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.

  • $\begingroup$ Hint: Prove $\Rightarrow$ and $\Leftarrow$ separately. $\endgroup$ – 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: $$A\lor (A\land B) \iff A$$


Not the answer you're looking for? Browse other questions tagged or ask your own question.