TITLE: $\exists x Px \land \exists x Qx$ does not imply $\exists x (P x \land Q x)$
QUESTION [5 upvotes]: I am pretty confused by this. 
We know that $\phi : = \exists x Px \land \exists x Qx $ 
    does not imply  $\psi : = \exists x (P x \land Q x)$,
    as for the model $M$ with domain $\{0,1\}$ with $P := \{0\}$ and $Q := \{1\}$,
    we have that $M \models \phi$, $M \not \models \psi$. But, seemingly:
\begin{eqnarray}
 \exists x Px \land \exists x Qx \implies & \lnot ( \lnot ( \exists x Px \land \exists x Qx))\\
             \implies   & \lnot ( \lnot \exists x Px \lor \lnot \exists x Qx)\\
             \implies  &  \lnot (\forall x \lnot Px \lor \forall x \lnot Qx))\\
             \implies  &  \lnot (\forall x (\lnot Px \lor \lnot Qx ))\\
             \implies  &  \exists x \lnot ( \lnot P x \lor \lnot Q x) \\
             \implies  &  \exists x (P x \land Q x) ,            
 \end{eqnarray}
where $(3) \implies (4)$ by the schema 
    $\forall x A x \lor \forall x B x \implies \forall x (A x \lor B x)$;
    (2) $\implies (3), (4) \implies (5)$ by quantifier/negation relations; and 
    $(1) \implies (2), (5) \implies (6)$ by De Morgan's laws. 
    What went wrong here? Thanks!

REPLY [3 votes]: $$
\lnot (\forall x \lnot Px \lor \forall x \lnot Qx)) \Rightarrow \lnot (\forall x (\lnot Px \lor \lnot Qx ))\\$$
is not correct.
It is equivalent to
$$
\varphi ~ =: ~\forall x \lnot Px \lor \forall x \lnot Qx \Leftarrow \forall x (\lnot Px \lor \lnot Qx ) ~ := \psi\\$$
Consider your examplary $\{P,Q\}$-structure $A$ over the universe $\{0,1\}$ with $P=\{0\}$ and $Q=\{1\}$.
Now $A \models \psi$ but $A \not \models \varphi$.
This the schema 
    $\forall x A x \lor \forall x B x \Rightarrow \forall x (A x \lor B x)$ is correct but you have been using it the otherway around which is generally not a correct implication.