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 $r$

  • $\begingroup$ What is your actual question? What have you begun and where have you gone wrong? We're not going to answer the problem for you, but we will help if you show what you've tried (or even just thought about trying). $\endgroup$ – Graham Kemp Nov 2 '16 at 8:13
  • $\begingroup$ It's typed up on my computer which is having Internet issues right now, I'm using my phone and 4g. I'll add a picture, but where I ended is a contradiction to the original question. $\endgroup$ – MyNameIs Nov 2 '16 at 8:16
  • $\begingroup$ Unfortunately I can't post the picture unless I have 10 rep... Anyways I bluetoothed the file over, and that's where I'm stuck. $\endgroup$ – MyNameIs Nov 2 '16 at 8:26

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....

  • $\begingroup$ Thanks, I'm still a bit new to discrete math and learning the rules is proving to be difficult, Implication equivalence seems to be what I was missing, thank you again for the help! $\endgroup$ – MyNameIs Nov 2 '16 at 9:04

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 $A\to B, B\to C\vdash A\to C$.

  • $\begingroup$ I think I may have over thought this. I've proven s, but I'm not sure what law to state when proposing that (not)r implies q $\endgroup$ – MyNameIs Nov 2 '16 at 8:30
  • $\begingroup$ See the "further hints" :) $\endgroup$ – BrianO Nov 2 '16 at 19:15

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