# Question about details for a proof for “if a coNP-hard problem is in NP (NP-hard problem is in coNP), then NP=coNP” [closed]

I searched the internet for a proof of the statement: If a $$coNP$$-hard problem is in $$NP$$, then $$NP=coNP$$ (NP-hard problem is in coNP). I found a proof sketch here (page 3 in the pdf, slide 341) and I wanted to make sure I understand it correctly. The proof-sketch gives a proof for one direction (if a $$coNP$$-hard problem is in $$NP$$, then $$coNP \subseteq NP$$), but leaves the other direction ($$NP \subseteq coNP$$) to the reader.

Here is the proof for both directions:

$$coNP \subseteq NP$$ (taken from the slides)

• Let $$L \in NP$$ be coNP-hard.
• Let NTM $$M$$ decide $$L$$.
• Since $$L$$ coNP-hard, for any $$L' \in coNP$$ there is a reduction $$R$$ from $$L'$$ to $$L$$.
• But then $$L' \in NP$$ as it is decided by NTM $$M(R(x))$$.
• Thus, $$coNP \subseteq NP$$

Now my attempt at the other direction ($$NP \subseteq coNP$$):

• Let $$L' \in NP$$ and its complement $$\bar{L'} \in coNP$$.
• Thus there is a reduction from $$\bar{L'}$$ to $$L$$.
• Thus $$\bar{L'} \in NP$$, as it is decided by NTM $$M(R(x))$$.
• But then $$L' \in coNP$$
• Hence, $$NP \subseteq coNP$$

Is this correct?

For the statement that if an $$NP$$-hard problem is in $$coNP$$, then $$NP=coNP$$, the slides just state that the proof works similarly. However, I am struggling to work out the details of the proof. In particular, how can I leverage the reduction to choose that a language which is NP/coNP is also in coNP/NP respectively?

## closed as unclear what you're asking by vonbrand, Evil, David Richerby, Yuval Filmus, Discrete lizard♦Sep 12 at 17:43

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