Rewrite each of these statements so that negations appear only within predicates (that is, so that no negation is outside a quantifier or an expression involving logical connectives). a) ¬∃y∃xP (x, y) b) ¬∀x∃yP (x, y) c) ¬∃y(Q(y) ∧ ∀x¬R(x, y)) d) ¬∃y(∃xR(x, y) ∨ ∀xS(x, y)) e) ¬∃y(∀x∃zT (x, y, z) ∨ ∃x∀zU (x, y, z))

Follows are the solution to the given point:

Step-by-step explanation:

In point a:

¬∃y∃xP (x, y)  

∀x∀y(>P(x,y))  

In point b:

¬∀x∃yP (x, y)

 ∃x∀y  ¬P(x,y)

In point c:

¬∃y(Q(y) ∧ ∀x¬R(x, y))  ∀y(> Q(y) V ∀ ¬ (¬R(x,y)))

∀y(¬Q(Y)) V ∃xR(x,y) )

In point d:

¬∃y(∃xR(x, y) ∨ ∀xS(x, y))  

∀y(∀x>R(x,y)) ∃x>s(x,y))

In point e:

¬∃y(∀x∃zT (x, y, z) ∨ ∃x∀zU (x, y, z))

∀y(∃x ∀z)>T(x,y,z) ∀x ∃z> V (x,y,z))  

find the magnitude of the trig term  

csc

(

x

)

−

1

by taking the absolute value of the coefficient.

1

find the lower bound of the range.

tap for more

y

=

−

2

find the upper bound of the range.

tap for more

y

=

0

the range is  

y

≤

−

2

or  

y

≥

0

.

(

−

∞

,

−

2

]

∪

[

0

,

∞

)

{

y

|

y

≤

−

2

,

y

≥

0

}

step-by-step explanation: