Show that (p → r) ∧ (q → r) and (p ∨ q) → r are logically equivalent.
(p→r)∧(q→r)=(¬p∧r)∧(¬q∧r)=(¬p∧¬q)∧(r∧r)=(p \to r) \land (q \to r) = (\lnot p \land r) \land (\lnot q \land r) = (\lnot p \land \lnot q) \land (r \land r)=(p→r)∧(q→r)=(¬p∧r)∧(¬q∧r)=(¬p∧¬q)∧(r∧r)=
=¬(p∨q)∧r=(p∨q)→r=\lnot (p \lor q) \land r=(p \lor q) \to r=¬(p∨q)∧r=(p∨q)→r
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