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By definition:

A polynomial-time computable function $f:$ $\{$0,1$\}$$^*$$\{$0,1$\}$$^*$ is a one-way function if for every probabilistic polynomial time Turing Machine $PTM$ there is a neglegibible function $E$ $:$$\mathbb{N} → \mathbb{N} $ $ \ s.t. \ $ $Pr$ $x$ $∈$ $\{$0,1$\}$$^n$ $[PTM \ \ inverts \ \ f(x)] ≤ E(n)$.

Now, reading the definition above, the existence of one-way functions seems like an easy bet, however it is considered a stronger assumption than $P \neq NP$. Apparently proving the existence of these functions would be more difficult than proving $P = NP ?$ And I'd like to know the technical reasons for this, does it have something to do with the "for every" in the definition?

What steps would a mathematician or computer scientist take to prove that one-way functions exist? I believe that since proving its existence would also prove $P \neq NP$, this means that we cannot use the techniques that we know for sure does not work: Relativization, diagonalization, Natural Proof.

So, what form could such a proof take? Maybe something in second(+)-order logic?

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  • $\begingroup$ I'm not sure how this is a better question than asking what a $\mathsf{P} \neq \mathsf{NP}$ proof would look like. Perhaps the question is better poised as "what form could such a proof not take (in addition to relativization techniques, etc.)?" $\endgroup$
    – dkaeae
    Jul 15, 2019 at 7:37
  • $\begingroup$ This question should scrap all mention of P≠NP (which is separate question) and simply focus on why it's hard to prove one-way functions exist. $\endgroup$
    – Geremia
    Sep 16, 2023 at 16:55

1 Answer 1

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Apparently proving the existence of these functions would be more difficult than proving $P\not =NP$? And I'd like to know the technical reasons for this, does it have something to do with the "for every" in the definition?

The answer is in your question itself:

I believe that since proving its existence would also prove $P\not =NP$

This is indeed the case. Somebody that proves the existence of one-way functions automatically also proves $P\not = NP$. On the other hand, it is conceivable that you could prove $P\not = NP$ without proving that one way functions exist. Thus, proving the existence of one-way functions is at least as hard as proving $P\not = NP$.

What steps would a mathematician or computer scientist take to prove that one-way functions exist?

Coming up with an extremely clever and creative idea that nobody came up with before.

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  • $\begingroup$ "The answer is in your question itself[:] 'I believe that since proving its existence would also prove P≠NP'" You're asked "Why is it difficult to prove the existence of one-way functions?" and you 'answer' "Because I assume P≠NP." $\endgroup$
    – Geremia
    Sep 16, 2023 at 16:53
  • $\begingroup$ @Geremia No, that's not what my answer is. My "assumption" is that settling the P v.s. NP question is hard (since a lot of people have tried and nobody has managed). Now if you believe that settling P v.s. NP (either way) is hard, then proving that one-way functions exist must also be hard: if you managed to do so you'd also prove that P≠NP, but it is not impossible that P≠NP but no one-way functions exist. Therefore proving that one-way functions exist is a "strictly harder" task than proving that P≠NP. $\endgroup$ Sep 17, 2023 at 6:37

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