How to prove that $P \neq NP$?
To prove that $P = NP$ all we need to do is to solve one NP-Complete problem in polynomial time for any input, and because all the NP-Complete problems have reduction from one to each other we can say $P = NP$.
What would be if I be able to prove that one of the NP-Complete problems cannot be solved in polynomial time. Is there any problem that been proven to be in NP and not in P?
I did not proved for any problem that it belongs only to NP and not to P, I am just asking in general.
Edit: Some one stated:
If you could prove that there existed an NP-Complete problem that
cannot be solved in P, then it would imply that $P \neq NP$
I want to comment about it here. If you have problem $A$ that could be solved by polynomial reduction to problem $B$, you can say that $A$ is not harder than $B$. If you be able to solve $B$ in polynomial time, it will also apply to $A$, how ever it's the lower bounds. If you be able to prove that $B$ cannot be solved in polynomial time, it doe's not prove that $A$ cannot be solved in polynomial time. A good example for that is 2-sat problem that could be reduced to 3-sat. 2-sat has polynomial solution, we did not found any polynomial solution for 3-sat yet.
Best Answer
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.
Then you would prove that all the $\rm NP\text{-}complete$ problems have no deterministic polynomial-time algorithms.
Proof by contradiction:
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 $\rm NP$.
$\rm \mathbf{NP}$:
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:
So back to your question:
EDIT: Just to be clear, I understood this to mean:
The answer is NO, because that would imply $\rm P\ne NP$. Proof by contradiction: