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?

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    $\begingroup$ 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.


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