Please i need help with this long equation:
$\neg\forall x\neg(\neg p(x) \lor q(x)) \rightarrow \neg\exists x(\neg p(x) \land \neg q(x)) \models \forall x(p(x) \lor q(x))$.
I am solving with resolution method. Below is what i have done, and was stock.
$\neg\forall x(p(x) \lor q(x))$
$\neg\forall x\neg(\neg p(x) \lor q(x))$
$\forall x(\neg p(x) \land \neg q(x))$
$\neg p(a) \land \neg q(a)$
$\neg\forall x(p(x) \lor q(x))\lor \neg\exists x(\neg p(x) \land \neg q(x))$
$\exists x\neg(\neg p(x) \lor q(x)) \lor \neg\exists x(\neg p(x) \land \neg q(x))$
$\exists x(p(x) \land \neg q(x)) \lor \neg\exists x(\neg p(x) \land \neg q(x))$
$\exists x\exists y((p(x)\land \neg q(x)) \lor p(y) \land q(y)$
First I'm going to change a bit the left part ($p\rightarrow q\equiv \neg p \vee q$ and De Morgan's rule). So we have to show $$\forall x (p(x)\wedge q(x))\vee\forall x(p(x)\vee q(x))\models \forall x (p(x)\vee q(x)).\quad \quad \quad(1)$$ Using the resolution method, we have to negate the right part and arrive to a contradiction. So, negating the right part we have $$\exists x \neg( p(x)\vee q(x)).$$ With this we know there is an element $a$ such that $$\neg (p(a)\vee q(a)).\quad\quad\quad (2)$$ Now using the left part of $(1)$ we know that for $x=a$ this stands (for both quantifiers), so we have $$(p(a)\wedge \neg q(a))\vee p(a)\vee q(a),$$ which you can show that is equivalent to $$p(a)\vee q(a).$$ Combining the last equation with $(2)$ we get that $$p(a)\vee q(a)\wedge \neg(p(a)\vee q(a)),$$ which is a contradiction.