(∀x(P(x)⇒Q(x)) ⇒ ∀u(R(u)⇒S(u)))

The outline for this proof is simple:

Assume ∀x(P(x)⇒Q(x)). Prove ∀u(R(u)⇒S(u)), by assuming R(u) is true.

The actual proof is as follows:

Supposed f is a fuction R → R and let g(x) = f(2x).

Prove that is f is increasing, then g is also increasing.

Assume f: R → R and g(x) = f(2x). Assume f is increasing. This means for all x, f^I > 0. By the chain rule, the derivative of f(2x) is 2f^I(2x), and 2f^I(2x) > 0. This means f(2x) is also a strictly increasing function. Since we have already assumed f(2x) = g(x), and we have shown f(2x) is an increasing function, we have also shown that g(x) is increasing.