Simplify the Boolean expression AB + (AC) '+AB'C (AB+C)

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'