I came up with the following. Am I doing something wrong?Suppose $P=NP$. Let $A$ be an NP hard problem. Let $A'$ be the polynomial reduction of $A$. By the assumtions, the Halting problem holts for $A$. Let $T'$ be the running time of $A'$. Clearly, $T' \in O(P(n))$, with $n$ as input size for $A'$ and $P(n)$ representing the class of polynomial functions of $n$. By that, the running time of $A'$ is bounded above and since it is a polynomial reduction of $A$, the running time of $A$ should also be bounded above. By that, there exists an algorithm, namely $X():=$return $True;$ which is a solution to the Halting problem for $A$. By that, the Halting problem does not hold for $A$, which is a contradiction. So the assumption that $P=NP$ can not hold and therefore $P \neq NP$.
A proof that P != NP
algorithms;halting problem
Let $A$ be an NP hard problemThe halting problem, among others, is $NP$-hard. Did you mean to pick a $NP$-complete problem instead?Let $A'$ be the polynomial reduction of $A$.What is the polynomial reduction, where is it defined? Reduction to/from what? Do you mean that $A'$ is a polynomial algorithm that solves $A$? (Which would exist by virtue of your $P=NP$ assumption and $A$ being in $NP$.)the Halting problem holts for $A$What does it mean for the Halting problem to holt (hold?) for $A$? The halting problem isn't parameterized on any particular program.the running time of $A$ should also be bounded above$A$ does not have a running time. It is a problem, not an algorithm.There exists an algorithm [...] is a solution to the Halting problem for $A$For any program $P$ there is an algorithm that computes whether or not $P$ halts. Namely, the algorithm that trivially returns one of true/false. The Halting Problem doesn't state that there can be no algorithm that computes whether a particular program halts, but rather that there is no algorithm that is correct for all programs.