Negating a predicate with a biconditional Negate the following statement and make the negation appear immediately before the predicates
$$ \forall x \forall y(Q(x,y) \leftrightarrow Q(y,x)) $$
I did the following steps:
$$ \neg(\forall x \forall y(Q(x,y) \leftrightarrow Q(y,x))) $$
$$ \exists x  \neg \forall y(Q(x,y) \leftrightarrow Q(y,x)) $$
$$ \exists x   \exists y \neg(Q(x,y) \leftrightarrow Q(y,x)) $$
Now, I do not know what to do simply because of the notation. I assumed the following, knowing that:
$$ \neg(a \leftrightarrow b) \equiv \neg a \leftrightarrow b $$ 
Would it be enough/correct to leave the following as the answer? 
$$ \exists x   \exists y (\neg Q(x,y) \leftrightarrow Q(y,x)) $$
 A: First, let's note a couple of equivalences:
$$\begin{align}
p\leftrightarrow q &\equiv (p\land q)\lor (\neg p \land \neg q) \tag{1} \\
&\equiv (p\to q)\land (q \to p). \tag{2}
\end{align}
$$
The negation of $p\leftrightarrow q$ is therefore
$$\begin{align}
\neg(p\leftrightarrow q) &\equiv \neg((p\to q)\land (q\to p)) \tag{by (2)} \\
&\equiv \neg(p\to q)\lor \neg(q \to p) \tag{De Morgan} \\
&\equiv (p\land \neg q)\lor (q \land \neg p) \tag{$\neg(A\to B)\equiv (A\land \neg B)$} \\
&\equiv p\leftrightarrow \neg q \tag{by (1)} \\
&\equiv \neg p\leftrightarrow q. \tag{by (1)} \\
\end{align}$$
Now recall that $\neg\forall \equiv \exists\neg$. Thus the negation of $\forall x \forall y\,(Q(x,y)\leftrightarrow Q(y,x))$ is
$$\begin{align}
\neg \forall x \forall y\,(Q(x,y)\leftrightarrow Q(y,x)) &\equiv \exists x \exists y\,\neg\,(Q(x,y)\leftrightarrow Q(y,x)) \\
&\equiv \exists x \exists y\,(\neg\, Q(x,y)\leftrightarrow Q(y,x)) \\
&\equiv \exists x \exists y\,(Q(x,y)\leftrightarrow \neg\, Q(y,x)).
\end{align}$$
