I have a primitive $A$ which is impossible to prove under some hardness assumption in a black-box way. Now, if another primitive $B$ is stronger than $A$ - in other words $B$ implies $A$ - will it imply that $B$ is also impossible to prove under the same hardness assumption in a black-box way?

  • 1
    $\begingroup$ If B implies A in a black box way, then yes, this just follows immediately from what you stated and the transitivity of black box reductions. $\endgroup$ – Mikero Aug 10 '18 at 1:38

This seems like a logic question: $B$ implies $A$, therefore not $A$ implies not $B$

Proving $B$ would prove $A$, so if it is impossible to prove $A$ it must be impossible to prove $B$ in the same setup.

Alternatively lets assume $B$ is proven, since $B$ implies $A$, $A$ is proven. Yet $A$ is unproveable: a contradiction.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.