Having a lot of trouble with the $q$ in $p \oplus q$ being replaced with $(\lnot p \land q)$.
This is for my first unit of Discrete Mathematics, but it's a bit of a curve ball.
I've been picking at it for a few hours, and I think I've managed to reduce it to $(p \lor q) \land (p \land \lnot q)$ using double negation and reduction laws but I don't know how to prove that it is equal to $p\lor q$.
A truth table would do the job. One thing you should keep in mind to understand the table is that $ A\oplus B$ is exclusive or. So it would be zero if both $A$ and $B$ have truth value $1$.
$$\begin{array}{|c|c|c|c|c|c|} \hline p & q&p \lor q&\neg p&\neg p\land q&p \oplus(\neg p\land q) \\\hline 1 & 1 &1&0& 0&1\\\hline 1 & 0&1 &0&0&1\\\hline 0 & 1& 1& 1&1&1\\\hline 0 & 0 &0 & 1&0&0 \\\hline \end{array}$$