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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 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 $NP$.

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 $NP$.

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 takes $O(n^{r(L)})$ time to reduce language $L$ to $SAT$, where $r$ 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 $NP$.

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 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 $NP$.

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 takes $O(n^{r(L)})$ time to reduce language $L$ to $SAT$, where $r$ 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.

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 takes $O(n^{r(L)})$ time to reduce language $L$ to $SAT$, where $r$ 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 $NP$.