$X\equiv(P∨Q)→R$
$X\equiv\lnot(P∨Q)\vee R$
$X\equiv (\lnot P \wedge \lnot Q)\vee R$
$X\equiv \bar{P}.\bar{Q}+R$
$Y\equiv(P→R)∨(Q→R)$
$Y\equiv(\lnot P\vee R)∨(\lnot Q\vee R)$
$Y\equiv\bar{P}+R+\bar{Q}+R$
$Y\equiv\bar{P}+\bar{Q}+R$
How $X \equiv Y$ gives always true?
It is not possible to $X \equiv Y$ gives always true(Tautology).