How to prove that $P \neq NP$

there is some confusion in the question showing a lack of some basic understanding of the theory of NP completeness which is well documented in many places. particularly in this question:

Is there any problem that been proven to be in NP?

yes all NP complete problems are in NP. there are hundreds, maybe thousands of NP complete problems and minor variants. see eg Garey/Johnson, Computers & Intractability, theory of NP completeness

How to prove that P≠NP?

obviously to anyone with even cursory knowledge in the area, this is the Big Open Question that has eluded researchers for over four decades, the short answer is that nobody knows even after serious/intense effort by worlds leading experts. there are many ideas. a few leading ones:

  • circuit theory/complexity. prove that boolean circuits that compute NP complete problems must be "inherently large". some success [ie lower bounds] has been achieved with simpler models eg monotone circuits.
  • GCT is a newer idea using geometry ideas.
  • descriptive complexity. related to the power of proofs. Fagins theorem shows a correspondence with P sets and a specific proof system.
  • etc! many others!

What would be if I be able to prove that one of the NP-Complete problems cannot be solved in polynomial time.

I assume you mean "problems cannot be solved in polynomial time on a deterministic Turing machine". NP, after all, stands for "nondeterministic polynomial", and includes the decision problems that can be solved in polynomial time on a nondeterministic Turing machine.

What would be if I be able to prove that one of the NP-Complete problems cannot be solved in polynomial time [on a deterministic Turing machine].

Then you would prove that all the $\rm NP\text{-}complete$ problems have no deterministic polynomial-time algorithms.

Proof by contradiction:

  1. Imagine we have some algorithm, $A$, can solve some $\rm NP\text{-}complete$ problem, $S$.
  2. Let us call your "one of the $\rm NP\text{-}complete$ problems" $R$.
  3. Accordingly, $R$ cannot be solved in polynomial time (deterministically); because you have some given proof.
  4. Since they are both $\rm NP\text{-}complete$ problems, then by Cook reductions, we can reduce $R\stackrel{...}{\rightarrow} S$ in polynomial time (deterministically).
  5. Then we can use $A(S)$ and solve $S$. This is a contradiction to $(3)$.

Is there any problem that been proven to be in NP and not in P?

I am not sure what you are actually trying to ask here, but here is a summary on the complexity classes.

$\rm \mathbf P$: is the class of decision problems that can can be solved in polynomial time (deterministically).

There are several equivalent ways of looking at $\rm NP$.

$\rm \mathbf{NP}$:

  • is the class of decision problems that can be solved in polynomial time on a nondeterministic Turing machine.
  • is the class of decision problems for which one can easily verify the solutions of their corresponding optimization problems (which are also proofs of the decision).

The two explanations above are intuitively equivalent: The non-determinism of a NTM can use the verifiability of the answer to "try all solutions" and only keep the one that successfully verifies.

$\rm {NP}$ obviously includes all of $\rm P$.

$\rm \mathbf{NP\text{-}complete}$: is the class of decision problems that are the hardest problems in $\rm NP$. These problems have reductions each-other. Indeed, all of $\rm P$ and $\rm NP$ can be reduced to any $\rm NP\text{-}complete$ problem.

$\rm\mathbf{P}$ vs $\rm\mathbf{NP}$: is the question about whether the $\rm NP\text{-}complete$ class of problems can be solved as fast as the problems in $\rm P$; in other words, does nondeterminism make a machine "faster" (in the sense of polynomial-time=fast, superpolynomial-time=slow).

$\rm\mathbf{NP\text{-}hard}$: are the set of decision problems (some definitions include search and optimization problems) that are at least as hard as $\rm NP\text{-}complete$ problems. So, all $\rm NP\text{-}complete$ are also $\rm NP\text{-}hard$ problems.

A nice picture from wikipedia:

enter image description here

So back to your question:

Is there any problem that been proven to be in NP and not in P?

EDIT: Just to be clear, I understood this to mean:

Is there any problem that been proven to be in NP and [proven] not [to be] in P?

The answer is NO, because that would imply $\rm P\ne NP$. Proof by contradiction:

  1. Imagine $\rm P=NP$
  2. Imagine there is a problem, $R$ that is in $\rm NP$, but not in $\rm P$.
  3. $\rm NP\text{-}complete$ problems are the hardest in $\rm NP$ (all problems in NP can indeed be reduced to any $\rm NP\text{-}complete$ problem).
  4. All $\rm NP\text{-}complete$ problems must be at least as hard as $R$, which follows from $(2,3)$.
  5. All $\rm NP\text{-}complete$ problems are not in $\rm P$, as $R$ is not in $\rm P$, and $\rm NP\text{-}complete$ problems hare harder than $R$ ( from $(2,4)$ ). Therefore $\rm NP \not\in P$, thus $\rm P\ne NP$. This contradicts $(1)$.

If you find a general algorithm that solves a single NP-complete problem in polynomial time, then you will have effectively proved that $P=NP$.

Beware that solving one instance of a specific problem is not enough, you have to give an algorithm which solves any instances of the problem with any input data, and prove that the algorithm is polynomial in the size of the input.

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Np Complete

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