In this sentence:
"No hero is cowardly and some soldiers are cowards."
Assuming h(x) = x is a hero
s(x) = x is a soldier
c(x) = x is a coward.
So the sentence is like this i think:
($\forall x\ (h(x) \longrightarrow \neg C(x)) \land (\exists y\ (s(y) \land C(y))$
In this case, are prenex formulas bellow the same thing?
$\forall x\ \exists y\ (\neg h(x) \lor \neg C(x)) \land ((s(y) \land C(y))$
$\exists y\ \forall x\ (\neg h(x) \lor \neg C(x)) \land ((s(y) \land C(y))$
On
Quite generally, for any formulas $\alpha(x)$ and $\beta(y)$ with only the indicated free variables, the formula
$$
(\forall x\,\alpha(x))\land(\exists y\,\beta(y))
$$
is equivalent to both of
$$
\forall x\exists y\,(\alpha(x)\land\beta(y))
$$
and
$$
\exists y\forall x\,(\alpha(x)\land\beta(y)).
$$
To clarify the other responses here,
$\forall x\ (h(x) \longrightarrow \neg C(x)) \land (\exists y\ (s(y) \land \neg C(y))$ is equivalent to $\forall x\ \exists y\ (\neg h(x) \lor \neg C(x)) \land ((s(y) \land \neg C(y))$, but not equivalent to $\exists x\ \forall y\ (\neg h(x) \lor \neg C(x)) \land ((s(y) \land \neg C(y))$