A question concerning the hypothetical syllogism.

Question

I am working on a proof and have found that given the statement if $P(x)$, then $\lnot Q(x)$ is true. Suppose that I do not know the truth value of if $\lnot Q(x)$, then $R(x)$. Taking the conjunction of these two statements allows for the conclusion if $P(x)$, then $R(x)$ by hypothetical syllogism. Now, assuming that it is known that if $P(x)$, then $R(x)$ is always false when $P(x)$ is true, and that a conjunction can only be true when both statements are true, does it follow that if $\lnot Q(x)$, then $R(x)$ is always false whenever $\lnot Q(x)$ is true? I am assuming that it is, but I want to makes sure there is not something I am missing where there is a case where if $\lnot Q(x)$ could be true, but also $R(x)$ could be true.

Answer 1

I am working on a proof and have found that given the statement if $P(x)$, then $\lnot Q(x)$ is true. Suppose that I do not know the truth value of if $\lnot Q(x)$, then $R(x)$. Taking the conjunction of these two statements allows for the conclusion if $P(x)$, then $R(x)$ by hypothetical syllogism.

That inference is valid.

$$P(x)\to\neg Q(x)~,~ \neg Q(x)\to R(x)~\vdash~ P(x)\to R(x)$$

Now, assuming that it is known that if $P(x)$, then $R(x)$ is always false when $P(x)$ is true, and that a conjunction can only be true when both statements are true, does it follow that if $\lnot Q(x)$, then $R(x)$ is always false whenever $\lnot Q(x)$ is true?

No, that is not entailed.

$$P(x)\to\neg(P(x)\to R(x)),P(x)\to\neg Q(x)~,~ \neg Q(x)\to R(x)~\nvDash \neg Q(x)\to\neg(\neg Q(x)\to R(x))$$

I am assuming that it is, but I want to makes sure there is not something I am missing where there is a case where if $\lnot Q(x)$ could be true, but also $R(x)$ could be true.

Yes, suppose $\neg P(x)$, $\neg Q(x)$, and $R(x)$. Well, all three premises are satisfied by this model, but the conclusion $\neg Q(x)\to\neg(\neg Q(x)\to R(x))$ is not satisfied. So the conclusion is not entailed by the premises.