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?
Proving that $A \vee (\neg A \wedge B) \equiv A \vee B$
logic;arithmetic;boolean algebra
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.