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∀ x, if P(x) then Q(x) → ∀ a, if ¬P(a) then ¬Q(a). The previous proposition is...?
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∀ x, if P(x) then Q(x) → ∀ a, if ¬P(a) then ¬Q(a). The previous proposition is...?
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∀ x, if P(x) then Q(x) → ∀ a, if ¬P(a) then ¬Q(a). The previous proposition is...?truefake

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∀ x, if P(x) then Q(x) → ∀ a, if ¬P(a) then ¬Q(a). The previous proposition is...?
true
fake