Suppose, for the sake of argument, that it was proved that $P \not= NP$. Then, this would imply that for every $NP$-complete problem, there is a "hardest instance" of the problem that cannot be solved in polynomiall time. Then, if $P$ were to not equal $NP$, what properties would be known about the hardest instances of $NP$-complete problems?
In particular, my question is:
If $H$ is the set of all hardest instances of some $NP$-complete problem (take subset-sum for instance), thereby not solvable in polynomial time, then do we know if $H$ would contain both "yes" and "no" instances of the problem, or only "yes" or only "no" instances? (By "yes" instance I mean the answer to the decision problem is "yes"; i.e., "yes, there is a subset of some integer set that sums to a target value").
My gut tells me the set of all hardest instances of a problem must contain at least one "yes" instance, and not necessarily a "no" instance, but I do not have a proof and am wondering if someone has one.
My general thought for a proof of why there must be "yes" instances in the hardest set is as follows:
Suppose, for the sake of contradiction, that the set of problems that no algorithm could solve in polynomial time are all "no" instances. Then this implies there is an algorithm that can solve all "yes" instances in polynomial time, and hence for parameter $n$ is gauranteed to halt after $n^d$ iterations, where $d$ is a fixed positive integer. But then this implies that if the algorithm does not return "yes" after $n^d$ iterations, then it knows it must be a "no" instance, and hence returns "no" in polynomial time ($n^d$ iterations). But then this is an algorithm solving all instances of the problem in polynomial time, a contradiction.
