Contradiction proof for inequality of P and NP? I'm trying to argue that N is not equal NP using hierarchy theorems. This is my argument, but when I showed it to our teacher and after deduction, he said that this is problematic where I can't find a compelling reason to accept.


  We start off by assuming that $P=NP$. Then it yields that $\mathit{SAT} \in P$ which itself then follows that $\mathit{SAT} \in TIME(n^k)$. As stands, we are able to do reduce every language in $NP$ to $\mathit{SAT}$. Therefore, $NP \subseteq TIME(n^k)$. On the contrary, the time hierarchy theorem states that there should be a language $A \in TIME(n^{k+1})$,  that's not in $TIME(n^k)$. This would lead us to conclude that $A$ is in $P$, while not in $NP$, which is a contradiction to our first assumption. So, we came to the conclusion that $P \neq NP$. 


Is there something wrong with my proof?
Then it yields that $SAT \in P$ which itself then follows that $SAT \in TIME(n^k)$.


Sure.


  As stands, we are able to do reduce every language in $NP$ to $SAT$. Therefore, $NP \subseteq TIME(n^k)$.


No. Polynomial time reductions aren't free. We can say it takes $O(n^{r(L)})$ time to reduce language $L$ to $SAT$, where $r(L)$ is the  exponent in the polynomial time reduction used. This is where your argument falls apart. There is no finite $k$ such that for all $L \in NP$ we have $r(L) &lt; k$. At least this does not follow from $P = NP$ and would be a much stronger statement.

And this stronger statement does indeed conflict with the time hierarchy theorem, which tells us that $P$ can not collapse into $TIME(n^k)$, let alone all of $NP$.