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Sorry, I am just preparing some notes for my students and want to double check I have my facts right before I give the notes to them. So these are my premises:

  1. $\lnot p\rightarrow o$
  2. $s\rightarrow r$
  3. $\lnot (o\land r)$
  4. $\lnot p$

And I want : $\lnot s$

So, casually, we can get from $\lnot p$ to $o$ using Modus Ponens, proving $o$.

$\lnot(o\land r)$ is the same as $\lnot o\lor\lnot r$.

Using simplification we can assert from this that ¬r is true.

From this we can get to $\lnot s$ using Modus tollens.

Is this correct ? I am tired from writing ~40 pages of notes and am nearly positive I have missed something here.

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    what language is this?2011-08-10
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    Dave's question seems to be tongue-in-cheek, but there is a serious questions there. You could be using any of several axiomatics. What is an axiom in one is a theorem in the other.2011-08-10
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    I'm so happy I'm not in that class and also that I'm clueless about where to find all those characters on my keyboard. Shouldn't this question be better off at math.stackechange.com ?2011-08-10
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    you meant $(\neg o \vee \neg r)$2011-08-11
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    40 pages to get to this point?2011-08-11
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    @Asaf Karagila: Why did you reintroduce the typo $\neg o\wedge\neg r$ for $\neg o\vee\neg r$?2011-08-11
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    @Christian: I assume the correct answer is "by mistake"... :-) I have corrected that. Thanks!2011-08-11

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