How can I prove that (p→q)∧(p→r) ⇔ p→(q∧r) How can I prove that (p→q)∧(p→r) compound statements and
compound statement p→(q∧r) are logically equivalent?
And can I use logical equivalences on this proof?
 A: $$(p\implies q) \land (p\implies r)$$
is equivalent to
$$( q \lor \lnot p) \land ( r \lor \lnot p)$$
is equivalent to 
$$(q \land r) \lor \lnot p$$
is equivalent to 
$$p \implies (q\land r)$$
A: $(p→q)∧(p→r) $ is the same as
$(\overline{p} \vee q)\wedge   (\overline{p} \vee r)$ which is the same as
$(\overline{p}\vee(\overline{p}\wedge r)\vee(q\wedge\overline{p})\vee(q\wedge r))$
From here, it is clear that if both $\overline{p}$ and  $(q\wedge r)$ is false, the complete statement is false. If either is true, then the full statement is true.  And so the full statement is the same as the statement $p→(q∧r)$ because  $p→(q∧r)$ is the same as $\overline{p} \vee (q\wedge r)$.
A: The other answers showed how to use logical equivalences to prove the result. Here is a proof using natural deduction in a Fitch-style proof checker.

The inference rules are listed on the proof checker's page. 
They are 


*

*conjunction introduction ($\land$I), 

*conjunction elimination ($\land$E), 

*implication introduction or conditional proof ($\to$I), 

*implication elimination or modus ponens ($\to$E) and

*biconditional introduction ($\leftrightarrow$I).



Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
