Question

Boolean Algebra
Simplify AB'(D'+C'D)+B(A+A'CD)

Answer 1

A′B(D′+C′D)+B(A+A′CD)

=A′B(D′+D)(D′+C′)+B(A+A′)(A+CD)

=A′B(1)(D′+C′)+B(1)(A+CD)

=A′BD′+A′BC′+AB+BCD

=B(A+A′D′+A′C′+CD)

=B[(A+D′)+(A+C′)+CD]

=B[A+D′+C′+CD]

=B[A+D′+(C′+C)(C′+D)]

=B[A+D′+(1)(C′+D)]

=B[A+D′+C′+D]

=B[A+C′+1]

=B(1)

=B