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Suppose the logical statement $P \rightarrow \lnot Q$ is true whose inverse's truth value is unknown. Also, suppose that $\lnot R \rightarrow \lnot P$ is true and $\lnot R \rightarrow \lnot Q$ is true, with $\lnot R \rightarrow Q$ false. The inverse of $P \rightarrow \lnot Q$ is given by $\lnot P \rightarrow Q$. Not knowing the truth value of the inverse, it follows that $\lnot P \rightarrow Q$ is only true if the conjunction $(\lnot R \rightarrow \lnot P) \land (\lnot P \rightarrow Q)$ is true, since $\lnot R \rightarrow \lnot P$ is true. By hypothetical syllogism the false statement $\lnot R \rightarrow Q$ is produced. Therefore, we can conclude that $\lnot P \rightarrow Q$ is false. Is this a correct approach to proving this statement is false?

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First of all: yes, your argument appears valid.

Second: note that this doesn't really have anything to do with the "original" implication $P\rightarrow\lnot Q$, and that your argument doesn't use the assumption $\lnot R\rightarrow\lnot Q$. Really, you are just starting with the two assumptions $\lnot R\rightarrow\lnot P$ and $\lnot(\lnot R\rightarrow Q)$ and trying to deduce the statement $\lnot(\lnot P\rightarrow Q)$.

Equivalently, since the negation of $A\rightarrow B$ is $A\land(\lnot B)$, we are trying to start with the two statements $\lnot R\rightarrow\lnot P$ and $(\lnot R)\land(\lnot Q)$ and deduce the statement $(\lnot P)\land(\lnot Q)$, that is, to deduce the two statements $\lnot P$ and $\lnot Q$.

This is actually pretty straightforward:

  • The assumption $(\lnot R)\land(\lnot Q)$ implies $\lnot R$, which together with the assumption $\lnot R\rightarrow\lnot P$ implies $\lnot P$;
  • and the assumption $(\lnot R)\land(\lnot Q)$ directly implies $\lnot Q$.
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  • $\begingroup$ That makes sense, and it is clear that is the easier way. My main concern was the use of the hypothetical syllogism and assuming that since it was known that one statement was true, the conjunction is "only" true if the other statement was true. Since the hypothetical syllogism produced a false statement, one could conclude that the other was false. It sounds like that is a valid way, but just longer. $\endgroup$ – jmath Mar 13 '20 at 0:12
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I am very confused by your argument. First: what exactly are the givens? At first it seems like $P \to \neg Q$ is a given .... but you never use it ... In fact, you say that $\neg P \to Q$ is its inverse, and that seems to be the statement you ar trying to prove false. So .... are you trying to show that $\neg P \to Q$ is false on the basis of the truth of $P \to \neg Q$? In particlar, are you trying to show that $\neg P \to Q$ is False by showing that $P \to \neg Q$ is True? If so, please understand that the inverse of a statement can have the same truth-value as the statement itself, so that approach does not work. Firtunately, youm don't seem to reason that way ... indeed, ypu seem to be assuming that $P \to \neg Q$ is True .. but why? What is you your purpose for assuming this? And note ... you never really use it. So in the end, it's all just very, very confusing what you are trying to do, and how you are trying to go about it.

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  • $\begingroup$ All of the true and false statements at the beginning are givens. My point was to use the conjunction of one statement known to be true and the other in question. Since the hypothetical syllogism is false, I concluded that the statement in question must be false. Since the conjunction could only be true if the statement in question were true. $\endgroup$ – jmath Mar 13 '20 at 0:16

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