25. A
a a
B
ababab
AB
ab aab
2
2
3
= ∈
=
=
{ , , }
{
}
{ ,
}
26. A
a a
a n
B
ab n
B
ab n
n
n
n
*
*
{ , , ,
} { |
}
{( ) |
}
{( ) |
= ∈
=
≥
=
≥
=
≥
+
2
0
0
1
KK
}
28. Σ
Σ
+
=
− ∈
*
{ }
29. (a) L
L
ab bc aa ac cb
1
2
∪ = { , , , , }
(b) L
L
aa
1
2
∩ = { }
(c) L L
xy x L
y L
1
2
1
2
− =
∈ ∧ ∈
{ |
}
(d) L L
abaa abac abcb bcaa bcac bccb caaa caac cacb
1 2 = {
,
,
,
,
,
,
,
,
}
30. A
A
A
A
A
i
i
*
=
=
∪
∪
≥
0
1
2
0
KK
U
31. A
A
A
A
A
i
i
+
≥
=
=
∪
∪
1
2
3
1
KK
U
32. A = {a, aa}, B = {a} and C = {aa}, to prove that
A B C
AB AC
(
)
.
∩
⊂
∩
AB = {aa, aaa}, AC = {aaa, aaaa}
AB AC
aaa B C
A B C
A B C
AB AC
∩
=
∩ =
∩
=
⇒
∩
⊂
∩
{ },
, (
)
(
)
.
φ
φ
33. (a) 155 (b) 100
35. (a) We know, A B A
∩ ⊂ .
If A B
⊂ , then A B A
∩ = .
∴ ∪
∩
=
A
A B
A
(
)
.
(b) A
A B
A A
A B
A
A B
A
a
∩
∪
=
∩
∪
∩
= ∪
∩
=
(
) (
) (
)
(
)
( ( ))
by
(c) x A B
x A x B
x A x B
x A B
∈ −
⇔ ∈ ∧ ∉
⇔ ∈ ∧ ∈
⇔ ∈ ∩
(
)
(
)
∨
∈ −
= ∨
∈ ∩
∴ − = ∩
x
x
x A B
x A B
A B A B
[
(
)]
[
(
)]
.
(d) A
A B
A A
A B
U
A B
A B
∪
∧
=
∪
∩
∪
= ∩
∪
= ∪
(
) (
) (
)
(
)
.
(e) A
A B
A A
A B
A B
A B
∩
∪
=
∩
∪
∩
= ∪
∩
= ∩
(
) (
) (
)
(
)
.
φ
36. (a) No (b) No (c) No.
Answers to Exercises
305