(3) Express the negations of each of these statements so that all negation symbols immediately precede predicates. (a)∃z∀y∀xT(x, y, z) (b)∃x∃yP(x, y)^∀x∀yQ(x, y) (c)∃x∃y(Q(x, y)<=>Q(y,x)) (d)∀y∃x∃z(T(x, y, z)∨Q(x, y))

(a)$\forall z\exists y\exists x\neg T(x,y,z)$

(b)Obtain the negation by steps:

1)$\neg(\exists x\exists y P(x,y))\vee\neg(\forall x\forall y Q(x,y))$

2)$\forall x\forall y \neg P(x,y)\vee\exists x\exists y \neg Q(x,y)$

(c)By steps:

1)$\forall x\forall y\neg(Q(x,y)\leftrightarrow Q(y,x))$

2)$\forall x\forall y(Q(x,y) \oplus Q(y,x))$, where $\oplus$ is exclusive or

(d)By steps:

1)$\exists y\forall x\forall z\neg(T(x,y,z)\vee Q(x,y))$

2)$\exists y\forall x\forall z(\neg T(x,y,z)\wedge \neg Q(x,y))$

Answer:

(a)$\forall z\exists y\exists x\neg T(x,y,z)$

(b)$\forall x\forall y \neg P(x,y)\vee\exists x\exists y \neg Q(x,y)$

(c)$\forall x\forall y(Q(x,y) \oplus Q(y,x))$

(d)$\exists y\forall x\forall z(\neg T(x,y,z)\wedge \neg Q(x,y))$