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In my lecture we learned:

If X is NPC and X in Co-NP => NP = Co-NP

Would it be enough to prove NP = Co-NP if I reduce a problem just in one direction (from NPC to Co-NPC) in polynomial time?

Or would it be necessary to show this for both directions?

Edmundo Del Gusto
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2 Answers2

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In short

yes

Proof

Let's assume $X$ is NP-complete and $X$ is in co-NP.

We show that $NP \subseteq coNP$ and viceversa.

[$NP\subseteq coNP$]

Because $X$ is NP-complete $=>$ for each $L\in NP$ we can found a polytime function $f$ that $s\in L$ iff $f(s)\in X$.

But $X$ is in coNP $=>$ for the polityme reduction closure of coNP, $L\in coNP$ too $=>$ $NP\subseteq coNP$

[ $coNP \subseteq NP$ ]

let $B\in coNP$ $=>$ $B^c \in NP$ $=>$ $s\in B^c$iff $f(s)\in X$ $=>$ $s\in B$ iff $f(s) \notin X$ $=>$ $s\in B$ iff $f(s) \in X^c$

but because $X \in coNP$ $=> X^c \in NP$ and for polityme reduction closure of $NP$ $=> B \in NP$ $=>coNP\subseteq NP$

So $NP=coNP$ under this assumptions.

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user89276
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Ok I think I got it know.

If I could reduce a NP-complete problem to a coNP-complete problem every problem in NP can be reduced to every coNP-complete problem and it would show that NP is a subset of coNP.

Because every problem A in coNP has a complement co_A in NP

A in coNP => co_A in NP => co_A in coNP => A in NP

=> every Problem in coNP is in NP

=> NP = coNP

If there is a NPC problem that can be reduced to coNPC => NP = coNP

Edit: I wrote this while the other answer was given

Edmundo Del Gusto
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