Logical equivalence and rules of inference question

Question

I need to use the laws of equivalence and rules of inference to show that the statement: "$s\land(r\to\lnot q)$" using the following premises:

1. $(r\lor\lnot t)\to p$
2. $t \to s$
3. $p \to \lnot q$
4. $t$

So far I've worked out

1. From $t$ and $t \to s$, we can infer $s$ from modus ponen.
2. From $p \to \lnot q$ and $(r\lor\lnot t)\to p$ we can infer $(r\lor\lnot t)\to \lnot q$

But I'm not sure how to get from $r\lor\lnot t$ to just $r$

Answer 1

From $t$ and $t→ s$ we get s by modus ponens. From $(r \lor ¬t) → p$ and $p → ¬q$ we can infer that $(r \lor ¬t) → ¬q$ by specialization. By using conditional equivalence on $(r \vee ¬t) → ¬q$, we can say $(¬r \land t) \lor ¬q$. Using Identity on $t$ and $(¬r \land t) \lor ¬q$, we can say $¬r \lor ¬q$. by using demorgan’s law on $¬r \lor ¬q$, we can say $r \land q$.

Everything seems okay until the last step.   That's not what deMorgan's law says, nor is it what you should do.

1. By deMorgan's: $\neg r\lor\neg q~\equiv~\neg(r\land q)$

2. However, you want to use conditional equivalence. $\neg r \vee\neg q~\equiv~r\to\neg q$

3. So you have proven $s$ and proven $r\to \neg q$, then....

Answer 2

Hint: You have to prove two things:

• $s$
• $r\to \neg q$

To prove the first, look at 2. and 4. together; to prove the second, consider 4., 1. and 3.

Further hints: $(r \lor \text{False}) \equiv r$, and $A\to B, B\to C\vdash A\to C$.