# How do I solve this discrete math/quantifier problem?

So I have to "Use equivalences to rewrite the propositions into a logically equivalent proposition in which quantifiers only appear directly in front of predicate symbols (only one quantifier can appear directly in front of the predicate symbol". Here is the problem: $(\forall x)(P(x) \land (Q(b) \lor R(x))$. I have to solve several problems like this one, but I just don't understand where to start. I know we can't distribute a universal quantifier over disjunction, it can only distribute over injunction. So, I would assume that I can use the logical identities like the negation laws and De Morgan's Laws, is that correct? If someone could just drop some hints or point me in the right way, I would greatly appreciate it.

• "injunction" ??? Maybe conjunction. Sep 8 '17 at 5:56
• You can see here (page 3 and 5) the laws for quantifiers; but you have to take care of free variables... Sep 8 '17 at 6:00
• In a nutshell: $\forall$ distribute over $\land$, i.e. $\forall x \ (Px \land Qx) \equiv (\forall x \ Px \land \forall x \ Qx)$ [$\forall$ can be seen as an infinite conjunction] while $\exists$ distribute over $\lor$, i.e. $\exists x \ (Px \lor Qx) \equiv (\exists x \ Px \lor \exists x \ Qx)$ [$\exists$ can be seen as an infinite disjunction]. Sep 8 '17 at 6:07

The baisc rules are the following.

$∀$ distribute over $∧$, i.e.

$∀x \ (Px ∧ Qx) ≡ (∀x Px ∧ ∀x Qx)$,

and $∃$ distribute over $∨$, i.e.

$∃x \ (Px ∨ Qx) ≡ (∃x Px ∨ ∃x Qx)$.

$∀x \ (Px ∨ A) ≡ (∀x Px ∨ A)$, only if $x$ is not free in $A$,
and the same restriction hold for $∃$ with $∧$.
$(∀x) \ [P(x) ∧ (Q(b) ∨ R(x)] ≡ (∀x)P(x) ∧ (Q(b) ∨ (∀x)R(x))$,
because we can "safely" distribute $∀$ over $∧$ and in this case we can also distribute it over $∨$, because $x$ is not free in $Q(b)$.