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I have to show that :

((p => q) v (~(p ^ ~q) ^ T)) ≡ ~p v q 

I've gotten this far :

((p => q) v (~(p ^ ~q) ^ T)) ≡ ~p v q 
((p => q) v ((~p ^ q) ^ T)) ≡ ~p v q          by DeMorgan & Double Negation
((~p v q) v ((~p ^ q) ^ T)) ≡ ~p v q               by Conditional

I'm stuck on the last line. Am I going down the right pathway so far?

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    $\begingroup$ If I understand the question correctly, when applying De Morgans Law, you did not change the "and" to "or". Also, it is useful to note that $r\wedge T\equiv r$ and $r\vee r\equiv r$. $\endgroup$ – Jack Nov 3 '15 at 7:45
  • $\begingroup$ Ahhh! Nice catch $\endgroup$ – Xirol Nov 3 '15 at 7:46
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    $\begingroup$ Wait, (~p ^ q) can become just 'r' in a sense? $\endgroup$ – Xirol Nov 3 '15 at 7:49
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    $\begingroup$ Take $r$ to be $\neg (p\wedge \neg q)$ and see what you get. $\endgroup$ – Jack Nov 3 '15 at 7:55
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    $\begingroup$ $p\implies q$ is synonymous with $(\sim p)\vee q$. So for the expression $\sim (p\wedge \sim q)$ in the first line, we have $\sim(p\wedge \sim q) \iff ((\sim p)\vee (\sim \sim q))\iff ((\sim p)\vee q)\iff (p\implies q)$ $\endgroup$ – DanielWainfleet Nov 3 '15 at 9:01
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By DeMorgan & Double Negation \begin{align} ((p \implies q) \lor (\neg(p \land \neg q) \land T))&\equiv((p \implies q) \lor ((\neg p \lor q) \land T)) \\ &\equiv((\neg p \lor q) \lor ((\neg p \lor q) \land T))\quad\text{by }p\land T \equiv p \\ &\equiv(\neg p \lor q) \lor (\neg p \lor q)\quad\text{by idempotent law} \\ &\equiv(\neg p \lor q) \\ \end{align}

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  • $\begingroup$ What logical equivalence was used for the 2nd to last line? $\endgroup$ – Xirol Nov 3 '15 at 8:05
  • $\begingroup$ @Xirol - $p \land T \equiv p$ is called sometimes Identity law; see Table of Logical Equivalences. $\endgroup$ – Mauro ALLEGRANZA Nov 3 '15 at 10:23

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