Is $(p \land q) \to (p \oplus q)$ a tautology?

Question

I am trying to show if this well formed formula is a tautology $(p \land q) \to (p \oplus q)$, I have done the following but I am stuck.

\begin{align} (p \land q) \to (p \oplus q) &\equiv \lnot(p \land q) \lor (p \oplus q) &\text{By Implication} \\ &\equiv (\lnot p \lor \lnot q) \lor (p \oplus q) &\text{By De Morgan} \end{align}

But I dont know what to do with $p \oplus q$ part.

I know it is an exclusive or, and it is true only when one of them p or q is true, but not both.

What is the next thing to do?

Answer 1

Exclusive Or: $~p\oplus q ~{~\equiv~(p\vee q)\wedge(\neg p\vee\neg q)\\~\equiv~ (p\wedge\neg q)\vee(\neg p\wedge q)}$

Substitute the equivalence of your choice into your expression and continue from there.

Answer 2

It is not a tautology. If the antecedent $p\land q$ is true, $p$ and $q$ are both true and hence the consequent $p\oplus q$ is false. When in doubt, use a truth table.

Answer 3

$$(p \land q ) \rightarrow (p \oplus q) \overset{Def.\oplus}{=}$$

$$(p \land q ) \rightarrow ((p \land \neg q) \lor (\neg p \land q))\overset{Implication}{=}$$

$$\neg (p \land q ) \lor (p \land \neg q) \lor (\neg p \land q)\overset{DeMorgan}{=}$$

$$\neg p \lor \neg q \lor (p \land \neg q) \lor (\neg p \land q)\overset{Absorption}{=}$$

$$\neg p \lor \neg q$$

... which is clearly not a tautology nor a contradiction, but a contingency.