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!