F= ~B(A + ~(AB))=~B(A + ~A + ~B))=~B(1 + ~B)=~B
F= (A + C)(A + ~C)(~B + C)=(AA + AC+ A~C + C~C)(~B + C)=
(A + A(C+~C)+0)(~B + C)=(A + A(1))(~B + C)=A(~B + C)
F=~(AB)+A~B + AB + BC=~A+~B+A(~B+B)+BC=~A+~B+A*1+BC=
=~A+A+~B+BC=1+~B+BC=1
F=D+ABC(~B+~C)=D+ABC~B+ABC~C=D+ACB~B+ABC~C=D+AC0+AB0=D+0+0=D