Yes

Let x ∈ (A – B) ∩ (C – B)

⇒ x ∈ A – B and x ∈ C – B

⇒ (x ∈ A and x ∉ B) and (x ∈ C and x ∉ B)

⇒ (x ∈ A and x ∈ C) and x ∉ B

⇒ (x ∈ A ∩ C) and x ∉ B

⇒ x ∈ (A ∩ C) – B

So,

(A – B) ∩ (C – B) ⊂ (A ∩ C) – B ....(1)

Now, conversely

Let

y ∈ (A ∩ C) – B

⇒ y ∈ (A ∩ C) and y ∉ B

⇒ (y ∈ A and y ∈ C) and (y ∉ B)

⇒ (y ∈ A and y ∉ B) and (y ∈ C and y ∉ B)

⇒ y ∈ (A – B) and y ∈ (C – B)

⇒ y ∈ (A – B) ∩ (C – B)

So (A ∩ C) – B ⊂ (A – B) ∩ (C – B) ... (2)

From (1) and (2),

(A – B) ∩ (C – B) = (A ∩ C) – B