# Show that $(p ⊕ q) \wedge r = (p \wedge r) ⊕ (q \wedge r)$ using laws of boolean algebra

So recently I've been studying Discrete Mathematics using a book written by Susanna S. There's an exercise in which she asks whether $$(p ⊕ q) \wedge r = (p \wedge r) ⊕ (q \wedge r)$$ I've done it using a truth table and concluded that both sides are logically equivalent, by checking each entry, but I couldn't do it using laws of boolean algebra. I've just got stuck trying to manage one expression into the other

## 1 Answer

\begin{align} (p\land r)\oplus (q\land r)&=\big((p\land r)\land\lnot(q\land r)\big)\lor\big(\lnot(p\land r)\land(q\land r)\big)&\text{Definition of }\oplus\\ &=\big((p\land r)\land(\lnot q\lor \lnot r)\big)\lor\big((\lnot p\lor \lnot r)\land(q\land r)\big)&\text{De-Morgans}\\ &=(p\land r\land\lnot q)\lor(p\land r\land\lnot r)\lor(q\land r\land\lnot p)\lor(q\land r\land\lnot r)&\text{Distribute}\\ &=(p\land r\land\lnot q)\lor(q\land r\land\lnot p)&\text{Simplify}\\ &=\big((p\land\lnot q)\lor(q\land\lnot p)\big)\land r&\text{Factor out}\\ &=(p\oplus q)\land r&\text{Definition of }\oplus \end{align}