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Sep 30, 2023 at 20:55 comment added Henry Yuen No, it does not, and necessarily so :) The reasoning for this comes from computability theory: the class $MIP^*$ is contained in $RE$, which is known not to contain $coRE$ (the class for which determining which Turing machines do not halt is complete for).
Sep 30, 2023 at 16:57 comment added Scott McKuen If the claim is "M does not halt", does the interactive proof mechanism also give the provers a way to convince the verifier of that?
Sep 30, 2023 at 14:06 history answered Henry Yuen CC BY-SA 4.0