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