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I don't understand how to show the following: (!Q -> P) ∧ !P -> Q

I understand the answer is true as I did it with a truth table but how can I prove this using propositional logic?

Thanks!

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  • $\begingroup$ Welcome to Math StackExchange! In order to best help with your question, we need to know where you're getting stuck. Do you know how to prove things in propositional logic at all? If so, what have you tried here, and why doesn't it work? $\endgroup$ – user231101 Sep 26 '15 at 21:52
  • $\begingroup$ "(!Q -> P) and !P -> Q": what does this mean? Does this mean that "(!Q -> P)" is equivalent with "!P -> Q"? $\endgroup$ – zoli Sep 26 '15 at 21:55
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If you want to show that $$!Q \rightarrow P \ \equiv \ !P \ \rightarrow Q$$ then use the following argumentation:

In general $A\rightarrow B$ can be expressed as $!A\lor B$. Accordingly $$!Q \rightarrow P \equiv \ !!Q\lor P\ \equiv Q\lor P$$ and $$!P \rightarrow Q \equiv \ !!P\lor Q \ \equiv P \lor Q=Q\lor P.$$

EDITED

If you meant

$$[(!Q\rightarrow P)\land !P]\rightarrow Q \tag 1$$

then use again that in general $A\rightarrow B$ can be expressed as $!A\lor B$.

Regarding $(1)$

$$(!Q\rightarrow P)\land !P \equiv !!Q\lor P\land !P\equiv Q\lor P\land !P\equiv Q.$$

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  • $\begingroup$ Sorry, I meant (!Q -> P) ∧ !P -> Q $\endgroup$ – forgetaboutme Sep 27 '15 at 2:56
  • $\begingroup$ So, you meant the $[(!Q\rightarrow P)\land !P]\rightarrow Q$ ? $\endgroup$ – zoli Sep 27 '15 at 8:18
  • $\begingroup$ I've edited my answer accordingly. $\endgroup$ – zoli Sep 27 '15 at 8:29

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