Recall the following definition of $NP-$completeness.

A problem $x$ is $NP-$complete if and only if:

$x$ is in NP$,$ and

$x$ is NP$-$hard. $($Recall that $"$NP$-$hard" here means: any problem in NP is poly$-$time reducible to $x.$ I.e$.,$ for all YinNP,$Y{?}_{P}x.)$

$($a$)(5$ points$)$ Fill in the blanks below. Let $A$ be a known NP$-$complete problem, which means it is both in NP and it is NP$-$hard.

If we reduce $A$ to some problem $B$ in poly$-$time, in other words, if we show $q,{?}_{P}q,$ then it proves $q,$ is also NP$-$hard.

This technique suffices to show $q,$ is NP$-$hard, because: if $A$ is known to be NP$-$complete, then by part $[\frac{1}{2}($circle one$)$ of the above definition of NP$-$complete, we know that for all YinNP, $--{?}_{P}A$

If we then show $A{?}_{P}B,$ then by "transitivity" of poly$-$time reductions, we get $Y{?}_{P}B$ for all Yin $q,$ This means $q,$ is also NP$-$hard, by definition of NP$-$hard.

Poly$-$time reductions are "transitive" because running a poly$-$time algorithm followed by another poly$-$time algorithm is still altogether $q,$

$q,Pq,$

$($b$)(5$ points$)$ Once we show problem $B$ is NP$-$complete, it means we know it is at least as hard as any other problem in NP$.$ In other words, a fast algorithm for $B$ could be used to solve any problem in NP efficiently. But it also means problem $B$ can be easily solved by a fast algorithm for any other known NP$-$complete problem, like $A.$

Using only the two parts of the above definition of NP$-$complete, and nothing else, explain $($concisely$)$ why: if $B$ is NP$-$complete, then a fast algorithm for any other known NP$-$complete problem, like $A,$ could be used to solve $B.$