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I'm starting a course in proofs and our first two chapters have been on set theory and logic, both of which I have understood to an extent. However, one of my homework questions asks me the following and I'm not quite sure what I'm being asked to do.

Suppose the statement $((P\land Q)\lor R)\Rightarrow(R\lor S)$ is false. Find the truth values of said variables.

In English, I believe the above claims that to say that "if either P and Q or R, then either R or S" is false. Am I being asked to find which combinations of P, Q, R, and S satisfy the above statement, or something else entirely?

I apologize in advance for resorting to asking a homework question on MSE but it's the only one I haven't finished and I'd like to know how to approach a question like this if it were to be on a quiz tomorrow.

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    Your translation in English is correct. Now, is it possible for $R$ to be true and the implication be false? – 2017-01-20
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    You need to assign truth values to $P,Q,R,S$ such that the statement $((P\land Q)\lor R)\Rightarrow(R\lor S)$ is false. In fact there is only one possible assignment. – 2017-01-20
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    Oh okay, so the implication's falsity holds if R is false, and the rest is gravy? – 2017-01-20
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    Yes, $R$ must be false, but, as @EclipseSun suggests, there is only one choice for each variable. What happens now if $P$ is false? – 2017-01-20

2 Answers 2

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Don't worry, I struggled with this myself and Math SE has offered me good solutions, so I'll give the same help to you.

$((P \wedge Q) \vee R) \Longrightarrow (R \vee S)$ is false when $((P \wedge Q) \vee R)$ is true and $(R \vee S)$ is false. And $((P \wedge Q) \vee R)$ is true when P and Q are true and when R is either true or false. So if the whole statement is false, then $(R \vee S)$ is false, and so $R$ must be false and $S$ is false as well.

and you get

   T                  F

$((P \wedge Q) \vee R) \Longrightarrow (R \vee S)$

The whole statement is false

Because:

\begin{array}{C|C} p & q & p \Longrightarrow q \\ \hline T & T & T\\ T & F & F\\ F & T & T\\ F & F & T \end{array}

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    Note that you can use `\land` and `\lor` to get "$\land$" and "$\lor$" in LaTeX. – 2017-01-20
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    Thank you, and thank you for the edit, yep I noticed that, but I don't know why for me \vee and \wedge stick more lol – 2017-01-20
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Since you're doing proofs:

$((P \land Q) \lor R) \to (R \lor S) \equiv$

$(R \lor S) \lor \lnot ((P \land Q) \lor R) \equiv$

$(R \lor S) \lor (\lnot (P \land Q) \land \lnot R) \equiv$

$(R \lor S) \lor ((\lnot P \lor \lnot Q) \land \lnot R) \equiv$

$(R \lor S \lor \lnot P \lor \lnot Q) \land (R \lor S \lor \lnot R) \equiv$

$(R \lor S \lor \lnot P \lor \lnot Q) \equiv$

$\lnot (\lnot R \land \lnot S \land P \land Q)$

Is false precisely when $\lnot R$, $\lnot S$, $P$ and $Q$. I find that there is frequently an easily-interpretable form for most logic statements somewhere along the way to converting to either CNF or DNF.