About a ℙ≠ℕℙ proof (v3.2)

[wij]

Basic proof (using C-like notation):
Define: bool S(Prog,Arg): S(c,n)==true iff ∃x, c(x) returns true in P-time.

S is a language interpreter. Prog and Arg are pass-by-value arguments. E.g.
scripts, TM description, CNF evaluator,... and their possible solutions for
Prog to check.

From definition of S, S computes a problem in ℕℙ.
If Problem(S)∈ℙ, there exists a P-time Prog f defined as follow:
bool f(Arg x) {
return !S(f,x);
}

From Liar's paradox and the HP proof, f is an undecidable instance for S. And,
f and S cannot both be in the same P-time convertible set. Therefore, ℙ≠ℕℙ.

Note: The definition of S (and ℕℙ) only requires the certifying instance
"c(x)==true" run in P-time. Complexity of c itself is not an issue, c can even
never terminate. These 'timed-out' cases should return false, but the result of
the S(f,n) case always contradicts the definition of S.
QED.

Prog f exists in various forms, shown as a function specification to make the
basic proof succinct by avoiding lots of symbol/definition/axiom/lemma...coding issues stuff.
If the above is correct, then, we change to use 'executable' to demonstrate,
because 'executable' is easier to understand and VERIFIABLE by real TM.

// The folowing proves S2,f2 both exist in the language of a real TM.
Executable: S2 <prog> <uint>
Definition: The exit status of "S2 <prog> <uint>" is 1 iff there exists an
"p x" instance, where x<=n, and the exit status of "p x" is 1.

// From the specification, S2 can be built from such a C program (error checks
// are omitted).
//
int main(int argc, char* argv[]) { // main() takes exactly two arguments
const char* FuncName= argv[1]; // argv[1] is <prog>
const UInt MaxN= atoi(argv[2]); // convert argv[2] to UInt

for(Uint x=0; x<=MaxN; ++n) {
const char* cmd= make_cmd(FuncName,x); // make command string for system(..)
if(system(cmd)==1) { // exec: S2(f,x), may be assumed a pseudo-parallel
return 1; // system call to simulate NDTM
}
}
return 0;
}

// As S2 exists, an executable f2 can be built from such a C program (error checks are omitted).
//
int main(int argc, char* argv[]) {
const char* cmd= make_cmd("S2 f2 ", argv[1]); // make command string for system(..)
return !system(cmd); // exec: S2(f,n)
}

// After S2 and f2 are built, test with "S2 f2 8"
[] S2 f2 8 // execute from terminal
...
Aborted (core dumped) // infinite recursive call, expression of undecidability

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I am not familiar with complexity theory. For such a proof to be really
convincing, one needs to explain existing ℕℙℂ examples as a double check and
possibly some others I may be unaware of. So, I should stop here and ask
question for possible flaws of this basic proof.