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To show A' - B' = B - A
We show that A' – B' = ⊆ B - A and vice versa
Let, x ϵ A' – B'
⟹ x ϵ A' and x ∉ B'
⟹ x ∉ A and x ϵ B [∵ A ∩ A' = ∮ and B ∩ B' = ∮]
⟹ x ϵ B and x ∉ A
⟹ x ϵ B - A
This is true for all x ϵ A' - B'
Hence A' - B' ⊆ B - A
Conversely,
Let, x ϵ B - A
⟹ x ∉ B' and x ϵ A'
⟹ x ϵ A' and x ∉ B' [∵ B ∩ B' = ∮ and A ∩ A' = ∮]
⟹ x ϵ A' - B'
This is true for all x ϵ B - A
Hence B- A ⊆ A' - B'
∴ A' - B' = B - A Proved.
LHS = A ∩ (A' ∪ B)
= (A ∩ A') ∪ (A ∩ B) [∴ ∩ distributes over (i)]
= ∮ ∪ (A ∩ B) [∵ A ∩ A' = ∮]
= A ∩ B [∵ ∮ ∪ x = x for any set x]
= RHS
∴ LHS = RHS Proved.
For any sets A and B we have by De-morgan's laws
(A ∪ B)' = A' ∩ B', (A ∩ B)' = A' ∪ B'
Also
LHS = A - (A - B)
= A ∩ (A - B)'
= A ∩ (A ∩ B')'
= A ∩ (A' ∩ (B')') [By De-morgan's law]
= A ∩ (A' ∪ B) [∵ (B')' = B]
= (A ∩ A') ∪(A ∩ B)
= A ∩ B [∵ ∮ ∪ x = x, for and set x]
LHS = A ∩ (A ∪ B')
= A ∩ (A' ∩ B') [By De-morgan's law]
= (A ∩ A') ∩ B' [By associative law]
= ∮ ∩ B' [∵ A ∩ A' = ∮]
= ∮
RHS = A ∆ (A ∩ 8)
= (A - (A ∩ B)) ∪ (A ∩ B - A) [∵ E∆F = (E - F) ∪ (F - E)]
= (A ∩ (A ∩ B)') ∪ (A ∩ B ∩ A') [∵ E - F = E ∩ F']
= (A ∩ (A' ∪ B')) ∪ (A ∩ A' ∩ B) [By de-morgan's law & associative law]
= (A ∩ A') ∪ (A ∩ B') ∪ (∮ ∩ B) [∵ ∩ distributes over ∪ and A ∩ A' = ∮]
= ∮ ∪ (A ∩ B') ∪ ∮ [∵ ∮ ∩ B = ∮]
= A ∩ B' [∵ ∮ ∪ x = x for any set x]
= A - B [∵ A ∩ B' = A - B]
= LHS
We have, ACB
To show: C - B ⊂ C - A
Let, x ϵ C - B
⟹ x ϵ C and x ∉ B
⟹ x ϵ C and x ∉ A [∵ A ⊂ B]
⟹ x ϵ C - A
Thus, x ϵ C - B ⟹ x ϵ C- A
This is true for all x ϵ C-B
∴ C - B ⊂ C- A
(A ∪ B) – B = (A - B) ∪ (A - B)
= ∮ ∩ (A - B)
= A - B
A – (A ∩ B) = (A - A) ∩ (A - B)
= φ ∩ (A – B)
= A – B
Let x ϵ A - (A - B) ⟺ x ϵ A and x ∉ (A – B)
⟺ x ϵ A and X ϵ (A ∩ B)
⟺ x ϵ A ∩ (A ∩ B)
⟺ x ϵ (A ∩ B)
∴ A - (A – B) = (A ∩ B)
Let x ϵ A ∪ (B - A) ⟹ x ϵ A or x ϵ (B - A)
⟹ x ϵ A or x ϵ B and x ∉ A
⟹ x ϵ B
⟹ x ϵ (A ∪ B) [∵ B ⊂ (A ∪ B)]
This is true for all x ϵ A ∪ (B - A)
∴ A ∪ (B - A) ⊂ (A ∪ B) .........(1)
Let, x ϵ (A ∪ B)
⟹ x ϵ A or x ϵ B
⟹ x ϵ A or x ϵ (B – A) [∵ B ⊂ (B – A)]
∴ (A ∪ B) ⊂ A ∪ (B - A) ..... .(2)
From (1) and (2), we get
A ∪ (B - A) = (A ∪ B)
Let x ϵ A.
Then either x ϵ (A- B) or x ϵ (A ∩ B)
⟹ x ϵ(A – B) ∪ (A ∩ B)
∴ A ⊂ (A – B) ∪ (A ∩ B)..........(1)
Let x ϵ (A - B) ∪ (A ∩ B)
⟹ x ϵ (A – B) or x (A ∩ B)
⟹ x ϵ A and x ∉ B or x ϵ A and x ϵ B
⟹ x ϵ A
∴(A-B) ∪ (A ∩ B) ⊂ A ......(2)
(A - B) ∪ (A ∩ B) = A
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