Logical Equivalences

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?

• 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 $r\vee r\equiv r$. – Jack Nov 3 '15 at 7:45
• Ahhh! Nice catch – Xirol Nov 3 '15 at 7:46
• Wait, (~p ^ q) can become just 'r' in a sense? – Xirol Nov 3 '15 at 7:49
• Take $r$ to be $\neg (p\wedge \neg q)$ and see what you get. – Jack Nov 3 '15 at 7:55
• $p\implies q$ is synonymous with $(\sim p)\vee q$. So for the expression $\sim (p\wedge \sim q)$ in the first line, we have $\sim(p\wedge \sim q) \iff ((\sim p)\vee (\sim \sim q))\iff ((\sim p)\vee q)\iff (p\implies q)$ – DanielWainfleet Nov 3 '15 at 9:01

• @Xirol - $p \land T \equiv p$ is called sometimes Identity law; see Table of Logical Equivalences. – Mauro ALLEGRANZA Nov 3 '15 at 10:23