# Prove A or (A and B) is equivalent to A [duplicate]

Prove $A \lor (A \land B) \Leftrightarrow A$ without using truth table.

The proof may involve expanding $B$ into $B \land B$ or possibly $B \lor B$. I am stuck after playing with distributive law several times. Thanks for any help.

• Hint: Prove $\Rightarrow$ and $\Leftarrow$ separately. – hmakholm left over Monica Jul 5 '14 at 18:17

$\Rightarrow:\;\;(1)$ Distribute. $\;\;(2) A\lor A \equiv A\land A \equiv A$. (Simplification.) $\;\;(3)\land$-Elimination.
$$A\lor (A\land B) \overset{(1)}{\iff} (A \lor A)\land (A\lor B) \overset{(2)}\iff A \land (A\lor B) \overset{(3)}\implies A$$
$\Leftarrow:\;\;$ We use disjunction-Introduction.
$$A \implies A\lor (\text{anything}.)\;\;\text{So,}\; A \implies A\lor (A\land B)$$
Therefore: $$A\lor (A\land B) \iff A$$