If G is a bipartite Euler and Hamiltonian graph, prove that complement of G, G is not Eulers.As G is a bipartite graph, it has two sets X and Y. Using the condition G is Hamiltonian, then |X|=|Y|. As G is also Eulerian, stands dG(upsilon) is even AA upsilon inV(G), where V(G) is set of vertices of the graph G.

If G is a bipartite Euler and Hamiltonian graph, prove that complement of G, $\stackrel{¯<!--\; ¯\; -->}{G}$ is not Eulers.
I would like to know if my proof of the statement in the title is correct.
So, I started like this:
As G is a bipartite graph, it has two sets X and Y. Using the condition G is Hamiltonian, then |X|=|Y|.
As G is also Eulerian, stands $d_G(v)$ is even $\mathrm{\forall <!--\; \forall \; -->}v\in <!--\; \in \; -->V(G)$, where V(G) is set of vertices of the graph G.
Now, let's look at some vertex $v\in <!--\; \in \; -->X(G)$. As stated above, it's degree is even. Let's look at two possible cases:
If |X|=|Y| is even too, then in $\stackrel{¯<!--\; ¯\; -->}{G}$, v will be connected with all the remaining vertices from X. That vertex v will also be connected with remaining vertices from Y, with which it wasn't connected in the graph G. That is, $d_{\stackrel{¯<!--\; ¯\; -->}{G}}(v)=|X|-<!--\; -\; -->1+m$, where m represents number of remaining vertices from Y. As |X| is even, then |X|−1 is odd, and m is also even, because $d_{\stackrel{¯<!--\; ¯\; -->}{G}}(v)$ is even and |Y| is even, so the remaining number of vertices, m is even too. Sum of an even and an odd number is odd, so $d_{\stackrel{¯<!--\; ¯\; -->}{G}}(v)$ is odd. That means $\stackrel{¯<!--\; ¯\; -->}{G}$ is not Eulerian;
If $|X|=|Y|$ is odd, in $\stackrel{¯<!--\; ¯\; -->}{G}$, v will be connected with all the remaining vertices from X and all the remaining vertices from Y, and let's call the number of Y remaining vertices m. As $|Y|$ is odd and $d_G(v)$ is even, then m is odd. Degree of v in $\stackrel{¯<!--\; ¯\; -->}{G}$ is once again $d_{\stackrel{¯<!--\; ¯\; -->}{G}}(v)=|X|-<!--\; -\; -->1+m$, but this time $|X|-<!--\; -\; -->1$is even, and m is odd. $d_{\stackrel{¯<!--\; ¯\; -->}{G}}(v)$ is odd and that means $\stackrel{¯<!--\; ¯\; -->}{G}$ is not Eulerian.
Thank You!