Answer:
a. Negation of ∃x ∀y(P(x,y) → Q(x,y)) = ∀x ∃y P(x,y) ∧ ¬Q(x,y) ]
b. Negation of ∃x ∀y(P(x,y) → P(y,x)) = ∀x ∃y [ ¬P(x,y) ∨ ¬P(y,x) ] ∧ [P(x,y) ∨ P(y,x)]
c. Negation of ∃x ∃y P(x,y) ∧ ∀x ∀y Q(x,y) = ∀x ∀y ¬ [P(x,y)] ∨ ∃x ∃y ¬ [Q(x,y)]
Step-by-step explanation:
a.∃x ∀y(P(x,y) → Q(x,y))
Negation = ¬ [ ∃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) ]
Negation of ∃x ∀y(P(x,y) → Q(x,y)) = ∀x ∃y P(x,y) ∧ ¬Q(x,y) ]
b. ∃x ∀y(P(x,y) → P(y,x))
Negation = ¬ [ ∃x ∀y(P(x,y) → P(y,x)) ]
= ∀x ¬ [ ∀y(P(x,y) → P(y,x)) ]
= ∀x ∃y ¬ [ (P(x,y) → P(y,x)) ]
= ∀x ∃y ¬ [ ( P(x,y) ∧ P(y,x) ) ∨ ( ¬P(x,y) ∧ ¬P(y,x) )]
= ∀x ∃y ¬ [ P(x,y) ∧ P(y,x) ] ∧ ¬[ ¬P(x,y) ∧ ¬P(y,x) ]
= ∀x ∃y [ ¬P(x,y) ∨ ¬P(y,x) ] ∧ [ P(x,y) ∨ P(y,x) ]
∴ we get
Negation of ∃x ∀y(P(x,y) → P(y,x)) = ∀x ∃y [ ¬P(x,y) ∨ ¬P(y,x) ] ∧ [P(x,y) ∨ P(y,x)]
c. ∃x ∃y P(x,y) ∧ ∀x ∀y Q(x,y)
Negation = ¬ [ ∃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) ]
∴ we get
Negation of ∃x ∃y P(x,y) ∧ ∀x ∀y Q(x,y) = ∀x ∀y ¬ [ P(x,y) ] ∨ ∃x ∃y ¬ [ Q(x,y) ]
