Reducing from NPC to Co-NPC => NP = Co-NP?
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
$=&gt;$ 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
$=&gt;$ for the polityme reduction closure of coNP, $L\in coNP$ too
$=&gt;$ $NP\subseteq coNP$
[ $coNP \subseteq NP$ ]
let $B\in coNP$
$=&gt;$ $B^c \in NP$
$=&gt;$ $s\in B^c$iff $f(s)\in X$
$=&gt;$ $s\in B$ iff $f(s) \notin X$
$=&gt;$ $s\in B$ iff $f(s) \in X^c$
but because $X \in coNP$
$=&gt; X^c \in NP$ and for polityme reduction closure of $NP$
$=&gt; B \in NP$
$=&gt;coNP\subseteq NP$
So $NP=coNP$ under this assumptions.