Need Help with Propositional Logic I am stuck with this proof. I am trying to use deduction (or induction I think) to prove for a tautology with logic laws like De Morgan's, distributive , and implication law etc. 
Note: I am not allowed to use truth tables.
Here it is:
$((p \vee q) \wedge (p \rightarrow r) \wedge (q \rightarrow r)) \rightarrow r$
I have tried using a condition/implication law where $p \rightarrow r$ becomes $\neg p \vee r$ to change the last to compound statements but I got stuck.
Next I tried:
$((p \vee q) \wedge (p \rightarrow r) \wedge (q \rightarrow r)) \rightarrow r \\ \equiv [(p \vee q) \wedge ((p \vee q) \rightarrow r)] \rightarrow r$
But I don't know where to go from here.
Need some guidance guys.

What you want to show is equivalent to $((p\vee q)\wedge(p\to r)\wedge(q\to r))$ implies $r$. 
Now we can work with $(p\vee q)$ separately since we have a string of $\wedge$'s, and a case by case analysis of this shows that $r$ is implied by our hypothesis.
Is this similar to what you have done in class?
