If then find using De Morgan’s Law.

$F(A,B,C) = A'.B'.C' + A'.B.C'$ By De Morgan's Law, $F' = (A'.B'.C')'.(A'.B.C')'$ Applying De Morgan's Law again to each term: $(A'.B'.C')' = A+B+C$ and $(A'.B.C')' = A+B'+C$ $F' = (A+B+C).(A+B'+C)$ (this can be further simplified using the distributive law: $(A+C)+B.B' = (A+C)+0 = A+C$, so $F' = A+C$)