Help with logical equivalence/proving not a contradiction

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

Just an add-on to William Elliot's answer:

$$ \begin{array}{c|c|c|c|c} p & q & p \rightarrow q & p \rightarrow \neg q & (p \rightarrow q) \wedge (p \rightarrow \neg q) \\ \hline 0 & 0 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 \end{array} $$

Answer 4

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.