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Logical equivalence

I'm only able to get 3.

≡  ¬p  ⋁  (q ⋁ s) ... by p  **→** q ≡ ¬p  ⋁ q

≡  ¬(p **⋀** ¬(q ⋁ s))  ...by De Morgan's First Law

≡ ¬(p **⋀**  (¬q **⋀** ¬s)) ...by De Morgan's Second Law

Any methods to get 2 more? I am having trouble.
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Try replacing q ⋁ s with an implication.
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Might I suggest a hearty portion of cheese?

p→(p → (q ⋁ s))

p→(p→(p → (q ⋁ s)))

p→(p→(p→(p → (q ⋁ s))))

p→(p→(p→(p→(p → (q ⋁ s)))))

...

Alternatively,

(p → (q ⋁ s)) \^ T

or

(p → (q ⋁ s)) ⋁ F

where T is any tautology and F is any contradiction.

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