Answer:
a) ∀x∃y ¬∀zT(x, y, z)
∀x∃y ∃z ¬T(x, y, z)
b) ∀x¬[∃y (P(x, y) ∨ Q(x, y))]
∀x∀y ¬ [P(x, y) ∨ Q(x, y)]
∀x∀y [¬P(x, y) ^ ¬Q(x, y)]
c) ∀x ¬∃y (P(x, y) ^ ∃zR(x, y, z))
∀x ∀y ¬(P(x, y) ^ ∃zR(x, y, z))
∀x ∀y (¬P(x, y) v ¬∃zR(x, y, z))
∀x ∀y (¬P(x, y) v ∀z¬R(x, y, z))
d) ∀x¬∃y (P(x, y) → Q(x, y))
∀x∀y ¬(P(x, y) → Q(x, y))
∀x∀y (¬P(x, y) ^ Q(x, y))
Answer:
a) ∃x∀y∃z~T(x, y, z)
b) ∃x∀y~P(x, y) ∧ ∃x∀y~Q(x, y)
c) ∃x∀y(~P(x, y) ∨ ∀z~R(x, y, z))
d) ∃x∀y(P(x, y) → ~Q(x, y))
Step-by-step explanation:
The negation of a is written as ~a.
Note the following properties that are going to be applied in the problems here :
~(P → Q) = P → ~Q
De Morgan's Laws
~(P ∨ Q) = ~P ∧ ~Q
~(P ∧ Q) = ~P ∨ ~Q
~∃xP = ∀xP
~∀xP = ∃xP
So back to the original problem.
a) ∀x∃y∀zT(x, y, z)
We have the negation as
~[∀x∃y∀zT(x, y, z)]
= ∃x~∃y∀zT(x, y, z)
= ∃x∀y∀~zT(x, y, z)
= ∃x∀y∃z~T(x, y, z)
b) ∀x∃yP(x, y) ∨ ∀x∃yQ(x, y)
Negation is:
~[∀x∃yP(x, y) ∨ ∀x∃yQ(x, y)]
= ~∀x∃yP(x, y) ∧ ~∀x∃yQ(x, y)
= ∃x~∃yP(x, y) ∧ ∃x~∃yQ(x, y)
= ∃x∀y~P(x, y) ∧ ∃x∀y~Q(x, y)
c) ∀x∃y(P(x, y) ∧ ∃zR(x, y, z))
Negation is:
~[∀x∃y(P(x, y) ∧ ∃zR(x, y, z))]
= ~∀x∃y(P(x, y) ∧ ∃zR(x, y, z))
= ∃x~∃y(P(x, y) ∧ ∃zR(x, y, z))
= ∃x∀y~(P(x, y) ∧ ∃zR(x, y, z))
= ∃x∀y(~P(x, y) ∨ ~∃zR(x, y, z))
= ∃x∀y(~P(x, y) ∨ ∀z~R(x, y, z))
d) ∀x∃y(P(x, y) → Q(x, y))
Negation is:
~[∀x∃y(P(x, y) → Q(x, y))]
= ~∀x∃y(P(x, y) → Q(x, y))
= ∃x~∃y(P(x, y) → Q(x, y))
= ∃x∀y~(P(x, y) → Q(x, y))
= ∃x∀y(P(x, y) → ~Q(x, y))
