TITLE: $(p \implies q) \wedge (p \implies r) = p \implies (q \wedge r)$ PROOF
QUESTION [5 upvotes]: $\def\implies{\to}$I am trying to prove the following: $$(p \implies q)\wedge (p\implies r) \equiv (p\implies q \wedge r) $$
I did this, although I am second-guessing because of the parenthesis placement and the order of operations:
$$ \begin{align}(p \implies q) \wedge (p \implies r) &\equiv (\neg p \vee q)\wedge (\neg p \vee r) \\[1ex] &\equiv \neg p \vee (q \wedge r) \\[1ex] &\equiv p \implies (q\wedge r)\end{align}$$ 
Any help/insight on this is appreciated!

REPLY [4 votes]: Your work looks good.  
You used that $a\implies b$ is equivalent to $\lnot a \lor b$.  
You also used the distributive property of disjunction over conjunction:  
$(a \lor b) \land (a \lor c)$ is equivalent to $a \lor (b \land c)$.  
(There is also a distributive property of conjunction over disjunction, but you didn't need that.)