Rewrite $((p \land q)\lor(q \rightarrow(p\land \lnot r))$ to DNF

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So, I'm a bit stuck here and I don't see how I can continue. I needed to write the following expression to DNF:

$$((p \land q)\lor(q \rightarrow(p\land \lnot r))$$

What I tried:

$$((p \land q) \lor (\lnot q \lor (p \land \lnot r)))$$

$$((p\land q)\lor((\lnot q\lor p) \land (\lnot q \lor \lnot r)))$$

$$((p \land q) \lor ((\lnot q\lor p) \land \lnot(q \land r)))$$

And I'm stuck here. The answer is supposed to be $p\lor \lnot q$, which is what has gotten me surprised. What am I doing wrong?

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((p ∧ q) ∨ (¬q ∨ (p ∧ ¬r))) - this is what you had

(p ∧ q) ∨ (¬q ∨ (p ∧ ¬r)) - you don't need the most outer brackets

((p ∧ q) ∨ ¬q) ∨ (p ∧ ¬r) - disjunction association, dropping the brackets

((p ∨ ¬q) ∧ (q ∨ ¬q)) ∨ (p ∧ ¬r)

((p ∨ ¬q) ∧ true) ∨ (p ∧ ¬r)

(p ∨ ¬q) ∨ (p ∧ ¬r)

p ∨ ¬q ∨ (p ∧ ¬r)

¬q ∨ p ∨ (p ∧ ¬r)

¬q ∨ (p ∨ (p ∧ ¬r))

Here I use absorption rule : A + AB = A(true + B) = A true = A (can also be checked by using truth table)

¬q ∨ (p ∧ (true ∨ ¬r))

¬q ∨ p