Q27

$(\forall _{x} p(x) \implies \forall _{x} q(x)) \implies (\exists _{x} \neg p(x) \vee \forall _{x} q(x))$

$(\exists x p(x) \vee \exists x q (x)) \implies \exists x (p(x) \vee q (x))$

$\exists x (p(x) \wedge q(x)) \implies (\exists x p(x) \wedge \exists x q(x))$

$\forall x (p(x) \vee q(x)) \implies (\forall x p(x) \vee \forall x q(x))$