I am happy to announce the ℙ≠ℕℙ problem can be succinctly proved.

Executable: Sat <prog>
Definition: The exit status of "Sat <prog>" is 1 iff there exists an instance
<prog> <arg> which runs in P-time and the exit status is 1.

From the definition, Sat computes a ℕℙ problem.
If ℕℙ=ℙ, then, an executable f exists that "Sat f" should run in P-time:

// C source of executable f
#include <stdlib.h>
int main(int, char* []) {
return system("Sat f")? 0:1;
}

Short analysis and explanation:
If Sat(f)==1, from definition of f, f(*)==0. But, from definition of Sat, ∃x,f(x)==1 ... Contradiction
If Sat(f)==0, from definition of f, f(*)==1. But, from definition of Sat, ∀x,f(x)==0 ... Contradiction (see note)

Conclusion: If ℙ=ℕℙ, there exists instances like f that Sat cannot decide.
Therefore, ℙ≠ℕℙ.

note: Exact negation should read: For all <prog> <arg> instance, <prog> <arg>
does not run in P-time or the exit status of <prog> <arg> is not 1.

I am not familiar with complexity theory. Such proof does not look like a
traditional theoretical proof, nonetheless, it is a valid proof. Because the
argument is supported by 'real programs', what can go wrong?

wij <wyniijj5@gmail.com> writes:

> I am happy to announce the ℙ≠ℕℙ problem can be succinctly proved.

Do you plan just keep posting this sort of stuff without engaging with
any one the comments?

If so, I wonder why, since the "proof" is so simple, you don't publish
it in a proper journal. My guess is that you know you have proved
nothing, but there may be some other explanation.

--
Ben.