If universal set = {x : x ∈ Z, -2 ≤ x < 4},

A = {x : -1 ≤ x < 3}, B = {x : 0 < x < 4} and

C = {x : - 2 ≤ x ≤ 0}; show that :

A - (B ∪ C) = (A - B) ∩ (A - C)

Universal set = {x : x ∈ Z, -2 ≤ x < 4}

Universal set = {-2, -1, 0, 1, 2, 3}

A = {x : -1 ≤ x < 3}

A = {-1, 0, 1, 2}

B = {x : 0 < x < 4}

B = {1, 2, 3}

C = {x : - 2 ≤ x ≤ 0}

C = {-2, -1, 0}

To prove:

A - (B ∪ C) = (A - B) ∩ (A - C)

Taking LHS:

A - (B ∪ C)

B ∪ C - contains all the elements in set B and C.

B ∪ C = {1, 2, 3} ∪ {-2, -1, 0}

B ∪ C = {-2, -1, 0, 1, 2, 3}

A - (B ∪ C) - contains all the elements in set A but not in B ∪ C.

A - (B ∪ C) = {-1, 0, 1, 2} - {-2, -1, 0, 1, 2, 3}

A - (B ∪ C) = { }

Taking RHS:

(A - B) ∩ (A - C)

A - B - contains all the elements in set A but not in B.

A - B = {-1, 0, 1, 2} - {1, 2, 3}

A - B = {-1, 0}

A - C - contains all the elements in set A but not in C.

A - C = {-1, 0, 1, 2} - {-2, -1, 0}

A - C = {1, 2}

(A - B) ∩ (A - C) - contains all the common elements in set (A - B) and (A - C).

(A - B) ∩ (A - C) = {-1, 0} ∩ {1, 2}

(A - B) ∩ (A - C) = { }

∴ LHS = RHS

∴ A - (B ∪ C) = (A - B) ∩ (A - C)