is false.'" and "problem(2n+1) := 'There is a proof of length n that the original statement is true.'"? It sure seems like proof size to me.
Dec 2, 2010 at 22:08 comment added Adam "and my proof size is..." not listed as an input to your algorithm.
Dec 2, 2010 at 14:51 comment added arsmath Adam, the vast majority of proofs found by humans are short. There's the occasional gigantic 100-page proof, but if the vast majority of proofs could be generated by machine, then the impact on mathematical practice would be large.
Dec 2, 2010 at 5:24 comment added user5810 "Choose your axioms to be ZFC and your proposition to be the Continuum Hypothesis" and my proof size as ... ? Once it checks all potential proofs of the given size, it will halt on "no short proof" (if ZF is consistent).
Dec 2, 2010 at 4:08 comment added Adam Ricky, your algorithm will not always halt. Choose your axioms to be ZFC and your proposition to be the Continuum Hypothesis. Your algorithm will run forever since there is no proof of either CH or $\neg$CH from ZFC, let alone one which is polynomial in the length of CH! Regarding your other question: if the size $|p|$ of a proof $p$ is polynomial in $n$, then $|p|\leq\underset{0\leq i<k}\sum a_i\cdot n^{b_i}$; the $a_i$ are the coefficients and the $b_i$ are the exponents (I should have written "coefficients and exponents as inputs").
Dec 2, 2010 at 3:26 comment added user5810 What are "the proof-size's coefficients"? The algorithms, including mine, take the proof's size as an additional input and always halt, with the answer "True", "False", or "no short proof" (or "inconsistency").
Dec 2, 2010 at 2:53 comment added Adam I should also add that any algorithm which always halts would necessarily need knowledge of the proof-size's coefficients as an additional input (the algorithm in Ricky's answer may fail to halt, which is why it doesn't need the coefficients as inputs).
Dec 2, 2010 at 2:51 comment added Adam David, the algorithm would only work if a polynomial-sized proof existed. If one didn't, the algorithm would simply fail to terminate, but you'd never know the difference between "will never terminate" and "needs to run just a bit longer". So the algorithm would be useless without prior knowledge of the proof size. It still appears that the ramifications are being over-hyped.
Dec 2, 2010 at 0:37 history answered David Harris CC BY-SA 2.5