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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?

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    $\begingroup$ Please, write something like $\mathit{SAT}$ instead of $SAT$. As Leslie Lamport wrote in his original LaTeX book, the latter stands for S times A times T. $\endgroup$ – Oliphaunt Apr 25 at 22:31
  • $\begingroup$ Better yet, use the complexity package and simply write \SAT. (I guess that's not available on this stack, though.) $\endgroup$ – Oliphaunt Apr 25 at 22:39
  • $\begingroup$ @Oliphaunt Why not suggest an edit when you can improve the post? Although I must say that here the difference (if any) is a lot more subtle than I'd expect. $\endgroup$ – Discrete lizard Apr 26 at 7:29
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    $\begingroup$ @Discretelizard I often do, but it was "too much work" this time (i was / am on mobile). Entering all those $ and \ is finicky work. I chose to educate instead. (This decision may not have been entirely rational.) $\endgroup$ – Oliphaunt Apr 26 at 11:38
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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 $NP$.

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    $\begingroup$ It's not only the time for the reduction itself. You could reduce to a make larger problem. If I can solve X in O (n^5), and I can reduce a problem in Y in O (n^6) to a O(n^3) sized instance of X, then I need O (n^15) in total. $\endgroup$ – gnasher729 Apr 27 at 11:21
  • $\begingroup$ Amusingly, this argument applies to PTIME-complete problems as well, e.g. HORNSAT, which is solvable in linear time (but not all problems in P are linear time). $\endgroup$ – cody Apr 30 at 19:40
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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 $k''>k'$.

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