If P = NP, then how not cannot solve NP-hardness (the one that doesn't intersect with NP-complete) in polynomial-time? My question is that if P = NP, then we can solve any NP-hard problems (the one which is NP-complete and the one which is not-NP-complete) by saying that since we have a polynomial time algorithm to solve the NP problems and since the reduction works in polynomial time. Therefore, polynomial to polynomial is polynomial, and so we can solve any NP-hardness problem. For example, we can solve the TSP by calling to an NP-complete problem which have polynomial time since P=NP, and since the reduction is polynomial time, therefore we can solve TSP in polynomial time.


  How could disprove this argument? What is the clue to say that NP-hardness(the one which doesn't intersect with NP-complete) can not be solved in polynomial time even if P=NP.


I know that if P = NP, then we can only solve any problem in NP (but not in NP-hard). But I want to know what it makes NP-hard (the one that have no intersection with NP-complete) is not solvable in polynomial time even if P=NP since we have polynomial time reduction and polynomial time algorithm given P=NP.

Thank you
You have your reductions the wrong way around. A problem&nbsp;$X$ is NP-hard if every problem in&nbsp;NP can be reduced to&nbsp;$X$. That is, saying that $X$&nbsp;is NP-hard means that, if we had an efficient algorithm for&nbsp;$X$, we would have an efficient algorithm for every problem in&nbsp;NP: $X$&nbsp;is easy implies all of&nbsp;NP is easy.

So, suppose that P$\,=\,$NP. Then, all of NP is easy, but that tells us nothing about&nbsp;$X$, because the implication goes the wrong way. Maybe $X$&nbsp;is easy; maybe it's hard. In fact, some NP-hard problems are definitely hard: the halting problem is NP-hard but it's undecidable, regardless of what the relationship between P and&nbsp;NP turns out to be.