1. Let A and B be two sets. If A CB, then A u B is equal to
Answers
Answer:
identity 1:- Let A and B be sets. Show that
A ∪ (B − A) = A ∪ B
Proof.
A ∪ (B − A) = A ∪ (B ∩ A
c
) set difference
= A ∪ (A
c ∩ B) commutative
= (A ∪ A
c
) ∩ (A ∪ B) distributive
= U ∩ (A ∪ B) complement
= A ∪ B identity.
Proof. Let x ∈ A ∪ (B − A). Then x ∈ A or x ∈ (B − A) by definition of
union. So x ∈ B and x 6∈ A (by set difference). But x ∈ A by previous
statement, so x ∈ A or x ∈ B. By definition of union, x ∈ (A ∪ B).
Step-by-step explanation:
Identity 2:- Let A and B be sets. Show that
(A ∩ B
c
)
c ∪ B = A
c ∪ B
Proof.
(A ∩ B
c
)
c ∪ B = (A
c ∪ (B
c
)
c
) ∪ B de Morgan’s
= (A
c ∪ B) ∪ B double complement
= A
c ∪ (B ∪ B) associative
= A
c ∪ B idempotent
Identity 3:-
. Let A, B and C be sets. Show that
(A − B) − C = A − (B ∪ C)
Proof.
(A − B) − C = (A ∩ B
c
) − C set difference
= (A ∩ B
c
) ∩ C
c
set difference
= A ∩ (B
c ∩ C
c
) associative
= A ∩ (B ∪ C)
c de Morgan’s
= A − (B ∪ C) set difference
Proof. Let x ∈ (A − B) − C. Then x ∈ (A − B) and x 6∈ C by definition of
set difference. Further, x ∈ A and x 6∈ B also by definition of set difference.
Thus x ∈ A and x 6∈ B and x 6∈ C, which implies x 6∈ (BorC). Hence,
x 6∈ (B∪C) by definition of union. Thus, given x ∈ A we have x ∈ A−(B∪C)
by definition of set difference.
1
Identity 4:-
gh h Let A, B and C be sets. Show that
(B − A) ∪ (C − A) = (B ∪ C) − A
Proof.
(B − A) ∪ (C − A) = (B ∩ A
c
) ∪ (C ∩ A
c
) set difference
= (B ∪ C) ∩ A
c distibutive
= (B ∪ C) − A set difference
Proof. Let x ∈ (B − A) ∪ (C − A). Then x ∈ (B − A) or x ∈ (C − A).
If x ∈ (B − A), then x ∈ B and x 6∈ A by definition of set difference. If
x ∈ (C − A), then x ∈ C and x 6∈ A by definition of set difference. Thus,
x ∈ B or x ∈ C and x 6∈ A. By definition of union, x ∈ (B∪C). By definition
of set difference, if x ∈ (B ∪ C) and x 6∈ A then x ∈ (B ∪ C) − A.