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Let $\mathcal{C}$ be a basic complexity class (such as $\mathrm{NP}, \mathrm{PSPACE}$). And $\mathcal{C}$ is closed under a reduction "$\leq$" (such as polynomial time many-one reduction "$\leq_{m}^{p}$", polynomial time Turing reduction "$\leq_{T}^{p}$").

Suppose that a decision problem $L \subseteq \Sigma^{*}$ is complete for $\mathcal{C}$ under reduction "$\leq$". Let $[L]^{\leq}$ denote the closure of $L$ under reduction "$\leq$", which means that $[L]^{\leq}$ is a class consists all the languages that can be reduced to $L$. $$[L]^{\leq} = \{ L' \in \Sigma^{*}| L' \leq L \}$$

Does $\mathcal{C} = [L]^{\leq}$ hold true?

Here is my proof. Firstly, $L \in \mathcal{C}$, thus for every $L' \in [L]^{\leq}$, we have $L' \in \mathcal{C}$ which leads that $[L]^{\leq} \subseteq \mathcal{C}$.

In the other hand, since $L$ is complete, for every $L' \in \mathcal{C}$, $L' \leq L$. We conclude that $L' \in [L]^{\leq}$ which means that $\mathcal{C} \subseteq [L]^{\leq}$.

Now, I have proved that $\mathcal{C} = [L]^{\leq}$.

If this is correct, I can say that $$[\text{SAT}]^{\leq_{m}^{p}} = [\text{3-SAT}]^{\leq_{m}^{p}} = \mathrm{NP}$$ Am I right?

Blanco
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  • What are $\leq^p_m$ and $\leq^p_T$? – Pål GD Nov 18 '20 at 10:54
  • The answer depends on the complexity class. Some complexity classes are actually defined this way, a notable example being LogCFL. – Yuval Filmus Nov 18 '20 at 11:28
  • @PålGD They are polynomial time many-one reduction and polynomial time Turing reduction. – Blanco Nov 18 '20 at 11:49
  • @YuvalFilmus Yes, even if the original definition does not say in this way, I want to know whether they are equilent. E.g. $[\text{SAT}]^{\leq_{m}^{p}} = [\text{SAT}]^{\leq_{T}^{p}} = [\text{3-SAT}]^{\leq_{T}^{p}} = \mathrm{NP}$ – Blanco Nov 18 '20 at 11:54
  • @TeamBright Please do not delete your question after you have received useful replies. We want the questions and answers to not just help you, but also others who have similar questions. Additionally, others may find it easier to answer your questions that are similar to this one if they know you've asked something similar earlier. – Discrete lizard Nov 22 '20 at 10:20
  • @Discretelizard Sorry for that, I asked the same question on the TCS site, so I have to delete this question. The question asked on the TCS site is closed later, and I have no idea how to undelete the original question. I had no intention of offending all the readers and respondents. – Blanco Nov 22 '20 at 14:55
  • @TeamBright You should ask a question on only one site at the same time. If you violate that rule, we will usually close (this is something different than deletion) the question on our site. An advantage of closing questions is that any replies that have been made are still visible. So, when you have received useful replies, please do not delete the question, but flag the question for moderator attention, so we can close it. If the question has been closed on TCS, then the question here can be open, so we do not need to do anything now. – Discrete lizard Nov 22 '20 at 15:20

1 Answers1

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If $X$ is any NP-complete decision problem, then NP consists of all decision problems which are polytime many-one reducible to $X$.

We cannot replace polytime many-one reductions with polytime Turing reductions, however (unless NP=coNP), since coSAT is polytime Turing reducible to SAT.

Yuval Filmus
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  • Can I say: If $X$ is any $\mathrm{NP}$-complete decision problem, then $\mathrm{NP}$ consists and only consists of all decision problems which are polytime many-one reducible to $X$ – Blanco Nov 18 '20 at 13:35
  • You can say it if you can prove it using your favorite definition of NP. – Yuval Filmus Nov 18 '20 at 15:00
  • That's easy, I think. Since $\mathrm{NP}$ is closed under the polytime many-one reduction, every decision problem that is reducible to $X$ is in $\mathrm{NP}$. – Blanco Nov 19 '20 at 02:00