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How would i solve the following.

Use the following premises to show the conclusion is t.

$p\vee q$

$q-r$

$p\wedge s-t$

$\neg R$

$\neg Q-U \wedge S$

$-$ for if then in this question.

I did the following

$q-r$

$\neg r$

$\neg q$ modus tollens

$p \vee q$

$\neg q$

$p$ elimination

$\neg q-u\wedge s$

$\neg q$

$u\wedge s$ modus ponens

$u \wedge s$

$s$ specialization

$p$

$s$

$p\wedge s$ conjuction

$p \wedge s-t$

$p \wedge s$

$t$ modus ponens.

But would my reasoning be correct .

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  • $\begingroup$ - would mean if then ... $\endgroup$ – Fernando Martinez Sep 28 '13 at 19:45
  • $\begingroup$ You can format the implication (if-then connective) by using \rightarrow = $\rightarrow$ $\endgroup$ – Namaste Sep 28 '13 at 19:50
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UPDATED POST: All is now good.


Initial post:

But where did you get the premise $\lnot q$? You'd need to also pursue what happens when $\lnot p$ so from $p\lor q$, we have $\lnot p$ which gives us $q$. If you had as a premise $\lnot q$ (and forgot to add that in your post), then your proof would be fine. I'm referring to your conclusion immediately following: "I did the following..."

I'd suggest starting from the premises $$q\rightarrow r$$ $$\lnot r$$ $$\therefore \lnot q$$, by modus tollens, as you note.

NOW we can use $\lnot q$: $\quad p \lor q$ and $\lnot q$ give us $p$, (by disjunctive syllogism, or $OR$ elimination.) as desired.

So the order or the first two parts of the proof needs to be switched, then all will be good.

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  • $\begingroup$ Yes, I have not q when I did q-r then not r therefore q but in my post i put it out of order, but I think then the conclusion is ok $\endgroup$ – Fernando Martinez Sep 28 '13 at 19:59
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    $\begingroup$ Yes, if you swap the order, then all is good! $\endgroup$ – Namaste Sep 28 '13 at 20:01
  • $\begingroup$ I see that makes sense $\endgroup$ – Fernando Martinez Sep 28 '13 at 20:04
  • $\begingroup$ Yes, I like these sorts of questions, @Amzoti! $\endgroup$ – Namaste Sep 29 '13 at 1:26

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