[RiderOfGiraffes]

I believe you to be mistaken. Let me explain.

Proving that any single NPC problem is in P will be enough to prove that every NP problem is in P, and not just the NPC ones. Suppose A is is NP, B is in NPC, and further suppose that solving B is polynomial. Reduce A to B (a polynomial operation because B is in NPC), solve B (a polynomial operation by assumption), and convert the solution back to a solution of A (a polynomial operation because B is in NPC), and that gives a polynomial solution of A.

Proving that any single NPC problem is not in P is enough to prove that all NPC problems are not in P. It would will leave in question those problems in NP, but not NPC. Currently it is generally believed, but not known, that factoring is such a problem.

Added in edit, to reply to your edit:

    RoG> It would will leave in question those problems
    RoG> in NP, but not NPC. Currently it is generally 
    RoG> believed, but not known, that factoring is such
    RoG> a problem.

    dejb> This is exactly my point. P = NPC doesn't imply
    dejb> anything about factorisation. Since factorisation
    dejb> is in NP we haven't shown that P = NP

Read my entire comment again. In particular, the second paragraph. In particular:

    Proving that any single NPC problem is in P will be
    enough to prove that every NP problem is in P.

More specifically, proving 3-SAT is in P will prove that every NP problem, not just the NPC problems, but every NP problem, is in P. Including factoring.

Let me add a little more.

   1. Suppose 3-SAT is in P.
   2. Let A be any problem in NP.
   3. 3-SAT is in NPC.
   4. Therefore A can be converted to A', an instance of 3-SAT.
   5. By (1), A', as an instance of 3-SAT, can be solved in polynomial time
   6. Hence A has been solved in polynomial time.

Replace "A" with factoring, and thus we've shown that P=NPC implies that factorisation is in P.

[djtumolo]

Can you supply a link to the proof that all NPC problems are equivalent? I remember learning it, but I can't remember how it worked.

[rincewind]

This in the definition of NPC. What you have to prove is that a language is in NPC.

This was proved for SAT first: http://en.wikipedia.org/wiki/Cook%E2%80%93Levin_theorem

http://en.wikipedia.org/wiki/NP_complete

[RiderOfGiraffes]

By definition, a problem P is NPC if:

a. P is in NP, and

b. Every problem in NP can be converted to an instance of P.

Consider any two problems in NPC, P and Q. Then each is in NP, and each can be converted to the other. They are therefore equivalent.

QED.