How to prove that $(p\rightarrow q)\wedge(p\rightarrow r)$ and $p\rightarrow (q \wedge r)$ are logically equivalent? I am trying to prove that $(p\rightarrow q)\wedge (p\rightarrow r) = p\rightarrow (q \wedge r)$. 
This is my approach:
$(p\rightarrow q)\wedge(p\rightarrow r) = (-p \vee q) \wedge (-p \vee r)$
= ${[(-p \vee q) \wedge -p] \vee [(-p \vee q) \wedge r]}$
= ${[(-p \wedge -p) \vee (q \wedge -p)]  \vee [(-p \wedge r) \vee (q \wedge r)]}$
But I have been stuck here. 
How can I prove this ? 
Thank you. 
 A: As some others have pointed out, you used distributivity correctly but in a less than desired way (by complicating the expression further as opposed to simplifying it). As Git Gud notes, you want to "factor out the $\neg p$" where you, instead, expanded things further by distributivity. Finally, you will want to use material implication to simplify your expression to obtain the desired expression as JnxF notes. Your argument should look like so:
\begin{align}
(p\to q)\land(p\to r)&\equiv (\neg p\lor q)\land(\neg p\lor r)\tag{material implication}\\[1em]
&\equiv \neg p\lor(q\land r)\tag{distributivity}\\[1em]
&\equiv p\to(q\land r).\tag{material implication}
\end{align}
Of course, you could always create a truth table, as thanasissdr notes, but I would only ever do this as a last resort (that is where a computer would be useful if all you were interested in were the truth table).
A: Unless the connective $\to$ is defined in terms of $\vee$ and $\neg$, there is no need to go about this proof using the equivalence $p \to q \equiv \neg p \vee q$ or using de Morgan's laws or any of that jazz. You can do it directly.


*

*($\to$) Assume $(p \to q) \wedge (p \to r)$. We'll prove $p \to (q \wedge r)$. To do this, assume $p$; we'll prove $q \wedge r$. Since $p$ and $p \to q$ are true, so is $q$ by modus ponens. Since $p$ and $p \to r$ are true, so is $r$ by modus ponens. Since $q$ and $r$ are both true, so is $q \wedge r$. As this follows from the assumption $p$, we have that $p \to (q \wedge r)$ is true, as required.

*($\leftarrow$) Likewise, assume $p \to (q \wedge r)$ is true, and derive $(p \to q) \wedge (p \to r)$ from this assumption. You can do this.
You could turn this into a proof in natural deduction (or whatever formal deductive system you're using) if you wanted to be really formal about it.
