My first upper division logic class.

Use the deduction theorem to give a syntactic proof

$p\wedge q\vdash q\wedge p$

$\{(p\rightarrow (q\rightarrow \bot))\rightarrow \bot, q\rightarrow (p\rightarrow \bot)\}\vdash \bot$ from the deduction thm.

  1. $(p\rightarrow (q\rightarrow \bot))\rightarrow \bot$ (premise)
  2. $q\rightarrow (p\rightarrow \bot)$ (premise)
  3. $\bot$ (Modus ponens)
  • $\begingroup$ @Derek Elkins I think I got it. Ill edit $\endgroup$ – HiPolyEraser Feb 13 '17 at 6:08
  • $\begingroup$ For 3. : Modus ponens from what ? $\endgroup$ – Mauro ALLEGRANZA Feb 13 '17 at 7:02
  • $\begingroup$ @MauroALLEGRANZA from 1 and 2. Since $(p\rightarrow (q\rightarrow \bot))\equiv q\rightarrow (p\rightarrow \bot)$ $\endgroup$ – HiPolyEraser Feb 13 '17 at 7:14
  • $\begingroup$ Yes, but what are the rules you are allowed to use ? If you can use $(p→(q→⊥))≡q→(p→⊥)$ why not also $(p∧q)≡(q∧p)$ ? $\endgroup$ – Mauro ALLEGRANZA Feb 13 '17 at 8:03
  • $\begingroup$ @MauroALLEGRANZA I'm not sure. We jump around between 3 different languages. One of which has only $\bot$ and $\rightarrow$. So I've been following those examples. But he didn't state it in the question. $\endgroup$ – HiPolyEraser Feb 13 '17 at 17:56

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