Question

Using set theory laws show that (B − A) ∪ (C − A) = (B ∪ C) – A

Answer 1

For any x ∈ (B-A) or x ∈ (C - A)   or both

When x∈(B-A)       ⇒     (x∈B) ∧ (x∉A)

⇒ (x ∈ B ∪ C)∧ (x∉A)

⇒ x ∈ (B ∪ C)  - A

When x∈(C-A)       ⇒     (x∈C) ∧ (x∉A)

⇒ (x ∈ B ∪ C)∧ (x∉A)

⇒ x ∈ (B ∪ C)  - A

Therefore, LHS ⊆ RHS

AWhen x∈(B-A)       ⇒     (x∈B) ∧ (x∉A)

⇒ (x ∈ B ∪ C)∧ (x∉A)

⇒ x ∈ (B ∪ C)  - A

And

For any x ∈ RHS , x ∈ (B∪C) and x∉A

When x ∈ B   and x ∉ A

( x ∈ B) ∧ ( x ∉ A) ⇒ x ∈ B - A

( x ∈ B) ∧ ( x ∉ A) ⇒ x ∈ (B - A) ∪ (C - A)

When x ∈ C   and x ∉ A

( x ∈ C) ∧ ( x ∉ A) ⇒ x ∈ C - A

( x ∈ C) ∧ ( x ∉ A) ⇒ x ∈ (B - A) ∪ (C - A)

Therefore,   RHS ⊆ LHS

With LHS ⊆ RHS and RHS⊆ LHS,  we can conclude that LHS = RHS

I hope it will help you.