Question

using algebraic manipulation solve:
A(A'+B)(A'B'+C)(A'B'C'+D)=ABCD

Answer 1

Greetings!!

A(A’+B)(A’B’+C)(A’B’C’+D)

=(AA’+AB) (A’B’+C)(A’B’C’+D) //used distributive law

= AB(A’B’+C)(A’B’C’+D)              //since AA’=0.

=(ABA’B’+ABC)( A’B’C’+D)          //ABA’B’=0

=ABC(A’B’C’+D)

=ABC A’B’C’+ABCD                       //ABC A’B’C’=0

=ABCD

Hope this helps