Contradiction proof for inequality of P and NP?

Question

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?

Answer 1

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 $r(L) < k$. 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 $NP$.

Answer 2

Suppose that $\mathrm{3SAT}\in\mathrm{NTIME}[n^k]$. By the nondeterministic version of the time hierarchy theorem, for any $r$, there is a problem $X_r\in\mathrm{NTIME}[n^r]$ that is not in $\mathrm{NTIME}[n^{r-1}]$. This is an unconditional result that doesn't depend on any kind of assumption such as $\mathrm{P}\neq\mathrm{NP}$

Choose any $r>k$. Suppose we have a deterministic reduction from $X_r$ to $\mathrm{3SAT}$ that runs in time $n^t$. It produces a $\mathrm{3SAT}$ instance of size at most $n^t$, which can be solved in time at most $(n^t)^k=n^{tk}$. By our choice of $X_r$, we must have $tk>r-1$, so $t>(r+1)/k$. This function grows without bound with $r$.

This means that there is no bound on how long it can take to reduce an arbitrary $\mathrm{NP}$ problem to $\mathrm{3SAT}$. Even if $\mathrm{3SAT}\in \mathrm{P}$, there's still no bound on how long those reductions can take. So, in particular, even if $\mathrm{3SAT}\in\mathrm{DTIME}[n^{k'}]$ for some $k'$, we can't conclude that $\mathrm{NP}\subseteq\mathrm{DTIME}[n^{k'}]$, or even $\mathrm{NP}\subseteq\mathrm{DTIME}[n^{k''}]$ for some $k''>k'$.