Write the negation of each of the following logical expressions so that all negations immediately precede predicates. In some cases, it may be necessary to apply one or more laws of propositional logic. a. ∃x ∀y(P(x, y) → Q(x, y))
b. ∃x ∀y(P(x, y) → P(y, x))
c. ∃x ∃y P(x, y) ∧ ∀x ∀y Q(x, y)

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) ]

a

step-by-step explanation:

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752