[Solved]: Proving that $A \vee (\neg A \wedge B) \equiv A \vee B$

Problem Detail: 

I'm reading a book at the moment about logic gates and Boolean simplification. There is a part which I can't seem to follow.

I can easily work out that $A \vee (\neg A \wedge B) \equiv A \vee B$ using a truth table as it's easy to see.

However, I can't seem to turn $A \vee (\neg A \wedge B)$ into $A \vee B$ using steps such as distributive / absorption etc.

Can someone talk me through the steps that you would take to simplify this?

Asked By : user1480135

Answered By : Raphael

Note that

$\qquad A \lor (B \land C) \equiv (A \lor B) \land (A \lor C)$;

you can "multiply out". Add in

$\qquad (A \lor \lnot A) \land B \equiv B$

and you are done.

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Question Source : http://cs.stackexchange.com/questions/63769