How to prove that $(A \lor B) \land (\lnot A \lor B) = B$

Question

I know this is fairly basic, and I understand that it becomes $$ \begin{align} (A \land \lnot A) \lor B \\ F \lor B \\ B \end{align} $$

However, I can't work out how to prove that it becomes that first line. It seems intuitive, but I cannot work out which laws let me simplify it like that.

This is from a previous homework, and the above equation is the correction I was given, and I can't remember exactly how it was produced.

Answer 1

The first line follows from distributivity: $$(X \wedge Y) \vee Z = (X \vee Z) \wedge (Y \vee Z)$$ Think what $X$, $Y$ and $Z$ are in this scenario.

Answer 2

As Clive pointed out, this comes from an application of the distributive law of "disjunction" $(\lor)$ over "conjunction" $(\land)$:

$$(\color{red}{\bf P}\color{blue}{\bf \lor R}) \color{red}{\bf \land} (\color{red}{\bf Q} \color{blue}{\bf \lor R})\iff (\color{red}{\bf P \land Q}) \color{blue}{\bf \lor R}$$

In your problem, you have: $$(\color{red}{\bf A} \color{blue}{\bf \lor B}) \color{red}{\bf \land} (\color{red}{\bf \lnot A} \color{blue}{\bf \lor B}) \iff (\color{red}{\bf A \land \lnot A}) \color{blue}{\bf \lor B}$$

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You'll also want to become familiar with the other application of distributivity (which is not needed in your problem). It follows a similar pattern, with connectives interchanged, as follows:

$$(P\land R){\bf \lor }(Q\land R) \iff (P \lor Q) \land R$$