Why ZFC is placed in top of the proof system hierarchy? How it can p-simulate other systems?
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2$\begingroup$ I really would appreciate some definition links to "proof system hierarchy" and "p-simulation". $\endgroup$– Hans-Peter StrickerCommented Jul 23, 2011 at 15:18
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2$\begingroup$ @Hans Stricker, check the Wikipedia pages for "proof complexity" (en.wikipedia.org/wiki/Proof_complexity) and "propositional proof system" (en.wikipedia.org/wiki/Propositional_proof_system). $\endgroup$– KavehCommented Aug 2, 2011 at 6:19
1 Answer
It is an open problem if there is an optimal propositional proof system. Therefore we don't know if ZFC as a propositional proof system is optimal either.
ZFC as propositional proof system can p-simulate any propositional proof system whose soundness (if there is proof for a formula then the formula is true) as a propositional proof system is provable in ZFC. The trick (which I think is due to Steve Cook) is based on the fact that proofs are concrete finite objects, if there is a proof for a formula in a proof system, then ZFC can prove its existence (i.e. ZFC is $\Sigma_1$-complete), combining this with the provability of the soundness we derive the truth of the encoded formula in ZFC. The rest of the argument is translating this first order proof in ZFC to a propositional proof in ZFC as propositional proof system and proving the equivalence of a propositional formula with the formula itself in the proof system. This can be done in $\mathsf{TC^0}$-Frege and any system that contains it. See Logical Foundations of Proof Complexity, 2010 by Cook and Nguyen for the details.
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1$\begingroup$ Strictly speaking, it's the function $F$ that maps ZFC-proofs of $T$ to $T$ (where $T$ is a tautology) that is the propositional proof system, not ZFC itself. Note also that the assertion that $F$ is a propositional proof system isn't provable in ZFC itself; you need to assume the soundness of ZFC. This is why people put ZFC at the top; conceivably, there could be "stronger" propositional proof systems $F$ that ZFC can't prove are actually propositional proof systems. But if you think that ZFC represents the outer limit of mathematical knowledge, you might have qualms about such $F$. $\endgroup$ Commented Jul 22, 2011 at 22:24
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$\begingroup$ @Timothy, there are two equivalent definitions for a pps, one is what you described, the other one is a proof checker program (which the one I found more natural). Yes, strictly speaking, but I didn't want to go into too much details. One can add axioms to obtains a possibly stronger pps, e.g. large cardinal axioms. Note that we don't need the full soundness, we only need a very restricted form of it. In fact, as you know :), some proof complexity theorist conjecture that EF is a plausible candidate for the optimal proof system, in which case ZFC will be just equal to it. $\endgroup$– KavehCommented Aug 2, 2011 at 6:13