Propositional logic problem Show that [ (p ∨ q) ∧ (p → r) ∧ (q → r) ] → r is a tautology (without a truth table).
I am new to this, so I am not quite sure of how some rules can be used. Here is what I have so far:
= [(p ∨ q) ∧  (¬p v r)  ∧   (¬q v r) ] → r           by logical equivalence

=¬[(p ∨ q) ∧  (¬p v r)  ∧   (¬q v r) ] v r           by logical equivalence

=¬(p v q)   v ¬(¬p v r)  v  ¬(¬q v r)   v r           by DeMorgan’s law

= (¬p ^ ¬q) v  (p  ^ ¬r) v   (q ^ ¬r)   v r           by DeMorgan’s law

EDIT (does this work?)
=    r v (¬r ^ p) v  (¬r ^q) v (¬p ^ ¬q)            just switched things around 

=    r v (¬r  v (p ^ q)) v (¬p ^ ¬q)                distributive law
         ^             ^ are these parenthesis required? 
=    (r v (¬p ^ ¬q)) v (¬r  v (p ^ q))              just switched things around

=     T                                             by negation laws

First of all, please correct me if I have made any mistakes here and tell me if I am on the right track. Second, I am just not sure what I can do next. I wanted to try using distributive laws to break (q ^ ¬r)   v r up, and then use negation laws to finish, but I am not sure if that's legal. 
Also, my book states: ¬(p ∨ q) = ¬p ^ ¬q. Since the parenthesis are no longer there, does that mean I would be able to move ¬p and ¬q around afterwards? Or was it just a 'mistake' in the print?
 A: Assume everything in the antecedent true.  Then each of the conjuncts qualifies as true.  Suppose "p".  Then, by detachment and (p $\rightarrow$ r), we may infer "r".  Suppose "$\lnot$p".  Then by disjunctive syllogism and (p $\lor$ q), we may infer q.  Since we have (q $\rightarrow$ r) also, we may infer "r" by detachment.  By the principle of bivalence it follows that "r" holds.  By conditional introduction with our original assumption, and soundness, the above qualifies as a tautology.
A: Let $c$ denote contradiction, and $t$ denote tautology.
$$( (p \vee q) \wedge (p \rightarrow r) \wedge ( q \rightarrow r) ) \rightarrow r$$
$$\equiv ((p \vee q) \wedge (\neg p \vee r) \wedge (\neg q \vee r) ) \rightarrow r$$
$$\equiv (((p \vee q) \wedge \neg p) \vee ((p \vee q) \wedge r))) \wedge (\neg q \vee r)) \rightarrow r$$
$$\equiv (((p \wedge \neg p) \vee (q \wedge \neg p)) \vee ((p \vee q) \wedge r))) \wedge (\neg q \vee r)) \rightarrow r$$
$$\equiv (c \vee (q \wedge \neg p) \vee ( p \wedge r) \vee (q \wedge r)) \wedge (\neg q \vee r)) \rightarrow r$$
$$\equiv (((q \wedge \neg p) \wedge (\neg q \vee r)) \vee ((p \wedge r) \wedge (\neg q \vee r)) \vee ((q \wedge r) \wedge (\neg q \vee r))) \rightarrow r$$
$$\equiv ((q \wedge \neg p \wedge \neg q) \vee (q \wedge \neg p \wedge r) \vee (p \wedge r \wedge \neg q) \vee (p \wedge r \wedge r)) \vee ((q \wedge r) \wedge (\neg q \vee r))) \rightarrow r$$
$$\equiv ((q \wedge \neg p \wedge \neg q) \vee (q \wedge \neg p \wedge r) \vee (p \wedge r \wedge \neg q) \vee (p \wedge r \wedge r)) \vee ((q \wedge r \wedge \neg q) \vee (q \wedge r \wedge r))) \rightarrow r$$
$$\equiv c \vee (q \wedge \neg p \wedge r) \vee ((p \wedge r \wedge \neg q) \vee (p \wedge r)) \vee ( c \vee (q \wedge r)) \rightarrow r$$
$$\equiv \neg (((q \wedge \neg p \wedge r) \vee (p \wedge r \wedge \neg q) \vee (p \wedge r) \vee (q \wedge r)) \wedge \neg r)$$
$$\equiv \neg ( (q \wedge \neg p \wedge r \wedge \neg r) \vee (p \wedge r \wedge \neg q \wedge \neg r) \vee (p \wedge r \wedge \neg r) \vee (q \wedge r \wedge \neg r))$$
$$\equiv \neg (c \vee c \vee c \vee c)$$ $$\equiv t$$
One way to think about this is to try and make it not a tautology.  What would that look like? Well the only way an implication can be false is if the hypothesis is true, while the conclusion is false.  That would mean that $r$ is false and then $(p \vee q) \wedge (p \rightarrow r) \wedge ( q \rightarrow r)$ is true, but then for that to happen, since $r$ is false, $p, q$ have to be false (because of $p \rightarrow r$ and $q \rightarrow r$) but then $p \vee q$ is false, so the hypothesis is false.  Hence, we have a tautology.
