Does (∀x(r → p)) ∧ (∀x(p → q)) imply ∀x(r → q)? So the frustrating thing is that I am asked to decide whether the first implies the second, but I was given no deduction rules for predicate logic except the fact that $ ∀x(r(x) → p(x)) $ implies $ ∀x(r(x)) → ∀x(p(x)) $.
I came to the conclusion that, maybe I could reason on a meta-level. What I mean is that, if  $ (∀x(r(x) → p(x))) ∧ (∀x(p(x) → q(x))) $ does imply $ ∀x(r(x) → q(x))$, then I would have to show it for possibly an infinite number of variables $x$. But to show that $ (∀x(r(x) → p(x))) ∧ (∀x(p(x) → q(x))) $ doesn't imply $ ∀x(r(x) → q(x))$, I would only need to find one counter example. Since I was not given enough deduction rules, there is no way I could show it is true for all $x$, so my guess then is that the first does not imply the second and I'm assuming the book wants me to find a counter example.
However, I looked at the truth table for $ (r → p) ∧ (p → q) $ and this does imply $ r → q $. So which is right? Was my meta-reasoning incorrect, or is it possible that an implication in propositional logic may not have the same implication for the analogous statement in predicate logic?
 A: First, I'll assume that $p$, $q$, and $r$ are formulas that contain $x$ as a free variable. Because if not, then these statements are simply equivalent to $r \to p$, $p \to q$, and $r \to q$, and obviously the first two imply the third.
So, let's write these statements as $\forall x ( r(x) \to p(x))$, $\forall x ( p(x) \to q(x))$, and $\forall x ( r(x) \to q(x))$ respectively.
Now, $\forall x ( r(x) \to p(x))$ is not equivalent to $\forall x\ r(x) \to \forall x \ p(x)$. For example, take as the domain all real numbers, let $r(x)$ be '$x$ is a rational number', and $p(x)$ be '$x$ is a prime number'. Then $\forall x\ r(x)$ is obviously false, and hence $\forall x\ r(x) \to \forall x \ p(x)$ is true, but $\forall x ( r(x) \to p(x))$ is false.
OK, but do $\forall x ( r(x) \to p(x))$ and $\forall x ( p(x) \to q(x))$ imply$\forall x ( r(x) \to q(x))$?  Intuitively, yes, of course: the first premise states that every '$r$' is a '$p$', and the second that every '$p$' is a '$q$'. So, if you take any '$r$', then by the first permise that has to be a '$p$', and by the second premise, it therefore has to be a '$q$'. Hence, every '$r$' is a '$q$', which is exactly what the conclusion is saying. So, this is a valid inference.
