Lets have an arbitrary complexity class $C$.Does $C$-complete = co-$C$-complete imply that $C$ = co-$C$?I think that the answer is yes, but I am not sure whether my reasoning is correct. I tried to prove it like this:Let $L$ be a $C$-complete language, so all problems in $C$ are polynomially many-to-one reducible to it. If $L \in C$ and $\bar{L} \in C$, then $C-complete = co-C-complete$ (because all complements of problems in co-$C$ are reducible to both $L$ and $\bar{L}$). That implies that $C = co-C$ so $C$ is closed under complement. Is this a correct approach?
Does $C$-complete = co-$C$-complete imply that $C$ = co-$C$?
complexity theory