Prove: (p → ¬q) → (¬p → q) ≡ p ∨ q

Question

Are there any mistakes in this proof? Is the logic correct? Feedback is greatly appreciated.

(p → ¬q) → (¬p → q) ≡

(¬p ∨ ¬q) → (¬¬p ∨ q) ≡

¬(¬p ∨ ¬q) ∨ (¬¬p ∨ q) ≡

¬(¬p ∨ ¬q) ∨ (¬¬p ∨ q) ≡

¬(¬p ∨ ¬q) ∨ (p ∨ q) ≡

¬(¬p ∨ ¬q) ∨ (p ∨ q) ≡

(¬¬p ∧ ¬¬q) ∨ (p ∨ q) ≡

(p ∧ q) ∨ (p ∨ q) ≡

{(p ∧ q) ∨ p} ∨ q ≡

{p ∨ (p ∧ q)} ∨ q ≡

p v q

Answer 1

Your proof is correct, though steps 4 and 6 are repeated.

You might also want to add:

$$(p\lor (p\land q))\lor q\equiv (p\lor q) \lor (p\land q)\equiv ((p\lor q)\lor p)\land ((p\lor q)\lor q)\equiv(p\lor q)\land (p\lor q)\equiv p\lor q$$

after step 10.