Question #132116

(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))

Expert's answer

(a)∀z∃y∃x¬T(x,y,z)\forall z\exists y\exists x\neg T(x,y,z)

(b)Obtain the negation by steps:

1)¬(∃x∃yP(x,y))∨¬(∀x∀yQ(x,y))\neg(\exists x\exists y P(x,y))\vee\neg(\forall x\forall y Q(x,y))

2)∀x∀y¬P(x,y)∨∃x∃y¬Q(x,y)\forall x\forall y \neg P(x,y)\vee\exists x\exists y \neg Q(x,y)

(c)By steps:

1)∀x∀y¬(Q(x,y)↔Q(y,x))\forall x\forall y\neg(Q(x,y)\leftrightarrow Q(y,x))

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

(d)By steps:

1)∃y∀x∀z¬(T(x,y,z)∨Q(x,y))\exists y\forall x\forall z\neg(T(x,y,z)\vee Q(x,y))

2)∃y∀x∀z(¬T(x,y,z)∧¬Q(x,y))\exists y\forall x\forall z(\neg T(x,y,z)\wedge \neg Q(x,y))

Answer:

(a)∀z∃y∃x¬T(x,y,z)\forall z\exists y\exists x\neg T(x,y,z)

(b)∀x∀y¬P(x,y)∨∃x∃y¬Q(x,y)\forall x\forall y \neg P(x,y)\vee\exists x\exists y \neg Q(x,y)

(c)∀x∀y(Q(x,y)⊕Q(y,x))\forall x\forall y(Q(x,y) \oplus Q(y,x))

(d)∃y∀x∀z(¬T(x,y,z)∧¬Q(x,y))\exists y\forall x\forall z(\neg T(x,y,z)\wedge \neg Q(x,y))


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