Express the negations of each of these statements so that all negation symbols immediately precede predicates. a) ∀x∃y∀zT (x, y, z) b) ∀x∃yP (x, y) ∨ ∀x∃yQ (x, y) c) ∀x∃y (P (x, y) ∧ ∃zR (x, y, z)) d) ∀x∃y (P (x, y) → Q (x, y))

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))