Distributivity
: (
)
(
) (
)
(
)
(
) (
)
A B
C
A C
B C
A B
C
A C
B C
∪
∩ =
∩
∪
∩
∩
∪ =
∪
∩
∪
Absorp tion
: (
)
(
)
A B
A A
A B
A A
∪
∩ =
∩
∪ =
DeMorgan’s Laws
:
A B C
A B
A C
A B C
A B
A C
−
∪
=
−
∩
−
−
∩
=
−
∪
−
(
) (
) (
)
(
) (
) (
)
Ì Exam ple 0.1.1: Show that A B C
A B
A C
−
∪
=
−
∩
−
(
) (
) (
).
Solu tion
x A B C
x A
x B C
x A
x B
x C
x A
x B
∈ −
∪
⇒ ∈
∉ ∪
⇒ ∈
∉
∉
⇒ ∈
∉
(
)
(
)
and
and
and
and
and
and
and
(
)
(
) (
)
x A
x C
x A B
x A C
x A B
A C
∈
∉
⇒ ∈ −
∈ −
⇒ ∈ −
∩
−
There fore
A B C
A B
A C
−
∪
⊆ −
∩
−
(
) (
) (
)
(1)
Conversely,
x A B
A C
x A B
x A C
x A
x B
x A
x
∈ −
∩
−
⇒ ∈ −
∈ −
⇒ ∈
∉
∈
(
) (
)
(
)
(
and
and
and
and ∉
⇒ ∈
∉
∉
⇒ ∈
∉ ∪
⇒ ∈ −
∪
C
x A
x B
x C
x A
x B C
x A B C
)
(
)
(
)
and
and
and
Therefore, (
) (
)
(
)
A B
A C
A B C
−
∩
−
⊆ −
∪ .
Hence A B C
A B
A C
−
∪
=
−
∩
−
(
) (
) (
).
Ì Exam ple 0.1.2: Given sets A and B are the subsets of a universal set U,
prove that
(a) A B A B
− = ∩ ′
(b) A B A
− = , if and only if A B
∩ = ∅
(c) A B
− = ∅, if and only if A B
⊆ .
Solu tion
(a) Let x A B
∈ − . Then
x A B
x A
x B
x A
x B
x A B
∈ − ⇒ ∈
∉
⇒ ∈
∈
⇒ ∈ ∩ ′
and
and
A B A B
− ⊆ ∩ ′
(1)
Introduction
3