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I know how to show this via truth tables, but I’m confused over the formal proof.

Wikipedia tells me that:

$ (P \Rightarrow Q) \equiv (\neg P \lor Q) \equiv (Q \lor \neg P) \equiv (\neg Q \Rightarrow \neg P) $.

I don’t understand how we get from $ (P \Rightarrow Q) $ to $ (\neg P \lor Q) $. I also don’t understand how $ (Q \lor \neg P) $ implies $ (\neg Q \Rightarrow \neg P) $.

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  • $\begingroup$ "if it rains then the ground gets wet" ($P\implies Q$). Now can it happen that both following statemens are not true: 1)"it is not raining" and 2)"the ground is getting wet."? No it cannot, so at least one of them is true ($\neg P\vee Q$) $\endgroup$ – drhab Sep 11 '16 at 18:59
  • $\begingroup$ Before seeking a formal proof, you need to specify which proof calculus you wish to use. There are three well-known ones: Hilbert Calculus, Natural Deduction and Sequent Calculus. $\endgroup$ – Berrick Caleb Fillmore Sep 11 '16 at 19:05
  • $\begingroup$ Another way to convince yourself is to make truth tables for the two statements. There are two variables, $P$ and $Q$, so there are four possibilities for the truth values (both false, $P$ false and $Q$ true, $P$ true and $Q$ false, and both false). For each of the four possibilites, find the truth value of $P\implies Q$ and the truth value of $\neg Q\implies P$. You’ll find that in each of the four cases the truth value of $P\implies Q$ is the same as the truth value of $\neg Q\implies P$. Thus the two are logically equivalent. $\endgroup$ – Steve Kass Sep 11 '16 at 20:19
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$ (P \Rightarrow Q) \equiv (\neg P \lor Q) \equiv (Q \lor \neg P) \equiv (\neg \neg Q \lor \neg P) \equiv (\neg Q \Rightarrow \neg P) $.

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  • $\begingroup$ Thank you! But could you tell me the logic behind P⇒Q = ¬PvQ and ¬(¬Q)v(¬P) = ¬Q⇒¬P? $\endgroup$ – Nikitau Sep 11 '16 at 18:57
  • $\begingroup$ Its just manipulation.Which part you did not understand? $\endgroup$ – Ashar Tafhim Sep 11 '16 at 19:05
  • $\begingroup$ DrHab replied above so I now I understand P⇒Q = ¬PvQ . I'm confused over how ¬(¬Q)v(¬P) = ¬Q⇒¬P. I guess I should provide context by saying this is my first very rigorous proof-based class. $\endgroup$ – Nikitau Sep 11 '16 at 19:07
  • $\begingroup$ Let X=¬Q and Y=¬P then ¬(¬Q)v(¬P) =¬XvY = X⇒Y.Does this solve the problem? $\endgroup$ – Ashar Tafhim Sep 11 '16 at 19:09
  • $\begingroup$ OH, now I get it. Thank you so much! $\endgroup$ – Nikitau Sep 11 '16 at 19:11

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