> A plausible proof of "P != NP" won't be quite as simple to express, since it needs to prove that all such algorithms do not run in polynomial time.
That sounds hard but, If for any NP-Complete problem there exists no P solution then for all NP problems there is no P solution. So this proof sounds like it has the right shape.
Well yes! Two things: You have a successful algorithm that runs in P time that solves an NP Hard problem and 2) you can map other NP Hard problems to your problem. Without the second factor, it is only a demonstration of a "range" in the computational realm in question, where p = np or whatever the declaration. Being able to show that your pizza slice is actually an ocean of pizza, and also show that any other shape of pizza slice can be appropriately transformed into the shape you have means p = np. a "solution." Or perhaps better put, it is a funnel through which complex computational patterns can either be reduced or simplified or elegantly correlated, approaching absolutely perfect parallelization of operations. This is just one way to look at it, but essentially intractability is an interesting term to consider.
Please forgive me if my liberal use of the language is an offense
Your 2 is the easy part: "NP-hard" is exactly the set of problems to which any NP problems can be reduced in polynomial time, and there are many known existing examples, both within NP (aka NP-complete) and outside it.
So part of the solution set satisfies P=NP and some satisfy P!=NP?
No, some NP-hard problems are outside NP (i.e. harder than NP), while some are inside. The wikipedia article has a good explanation and a good diagram: https://en.wikipedia.org/wiki/NP-hardness
A plausible proof of "P != NP" won't be quite as simple to express, since it needs to prove that all such algorithms do not run in polynomial time.