Theorem
A\(B ∩ C) = (A\B) ∪ (A\C)
where A, B and C are sets.
Proof
If x ∈ A\(B ∩ C), then x is in A, but x is not in (B ∩ C). Hence, x ∈ A and both x ∉ B and x ∉ C. So either x ∈ A and x ∉ B or x ∈ A and x ∉ C ⇒ x ∈ (A\B) or x ∈ (A\C) ⇒ x ∈ (A\B) ∪ (A\C).
If x ∈ (A\B) ∪ (A\C), then x ∈ (A\B) or x ∈ (A\C).
If x ∈ (A\B), then x ∈ A and x ∉ B. But (B ∩ C) ⊆ B (proof). Therefore, x ∉ (B ∩ C) ⇒ x ∈ A\(B ∩ C).
Likewise, if x ∈ (A\C), then x ∈ A and x ∉ C. But (B ∩ C) ⊆ C (proof). Therefore, x ∉ (B ∩ C) ⇒ x ∈ A\(B ∩ C).
Since the sets A\(B ∩ C) and (A\B) ∪ (A\C) contain the same elements, A\(B ∩ C) = (A\B) ∪ (A\C).
$ \blacksquare $
Note
Using the above result, we can prove that (A ∩ B)' = A' ∪ B' because:
(A ∩ B)' = S\(A ∩ B) = (S\A) ∪ (S\B) = A' ∪ B'.
References
- R. G. Bartle, D. R. Sherbert, "Sets and Functions" in "Introduction to Real Analysis", 3rd Edition, John Wiley and Sons, Inc. 2000. ch 1, pp 3.
