Answer:
(X'U'Y) + (XUY')
Explanation:
Starting with [Xn(X'nY)]', we can use De Morgan's laws to simplify the expression:
[Xn(X'nY)]' = (Xn)' + (X'nY)'
Recall that Xn represents the logical operator "and", while X' represents "not X". Using these definitions, we can expand the expression:
(Xn)' + (X'nY)' = (X'U'Y) + (XUY')
where U represents the logical operator "or".
Therefore, [Xn(X'nY)]' simplifies to (X'U'Y) + (XUY').