B∩C={x,y}∪{a,c,y}={y}
A×(B∩C)={a,b}×{y}
={(a,y),(b,y)}
∴A×(B∩C)={(a,y),(b,y)} ..........(1)
A×B={a,b}×{x,y}={(a,x),(a,y),(b,x),(b,y)}
A×C={a,b}×{a,c,y}={(a,a),(a,c),(a,y),(b,a),(b,c),(b,y)}
(A×B)∩(A×C)={(a,x),(a,y),(b,x),(b,y)}∩{(a,a),(a,c),(a,y),(b,a),(b,c),(b,y)}
={(a,y),(b,y)} .........(2)
From (1) and (2) we have
A×(B∩C)=(A×B)∩(A×C)
Hence verified.