Help with logical equivalence/proving not a contradiction The question is this: 
Demonstrate using logical equivalences that $(p → q) ∧ (p → ¬q)$ is not
a contradiction. Identify all logical equivalences by name.
So far, I have
$(p → q) \land (p → ¬q)$
a.  $(¬p\lor q) \land (¬p\lor¬q)$ 
b.  $¬(¬p\lor q) \lor(¬(¬p\lor¬q)$ 
c.  $( p\land¬q) \lor (p \lor q)$ 
I can't figure out where to go from here. Any help would be appreciated. 
 A: It is not a contradiction becauss it is true when p is false.
Truth tables show this quickly.
a. perhaps is the best for showing that.
a. is equivalent to:  (not p) or (q and not q);
which in turn is equivalent to;  not p.
So there's the answer - to show the orginal statement is
equivalent to:  not p.  
Little surprise, as the orginal statement is basically
proof by contradiction.
A: From :
a) $(¬p \lor q) \land (¬p \lor ¬q)$
we have to apply the Distributive law to get :
b) $\lnot p \lor (q \land \lnot q)$.
Now we use the equivalence (often called : Negation law) : $(\alpha \land \lnot \alpha) \equiv \text F$ to get :
c) $\lnot p \lor \text F$.
Finally, we use the equivalence (often called : Identity law) : $\alpha \lor \text F \equiv \alpha$ to conclude with :

d) $\lnot p$

that is not a contradiction.
A: An easy way to say that is: since $p$ concludes $q$ and $\lnot q$, if it holds true then two controversy statements hold true which is a contradiction, but if $\lnot p$ is true, the contradiction is removed.
In a more formal way $$(p\to q)\land (p\to \lnot q)\iff p\to (q\land \lnot q)\iff p\to 0\iff \lnot p\to 1$$which means that $\lnot p$ is true.
A: Just an add-on to William Elliot's answer:
$$
\begin{array}{c|c|c|c|c}
p & q & p \rightarrow q & p \rightarrow \neg q & (p \rightarrow q) \wedge (p \rightarrow \neg q) \\ \hline
0 & 0 & 1 & 1 & 1 \\
0 & 1 & 1 & 1 & 1 \\
1 & 0 & 0 & 1 & 0 \\
1 & 1 & 1 & 0 & 0
\end{array}
$$
