Choose a random oracle $f : \{0,1\}^\ast \to \{0,1\}$, and define the logic $ZFC^f$ by adding a fresh symbol $g$, an axiom that $g$ has the correct type, and one axiom $g(s) = f(s)$ for each $s \in \{0,1\}^\ast$. This is a countably infinite set of axioms since $f$ is not a symbol in $ZFC^f$.

Within $ZFC^f$ we can define $g$-oracle Turing machines in the standard way, for example by using a second tape for the input to $g$ and letting the machine branch on the value of $g$ on the second tape. We then have the complexity classes $P^g$, $NP^g$, etc. Since $f$ was chosen at random in the metalogic, $P^g \ne NP^g$ is true with probability 1.

Question 1: Is $P^g = NP^g$ independent over $ZFC^f$?

The proof seems immediate: any finite proof in $ZFC^f$ can interrogate at most finitely many values of $g$. However, I am not confident the definitions are sensible.

Question 2: Is there a good reference exploring this type of logic + random oracle construction?

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    $\begingroup$ This will not do what you presumably intend. Any proof may include only finitely many axioms, hence it may only use information on a finite part of the oracle. In particular, it makes no difference whatsoever if you choose the oracle randomly or otherwise. Since interesting properties are invariant under a finite change in the oracle anyway, you may as well omit the $g(s)=\dots$ axioms entirely. $\endgroup$ Mar 5, 2018 at 8:08
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    $\begingroup$ For what its worth, relativization by means of adding an extra symbol in the language (with no specific axioms) is used extensibly in bounded arithmetic. $\endgroup$ Mar 5, 2018 at 8:11
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    $\begingroup$ For example, $P^g=BPP^g$ is still independent of $ZFC^g$, even though for a random oracle $f$, $P^f=BPP^f$ holds with probability 1. $\endgroup$ Mar 5, 2018 at 10:06

1 Answer 1


Yes on Question 1 (assuming ZFC is consistent). You don't need $f$ to be random exactly, any $f$ will do. And for the proof you need to also use the fact that there is an oracle $h$ with NP$^h=$P$^h$.


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