logicpropositional-calculus
Show that $(p \rightarrow q) \wedge (p \rightarrow r)$ and $p \rightarrow (q \wedge r)$ are logically equivalent.
I tried to do this making a truth table but I think my teacher wants me to solve it using the different laws of Logical Equivalences.
Can anyone help me?
Yes, for example you can use DeMorgan's laws: $$a \vee b = \neg((\neg a) \wedge (\neg b))$$ $$a \wedge b = \neg((\neg a) \vee (\neg b))$$
As well as distributivity over the operators:
$$a \vee (b \wedge c) = (a \vee b) \wedge (a \vee c) $$ $$a \wedge (b \vee c) = (a \wedge b) \vee (a \wedge c) $$
And there are a lot of more rules you can use to simplify expressions.
I managed to work out a solution, and I thought I would share even though it is rather hideous (picture posted to make formatting much more pleasant to read). 
Best Answer
Here is an approach
$$(p \to q) \wedge (p \to r) \equiv (\neg p \vee q) \wedge (\neg p \vee r) \equiv \neg p \vee (q \wedge r) \equiv p \to (q \wedge r)$$