$(\exists x)(\forall y)[(a(x, y) \wedge b(x, y)) \wedge ¬c(x, y)]$
$\quad \equiv ¬(\forall x)¬(\forall y)[(a(x, y) \wedge b(x, y)) \wedge ¬c(x, y)]$
$\quad (\because (\exists x) F(x) = \neg \forall x \neg F(x))$
$\quad \equiv ¬(\forall x)(\exists y)¬[(a(x, y) \wedge b(x, y)) \wedge ¬c(x, y)]$
$\quad (\because (\forall x) F(x) = \neg \exists x \neg F(x)), \neg \neg F(x) = F(x))$
$\quad \equiv ¬(\forall x)(\exists y)[¬(a(x, y) \wedge b(x, y)) \vee c(x, y)]$
$\quad \equiv ¬(\forall x)(\exists y)[(a(x, y) \wedge b(x, y)) → c(x, y)]$
(C) choice.