Re: Is this ℙ≠ℕℙ proof 'humiliating'?

Liste des GroupesRevenir à theory 
Sujet : Re: Is this ℙ≠ℕℙ proof 'humiliating'?
De : ben (at) *nospam* bsb.me.uk (Ben Bacarisse)
Groupes : comp.theory
Date : 10. Jun 2024, 21:45:58
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <87ikyg34s9.fsf@bsb.me.uk>
References : 1 2 3 4 5
User-Agent : Gnus/5.13 (Gnus v5.13)
wij <wyniijj5@gmail.com> writes:

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. 

Correct, though uselessness is not a property that is provable.

If p∉ℕℙℂ, then ℙ=ℕℙ will be a contradiction (leads to p∈ℕℙℂ), so ℙ≠ℕℙ
is true in this case.

No.  If p∉ℕℙℂ, then ℙ=ℕℙ is false, i.e. P≠ℕℙ.  There is no
contradiction.  You cannot conclude that this leads to p∈ℕℙℂ.  That's
why in your "less formal" argument you just stated it as an unproven
fact.

Summary: Because ℕℙℂ is considered not useless, therefore ℙ≠ℕℙ is
concluded.

ℕℙℂ is interesting only because it's that part of ℕℙ that might not be
in ℙ.  Once the question is settled, it stops being interesting.

--
Ben.

Date Sujet#  Auteur
8 Jun 24 * Is this ℙ≠ℕℙ proof 'humiliating'?17wij
8 Jun 24 +- Re: Is this ℙ≠ℕℙ proof 'humiliating'?1wij
9 Jun 24 `* Re: Is this ℙ≠ℕℙ proof 'humiliating'?15Ben Bacarisse
9 Jun 24  `* Re: Is this ℙ≠ℕℙ proof 'humiliating'?14wij
10 Jun 24   +* Re: Is this ℙ≠ℕℙ proof 'humiliating'?4Andy Walker
10 Jun 24   i`* Re: Is this ℙ≠ℕℙ proof 'humiliating'?3wij
10 Jun 24   i `* Re: Is this ℙ≠ℕℙ proof 'humiliating'?2Andy Walker
10 Jun 24   i  `- Re: Is this ℙ≠ℕℙ proof 'humiliating'?1wij
10 Jun 24   `* Re: Is this ℙ≠ℕℙ proof 'humiliating'?9Ben Bacarisse
10 Jun 24    +* Re: Is this ℙ≠ℕℙ proof 'humiliating'?6wij
10 Jun 24    i`* Re: Is this ℙ≠ℕℙ proof 'humiliating'?5Ben Bacarisse
11 Jun 24    i `* Re: Is this ℙ≠ℕℙ proof 'humiliating'?4wij
11 Jun 24    i  +* Re: Is this ℙ≠ℕℙ proof 'humiliating'?2Jeff Barnett
11 Jun 24    i  i`- Re: Is this ℙ≠ℕℙ proof 'humiliating'?1wij
11 Jun 24    i  `- Re: Is this ℙ≠ℕℙ proof 'humiliating'?1Ben Bacarisse
10 Jun 24    `* Re: Is this ℙ≠ℕℙ proof 'humiliating'?2wij
10 Jun 24     `- Re: Is this ℙ≠ℕℙ proof 'humiliating'?1Ben Bacarisse

Haut de la page

Les messages affichés proviennent d'usenet.

NewsPortal