Answer:
A+A'C'
Explanation:
let F= AB+(AC)'+AB'C(AB+C)
=AB +A'C'+AB'CAB+AB'CC
as we know that BB'=0 ( multiplicative inverse )
CC=C and AA=A
putting BB'=0 , CC=C and AA=A in the expression of F we can find
F= AB +A'C'+0+AB'
F=AB+A'C'+AB'
F=AB+AB'+A'C'
taking A common
F=A(B+B')+A'C'
as B+B'=1 (additive inverse )
F=A+A'C'
