Re: Is this ℙ≠ℕℙ proof 'humiliating'?

On Mon, 2024-06-10 at 00:36 +0100, Ben Bacarisse wrote:

wij <wyniijj5@gmail.com> writes:
 

On Sun, 2024-06-09 at 20:55 +0100, Ben Bacarisse wrote:

wij <wyniijj5@gmail.com> writes:
 

ℙ≠ℕℙ
Proved. https://sourceforge.net/projects/cscall/files/MisFiles/PNP-proof-en.txt/download
...[cut]
   Proof2: Let p="Given a number n, determine whether or not n is even". If
          ℙ=ℕℙ, then p∉ℕℙℂ is a false proposition because all ℕℙ problems
          including ℕℙℂ are mutually Ptime reducible. Since p∉ℕℙℂ is true,
          ℙ≠ℕℙ is concluded.

 
Where is your proof that p is not NP-complete?  Since you don't know
this subject very well, you would benefit more from asking people to
direct you to resources from which you could learn, rather than posting
provocative messages.

 
<silly insults deleted>
 

To be on topic, can you show us the p (as mentioned) is NPC or p is
not NPC, either will do, to prove how much you understand what you
talked about.

 
If I could do that I would be rich, quite literally.  Sadly, I can't and
neither can anyone else on the planet (so far).  But if you think you
can, head over to the Clay Mathematics Institute and persuade them to
give you a million dollars[1].
 
For the hard-of-understanding, a proof that p, which is obviously in P,
is also in NPC would immediately prove that P=NP.  Alternatively, a
proof that p is not in NPC would immediately prove that P=/=NP.
 
[1] https://www.claymath.org/millennium/p-vs-np/

Probably I should make the Proof2 more formal:

If p∈ℕℙℂ, then ℙ=ℕℙ and the concept of ℕℙℂ is useless. 
If p∉ℕℙℂ, then ℙ=ℕℙ will be a contradiction (leads to p∈ℕℙℂ), so ℙ≠ℕℙ is true in this case.
Summary: Because ℕℙℂ is considered not useless, therefore ℙ≠ℕℙ is concluded.