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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)
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    @Derek Elkins I think I got it. Ill edit – 2017-02-13
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    For 3. : Modus ponens from what ? – 2017-02-13
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    @MauroALLEGRANZA from 1 and 2. Since $(p\rightarrow (q\rightarrow \bot))\equiv q\rightarrow (p\rightarrow \bot)$ – 2017-02-13
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    Yes, but what are the rules you are allowed to use ? If you can use $(p→(q→⊥))≡q→(p→⊥)$ why not also $(p∧q)≡(q∧p)$ ? – 2017-02-13
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    @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. – 2017-02-13

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