Find the mistake that for all $A$ and $B$, $$A=\emptyset\land B=\emptyset\leftrightarrow A\times B=\emptyset.$$
The "proof":
$$ \begin{align*} A=\emptyset\land B=\emptyset&\leftrightarrow\neg(\exists x\,x\in A)\land\neg(\exists y\,y\in B)\tag{1}\\ &\leftrightarrow\forall x\,\neg(x\in A)\land\forall y\,\neg(y\in B)\tag{2}\\ &\leftrightarrow\forall x\,x\notin A\land\forall y\,y\notin B\tag{3}\\ &\leftrightarrow\forall(x,y)\,(x,y)\notin A\times B\tag{4}\\ &\leftrightarrow\neg(\exists(x,y)\,(x,y)\in A\times B)\tag{5}\\ &\leftrightarrow A\times B=\emptyset\tag{6} \end{align*} $$
I think the mistake is on $(3)\leftrightarrow(4)$, but I am not able to justify why it is a mistake. Could you help me?