User:Dfeuer/ConjDistDisLeftFor

Theorem
$p \wedge (q \vee r) \preceq (p \wedge q) \vee (p \wedge r)$

Proof
$p \wedge (q \vee r) \prec p$

$p \wedge (q \vee r) \prec (q \vee r)$

$(p \wedge (q \vee r)) \wedge q \prec q$

$(p \wedge (q \vee r)) \wedge q \prec p$

Thus $(p \wedge (q \vee r)) \wedge q \prec p \wedge q$

$(p \wedge (q \vee r)) \wedge r \preceq p \wedge (q \vee r) \prec p$

$(p \wedge (q \vee r)) \wedge r \prec r$

Thus $(p \wedge (q \vee r)) \wedge r \prec p \wedge r$.