Logic (philosophy)

Question

Let P(x), Q(x), and R(x) be open sentences containing the variable x. Prove that if the statements for all x (P(x) implies Q(x)) and for at least x (∃x) (Q(x) implies R(x)) are true, then for all x (P(x) implies R(x)) is true. How can I do this without using truth tables?

Answer 1

Please notice two differences between what you have written here and the question:

The question says $R(y)$, not $R(x)$.

And also, it has $(\exists x(Q(x)) \implies R(y)$, not $\exists x (Q(x) \implies R(y))$.

You can translate into "natural deduction" (which is a lousy name, because it's a very artificial way of thinking), but the reason is this:

For any $x$, either $P(x)$ is true, in which case $Q(x)$ is true, and therefore $\exists x Q(x)$ is true, so $R(y)$ is true, and therefore $P(x) \implies R(y)$ is true, or $P(x)$ is false, so $P(x) \implies R(y)$ is true. In either case $R(y)$ is true. So $\forall x(P(x) \implies R(y))$ is true.