Answer
a) ¬∃z∀y∀x T(x, y, z) ≡ ∀z¬∀y∀x T(x, y, z)
≡ ∀z∃y¬∀x T(x, y, z)
≡ ∀z∃y∃x¬T(x, y, z)
b) ¬(∃x∃y P(x, y) ∧ ∀x∀y Q(x, y)) ≡ ¬∃x∃y P(x, y) ∨ ¬∀x∀y Q(x, y)
≡ ∀x¬∃y P(x, y) ∨ ∃x¬∀y Q(x, y)
≡ ∀x∀y ¬P(x, y) ∨ ∃x∃y ¬ Q(x, y)
c) ¬∃x∃y(Q(x, y) ↔ Q(y, x)) ≡ ∀x¬∃y(Q(x, y) ↔ Q(y, x))
≡ ∀x∀y¬(Q(x, y) ↔ Q(y, x))
≡ ∀x∀y(¬Q(x, y) ↔ Q(y, x))
d) ¬∀y∃x∃z (T(x, y, z) ∨ Q(x, y)) ≡ ∃y¬∃x∃z (T(x, y, z) ∨ Q(x, y))
≡ ∃y∀x¬∃z (T(x, y, z) ∨ Q(x, y))
≡ ∃y∀x∀z ¬(T(x, y, z) ∨ Q(x, y))
≡ ∃y∀x∀z (¬T(x, y, z) ∧ ¬Q(x, y))
Work Step by Step
As we push the negation symbol toward the inside, each quantifier it passes must change its type. For logical
connectives we either use De Morgan’s laws or recall that ¬(p → q) ≡ p∧¬q (Table 7 in Section 1.3) and that
¬(p ↔ q) ≡ ¬p ↔ q (Exercise 21 in Section 1.3).
