Question

Prove A-(B-C) = (A-B) U (A intersect C) for any sets A,B and C.

How do we prove it without using the Venn diagram?

Answer 1

A - (B - C) = (A - B) U (A ∩ C)

Proof:

L.H.S. = A - (B - C)

= A - (B ∩ C'), since A - B = A ∩ B'

= A ∩ (B ∩ C')'

= A ∩ {B' U (C')'}, since (A ∩ B)' = A' U B'

= A ∩ (B' U C)

= (A ∩ B') U (A ∩ C)

= (A - B) U (A ∩ C) = R.H.S.

∴ A - (B - C) = (A - B) U (A ∩ C)

This completes the proof.

Some properties of set algebra:

1. A U Φ = A, A ∩ Φ = Φ

2. A U U = U, A ∩ U = A

3. A U A = A, A ∩ A = A

4. A U (B U C) = (A U B) U C

5. A ∩ (B ∩ C) = (A ∩ B) ∩ C

6. A U (B ∩ C) = (A U B) ∩ (A U C)

7. A ∩ (B U C) = (A ∩ B) U (A ∩ C)

8. A U A' = U

9. A ∩ A' = Φ

10. (A U B)' = A' ∩ B'

11. (A ∩ B)' = A' U B'

12. A - B = A ∩ B'

13. A - (B ∩ C) = (A - B) U (A - C)

14. A - (B U C) = (A - B) ∩ (A - C)

15. A Δ B = (A - B) U (B - A)

16. A Δ B = B Δ A

Answer 2

Answer:

Step-by-step explanation:

LHS = A − (B−C)

=A − (B∩C')

=A∩(B∩C')'

=(A∩B')∪(A∩C)

=(A−B)∪(A∩C)

=RHS