Section 1.4 Question 48
:
Show that
∀
xP
(
x
)
∨ ∀
xQ
(
x
) and
∀
x
∀
y
[
P
(
x
)
∨
Q
(
y
)], where all quantifiers
have the same nonempty domain, are logically equivalent.
Solution:
Since the scope of the second
∀
in
∀
xP
(
x
)
∨ ∀
xQ
(
x
) involves only
Q
(
x
), we
can rewrite
∀
xP
(
x
)
∨ ∀
xQ
(
x
) as
∀
P
(
x
)
∨ ∀
yQ
(
y
)
First, we assume that
∀
xP
(
x
)
∨ ∀
yQ
(
y
) is true.
(We want to show that
∀
x
∀
y
[
P
(
x
)
∨
Q
(
y
)]
is also true.)
Now either
∀
xP
(
x
) is true or
∀
yQ
(
y
) is
true. If
∀
xP
(
x
) is true, then
P
(
x
) is true for every
x
in the domain which
implies that
P
(
x
)
∨
Q
(
y
) is true for every
x
and
y
. So
∀
x
∀
y
[
P
(
x
)
∨
Q
(
y
)] is
true. If
∀
yQ
(
y
) is true, then
Q
(
y
) is true for every
y
in the domain which
implies that
P
(
x
)
∨
Q
(
y
) is true for every
x
and
y
. So
∀
x
∀
y
[
P
(
x
)
∨
Q
(
y
)] is
true.
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 Spring '10
 ADACHAN
 Logic, Logical connective, Jazz in the Domain, ∀xP

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