Simplify statement without using truth table

Question

I can't find an appropriate law that would lead to further shortening of the equation, what would be the simplest and shortest method?

Work

$((p ∧ q) ∧ r) ∨ ((¬r ∧ p) ∧ q) ∨ ¬q\tag*{Original}$

$((p\land q)\land r)\lor (¬r\land (p\land q))\lor ¬q \tag*{Associative law}$

$((p\land q)\land (r\lor ¬r))\lor ¬q \tag*{Dissociation law}$

Answer 1

Here is a simplification use Logical equivalence: \begin{align} &((p ∧ q) ∧ r) ∨ ((¬r ∧ p) ∧ q) ∨ ¬q\\ \equiv&((p ∧ q) ∧ r) ∨ (¬r ∧ (p ∧ q)) ∨ ¬q\tag*{Associative law}\\ \equiv&((p ∧ q) ∧ r) ∨ ((p ∧ q) ∧ ¬r) ∨ ¬q\tag*{Commutative law}\\ \equiv&(p ∧ q) ∧ (r∨ ¬r) ∨ ¬q\tag*{Distributive law}\\ \equiv&(p ∧ q) ∧ \top ∨ ¬q\tag*{Negation law}\\ \equiv&(p ∧ q) ∨ ¬q\tag*{Identity law}\\ \equiv&(p∨¬q)∧(q∨¬q)\tag*{Distributive law}\\ \equiv&(p∨¬q)∧\top\tag*{Negation law}\\ \equiv&~p∨¬q\tag*{Domination law}\\ \end{align}

Answer 2

Since the original expression is something$\vee\lnot q$, when evaluating something, you can assume $q$. Hence, the expression simplifies to $$(p\wedge r)\vee(\lnot r\wedge p)\vee\lnot q$$This simplifies to $$p\vee\lnot q$$or more simply, $$q\to p$$