A = ${\displaystyle \{y:y=\frac{\text{a}+1}{2},\text{a}\in \text{W}\text{and}\text{a}\le 5\}}$

a = {0, 1, 2, 3, 4, 5}

⇒ y = ${\displaystyle \frac{\text{a}+1}{2}=\frac{1}{2}}$

y = ${\displaystyle \frac{1+1}{2}=\frac{2}{2}}$ = 1

y = ${\displaystyle \frac{2+1}{2}=\frac{3}{2}}$

y = ${\displaystyle \frac{3+1}{2}=\frac{4}{2}}$ = 2

y = ${\displaystyle \frac{4+1}{2}=\frac{5}{2}}$

y = ${\displaystyle \frac{5+1}{2}=\frac{6}{2}}$ = 3

∴ A = ${\displaystyle \{\frac{1}{2},1,\frac{3}{2},2,\frac{5}{2},3\}}$

B = ${\displaystyle \{y:y=\frac{2\text{n}–1}{2},\text{n}\in \text{W}\text{and}\text{n}<5\}}$

n = {0, 1, 2, 3, 4}

⇒ y = ${\displaystyle \frac{2\times 0-1}{2}=\frac{-1}{2}}$

y = ${\displaystyle \frac{2\times 1-1}{2}=\frac{1}{2}}$

y = ${\displaystyle \frac{2\times 2-1}{2}=\frac{3}{2}}$

y = ${\displaystyle \frac{2\times 3-1}{2}=\frac{5}{2}}$

y = ${\displaystyle \frac{2\times 4-1}{2}=\frac{7}{2}}$

∴ B = ${\displaystyle \{-\frac{1}{2},\frac{1}{2},\frac{3}{2},\frac{5}{2},\frac{7}{2}\}}$

C = ${\displaystyle \{-1,-\frac{1}{2},1,\frac{3}{2},2\}}$

B ∪ C = ${\displaystyle \{-1,-\frac{1}{2},\frac{1}{2},1,\frac{3}{2},2,\frac{5}{2},\frac{7}{2}\}}$

A − (B ∪ C) = {3}    ...(1)

A − B = {1, 2, 3}

A − C = ${\displaystyle \{\frac{1}{2},\frac{5}{2},3\}}$

(A – B) ∩ (A – C) = {3}  ...(2)

From (1) and (2), it is verified that

A – (B ∪ C) = (A – B) ∩ (A – C).