How to prove two graphs are isomorphic if and only if their complements are isomorphic? How this can be proved that two graphs $G_1$ and $G_2$ are isomorphic iff their complements are isomorphic? 
 A: Let graph $G$ be isomorphic to $H$, and let $\overline G$, $\overline H$ denote their complements.
Since $G$ is isomorphic to $H$, then there exists a bijection $f: V(G) \to V(H)$, such that $uv \in E(G)$ if and only if $f(u)f(v) \in E(H)$. -> [this should be edge set]
Equivalently, there exists a bijection $f: V(G) \to V(H)$, such that $uv \notin E(G)$ if and only if $f(u)f(v) \notin E(H)$.  -> [this should be edge set]
Since the vertex set of $G$ and $\overline G$ are the same, therefore $f$ is a bijection from $V(\overline G)$ to $V(\overline H)$. Then suppose $uv \notin E(G)$, by definition of a complement, $uv \in E(\overline G)$. Likewise, if $f(u)f(v) \notin E(H)$, then $f(u)f(v) \in E(\overline H)$.
Hence $\overline G$ and $\overline H$ are isomorphic.
A: $G_1 \equiv G_2 \implies \exists$ a bijective function $f: V(G_1) \rightarrow V(G_2)$ such that $xy \in E(G_1) \iff f(x)f(y) \in E(G_2)$
Hints:


*

*Note since the vertex set of $G^c$ complement is the vertex set of $G$ itself. So $f$ can actually be considered a bijection from $V({G_1} ^ c ) \rightarrow V({G_2} ^ c )$

*How can the statement: $xy \in E(G_1) \iff f(x)f(y) \in E(G_2)$ be expressed in terms of a negative argument, (say $xy \not \in E(G_1)$)?? 

A: One direction is trivial because isomorphism conquers all. The other follows from $G^{cc}=G$.
