Generating set-checking reduces to membership testing

Membership testing problem (co-nondeterministic)
Recall that in the membership testing problem, we are given a subset $$A$$ of $$G$$ and are required to formulate a test that will take in an element of $$G$$ and output whether that element is in the subgroup $$H$$ generated by $$A$$.

The generating set-checking problem has a co-nondeterministic reduction to the membership testing problem in the following sense. If a given subset does not generate the whole group, then we can find a short proof of the fact invoking the membership testing problem. Namely, pick an element $$g \in G$$ that is not in the subgroup generated by the $$A$$, and then prove that it actually is not in the subset by invoking the membership testing problem.

Generating set-finding problem
If we can find another set $$B$$ that is also a generating set for the group, then the problem of checking whether $$A$$ is a generating set can be solved by invoking the membership testing problem a few times and taking the AND of all the outputs. This gives us a polynomial-time positive truth table reduction from generating set-checking to membership testing.

The idea is as follows: Check that every $$b \in B$$ is in the subgroup generated by $$A$$