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⇒ x ∈ A and x ∈ (B-C)

⇒ x ∈ A, and x ∈ B and x ∉ C

In either case, this implies x ∈ (A and B) but x ∉ (A and C)

⇒ x ∈ (A∩B) - (A∩C)

Hence, A ∩(B-C) ⊂ (A∩B) - (A∩C) ------- (1)

2) Again let y be an arbitrary element in (A∩B) - (A∩C)

y ∈(A∩B) but y ∉ (A∩C)

⇒ y ∈A, and y ∈ B but y ∈ A and y ∉ C

⇒ y ∈ A ∩(B-C)

So, (A∩B) - (A∩C) ⊂ A ∩(B-C) ---- (2)

Thus from (1) & (2), it is proved that

A ∩(B-C) = (A∩B) - (A∩C)

⇒ Assume

a= {1,2,3,4,5}

b= {3,4,5,6,7} c= {6,7,8,9}

LHS= a ∩(b-c) b-c= {3,4,5,6,7}-{6,7,8,9} b-c= {3,4,5} a ∩ (b-c)= {1,2,3,4,5} ∩{3,4,5}

= {3,4,5} (a∩b)- (a∩c)=∩{3,4,5} (a∩b)={1,2,3,4,5,}∩ {3,4,5,6,7} = {3,4,5}

a∩c= {1,2,3,4,5}∩{6,7,8,9} = {0} so - 2 (a∩b)-(a∩c) ={3,4,5}-{0} from 1

and 3 we get a∩(b-c)= (a∩b)-(a∩c) = {3,4,5}