# Why is $p\Rightarrow q$ equivalent to $\neg p\lor q$ and how to prove it

I don't know how to prove that $p\Rightarrow q$ is equivalent to $\neg p\lor q$ ,here is the link p=>q . And I don't know how wolframalpha generate "Minimal forms" .

Can you prove $p\Rightarrow q \equiv \neg p\lor q$, and explain how to get "Minimal forms" ?

Thanks!

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Okay, => means implication, $\Rightarrow$. I take it that $||$ means "or", $\lor$? – Arturo Magidin Aug 18 '11 at 3:39
yes thanks I'm new here – ifree Aug 18 '11 at 3:42
No problem. Want me to edit it to more standard mathematical notation? – Arturo Magidin Aug 18 '11 at 3:45
yes thanks and how about the Minimal forms – ifree Aug 18 '11 at 3:49
There are algorithms for transforming propositional formulas into normal forms; see for example Wikipedia. Presumably, wolframalpha has programmed the algorithms in order to find the normal minimal forms. – Arturo Magidin Aug 18 '11 at 3:52

By definition, $p\Rightarrow q$ is true if and only if the consequence $q$ is true, or the antecedent $p$ is false. You can see it in the truth table that defines the implication. That is, $p\Rightarrow q$ is true if and only if either $\neg p$ is true, or $q$ is true; i.e., if and only if $\neg p\lor q$ is true (what you write as $!p||q$).
Or you can simply look at the truth tables. The truth table of $\neg p\lor q$ is the same as the truth table of $p\Rightarrow q$: true if $p$ and $q$ are false; true if $p$ is false and $q$ is true; false if $p$ is true and $q$ is false; true if $p$ and $q$ are both true: $$\begin{array}{c|c||c} p & q & p\Rightarrow q\\ \hline 0 & 0 & 1\\ 0 & 1 & 1\\ 1 & 0 & 0\\ 1 & 1 & 1 \end{array}\qquad\qquad \begin{array}{c|c|c|c} p & q & \neg p & \neg p\lor q\\ \hline 0 & 0 & 1 & 1\\ 0 & 1 & 1 & 1\\ 1 & 0 & 0 & 0\\ 1 & 1 & 0 & 1 \end{array}.$$ The final columns are identical, so the two formulas take the same truth values given the same truth inputs: that is, they are propositionally equivalent.