Verified answer

We will use Extensionality. First, we must prove that every member of ( A - (B∩C) )

is a member of (A-B)∪(A-C) , and vice versa. Let D be an arbitrary object

(→) Assume D∈ ( A - (B∩C) ) . Then d∈A and D not∈ (B∩C).

So , ¬(D∈B&D∈C). By De Morgans law , it follows that D not∈ B or D not∈ C

Case 1. D not∈ B. In this case, D∈A and D not∈ B, so D ∈ A-B. By disjunction

introduction, ( D ∈ A-B∨D∈ A-C ). So, D ∈ (A-B)∪(A-C).

Case 2. D not∈ C. Similar to case 1.

Thus, D∈ (A-B)∪(A-C).

(←) Assume D ∈ (A-B)∪(A-C). So, either D∈ (A-B) or D ∈ (A-C).

Case 1. D∈ (A-B). Thus, D∈A and D not∈ B. Trivially, since D not∈ B.

¬(D∈B&D∈C). So, D not∈(B∩C). Since D∈A and D not∈(B∩C)

We have that D∈( A- (B∩C) )

Case 2. D∈ (A-C). Similar to case 1.

Thus, D∈ ( A- (B∩C) )