Inference Proof with Quantifiers I am trying to figure out this implication proof. Can any of you guys tell me how to prove this?
Prove ∀x((¬P(x) ∧ Q(x))  → R(x)) Implies ∀x(¬R(x) → P(x))

 A: $\forall x \left(\left(\neg P\left(x\right)\wedge Q\left(x\right)\right)\rightarrow R\left(x\right)\right)$
$\Leftrightarrow\forall x \left(R\left(x\right)\vee\left(P\left(x\right)\vee \neg Q\left(x\right)\right)\right)$
and
$\forall x \neg R\left(x\right)\rightarrow P\left(x\right)$
$\Leftrightarrow \forall x \left(R\left(x\right)\vee P\left(x\right)\right)$
this means without further information about $P$ and $Q$ your claim is true if and only if for any reasons $\forall x Q\left(x\right)$.
A: Your problem is :

prove that $\forall x ((\lnot P(x) \land Q(x)) \rightarrow R(x))$ implies $\forall x(\lnot R(x) \rightarrow P(x))$.

Taking benefit of the transformation of the first formula suggested by @Max in the first version of his answer, your problem becomes : "prove that $\forall x (\lnot R(x) \rightarrow (P(x) \lor \lnot Q(x)))$ ..."
So, the first formula says that, "for every $x$, if $R$ does not hold for $x$, then $P$ or not-$Q$ holds for it".
But the second formula assert that, under the same condition (i.e. when $R$ does not holds), $P$ holds.
Now, think about a box full of balls half of which are white and half black: every ball is black-or white, but it is not true that all balls are black.
Now, form the above "informal" argument, we move on to a more formal one; in order to show that $A \nvDash B$, it is sufficient to find an interpretation where $A$ is true and B is false.
We can use an interpretation with domain $D = \{ a \}$ such that : $P, Q$ and $R$ are predicates standing for "attributes" of the objects of the domain such that nor $P$, nor $Q$ nor $R$ hold for $a$; i.e., we have that :

$P(a), Q(a)$ and $R(a)$ are all false.

Now see what happens with this interpretation :

$(\lnot P(a) \land Q(a)) \rightarrow R(a))$ evaluates to : (not-false and false) $\rightarrow$ false , that is : false $\rightarrow$ false, that is true;
$(\lnot R(x) \rightarrow P(x))$, instead, evaluates to : not-false $\rightarrow$ false , that is : true $\rightarrow$ false, that is false.

In conclusion, we have found an interpretation where the premise is true and the conclusion is false : this it's enough to prove that the first does not imply the second.
