Equivalences

The truth tables for the equivalence $p \vee (q \wedge r) \Leftrightarrow (p \vee q) \wedge (p \vee r)$ are:

$p$$q$$r$$q \wedge r$$\boldsymbol{p \vee (q \wedge r)}$
$0$$0$$0$$0$$0$
$0$$0$$1$$0$$0$
$0$$1$$0$$0$$0$
$0$$1$$1$$1$$1$
$1$$0$$0$$0$$1$
$1$$0$$1$$0$$1$
$1$$1$$0$$0$$1$
$1$$1$$1$$1$$1$
$p$$q$$r$$p \vee q$$p \vee r$$\boldsymbol{(p \vee q) \wedge (p \vee r)}$
$0$$0$$0$$0$$0$$0$
$0$$0$$1$$0$$1$$0$
$0$$1$$0$$1$$0$$0$
$0$$1$$1$$1$$1$$1$
$1$$0$$0$$1$$1$$1$
$1$$0$$1$$1$$1$$1$
$1$$1$$0$$1$$1$$1$
$1$$1$$1$$1$$1$$1$

Because the sequences of bits that we got under the $p \vee (q \wedge r)$ and $(p \vee q) \wedge (p \vee r)$ columns are exactly the same: $(0, 0, 0, 1, 1, 1, 1, 1)$, this means that the expressions are equivalent!