Q4. (10 points) Prove the following:
∀x(x ∈ P → ¬Q(x)) is equivalent to ¬∃x(x ∈ P ∧ Q(x))



	∀x(x ∈ P → ¬Q(x))     conditional law
	∀x(x ∉ P → ¬Q(x))      double negation
	¬(¬∀x(x∉P ∨ ¬Q(x)))   quantifier negation laws
	¬(∃x¬(x∉P ∨ ¬Q(x)))   de morgan
	¬(∃x(x ∈ P ∧ Q(x)))
	¬∃x(x ∈ P ∧ Q(x)))



(b) ∃x ∈ A P(x) ∨∃x ∈ B P(x) is equivalent to ∃x ∈ (A ∪ B)


	∃x ((x ∈ A ∧ P(x)) ∨ (x ∈ B ∧ P(x))      distribution law
	∃x (P(x) ∧ ((x ∈ A) ∨ (x ∈ B))              distribution law
	∃x (P(x) ∧ (x ∈ (A ∪ B)))                     distribution law
	∃x ((x ∈ (A ∪ B)) ∧ P(x))
	∃x ∈ (A ∪ B)P(x)