Show that are logically equivalent Show that are logically equivalent (without truth table)
(p → r) ∧ (q → r) and (p ∨ q) → r
My solution:
(p→r)∧(q→r)
≡(¬p∨r)∧(¬q∨r) 
≡(¬p∧¬q)∨(r∨r)
≡(¬p∧¬q)∨r   
≡¬(p∨q)∨r
≡(p∨q)→r

(p → r) ∨ (q → r) and (p ∧ q) → r
(p→r)∨(q→r)
≡(¬p∨r)∨(¬q∨r)   
 ≡(¬p∨¬q)∨r
≡¬(p∧q)∨r  
 ≡(p∧q)→r

Is it correct ? 

Looks good! You just made a small mistake; idempotent law isn't necessary in the first solution. By distributive law, we have that:
$$
(\neg p \lor r) \land (\neg q \lor r) \equiv (\neg p \land q) \lor r
$$
However you do need to use idempotent law for the second solution:
$$
(\neg p \lor r) \lor (\neg q \lor r) \equiv \neg p \lor \neg q \lor r \lor r
\equiv \neg p \lor \neg q \lor r
$$
