Determining the truth value of certain quantifiers based on this proposition being false. Can you help me verify if I answered this question correctly?

Consider
$[(\forall x)(P(x)) \land (\exists x)(\lnot Q(x))] \implies \{(\forall x)(P(x)) \iff [\lnot(\forall x)(R(x)) \lor (\exists x)(S(x))]\}$
If you know that this proposition is false, then determine the
  truth values of the following:
$(\forall x)(P(x))$
$(\forall x)(Q(x))$
$(\exists x)(\lnot R(x))$
$(\exists x)(S(x))$

Since we know that this implication is false, it must be of the form $V_0\implies F_0$
Since the rightmost proposition is false, we should negate it. That is to say,
$\lnot \{(\forall x)(P(x)) \iff [\lnot(\forall x)(R(x)) \lor (\exists x)(S(x))]\}$
The negation of $P \iff Q$ is $(P \land \lnot Q) \lor (\lnot P \land Q)$, so the above becomes this:
$\{(\forall x)(P(x)) \land \lnot[\lnot(\forall x)(R(x)) \lor (\exists x)(S(x))]\} \lor \{\lnot(\forall x)(P(x)) \land [\lnot(\forall x)(R(x)) \lor (\exists x)(S(x))]\}$
$\{(\forall x)(P(x)) \land [(\forall x)(R(x)) \land \lnot (\exists x)(S(x))]\} \lor \{\lnot(\forall x)(P(x)) \land [\lnot(\forall x)(R(x)) \lor (\exists x)(S(x))]\}$
$\{(\forall x)(P(x)) \land [(\forall x)(R(x)) \land (\forall x)(\lnot S(x))]\} \lor \{(\exists x)(\lnot P(x)) \land [(\exists x)(\lnot R(x)) \lor (\exists x)(S(x))]\}$
Now, if you look at the rightmost part of this disjunction, you will see that we got a conjunction. One of the propositions in this conjunction is $(\exists x)(\lnot P(x))$, but that happens to be false because we know that $(\forall x)(P(x))$. So the whole conjunction is false, and by disjunctive syllogism we end up with only the leftmost part of the disjunction:
$\{(\forall x)(P(x)) \land [(\forall x)(R(x)) \land (\forall x)(\lnot S(x))]\}$
At the end, we got all of these facts:
$(\forall x)(P(x)) \land (\exists x)(\lnot Q(x)) \land (\forall x)(R(x)) \land (\forall x)(\lnot S(x))$
So the answers to the question would be:
$(\forall x)(P(x)) \equiv V_0$
$(\forall x)(Q(x)) \equiv  F_0$
$(\exists x)(\lnot R(x)) \equiv  F_0$
$(\exists x)(S(x)) \equiv  F_0$
 A: You can use Wolfram Alpha to generate a truth table, provided you know the correct syntax. Here is the table for your schema.
We identify $$ p = (\forall x) P(x)$$ $$q = (\forall x) Q(x)$$ $$ r = (\forall x) R(x) $$ $$s = (\exists x) S(x)$$
The table shows that there is only 1 assignment that makes the statement false, namely $$ p \equiv T, \ q \equiv F, \ r\equiv T, \  s \equiv F$$ which gives $$(\forall x) P(x) \equiv T$$ $$(\forall x) Q(x) \equiv F$$ $$\neg (\forall x) R(x) \equiv (\exists x)(\neg R(x)) \equiv F$$ $$(\exists x) S(x) \equiv F$$ the same as you surmised originally.
A: Your reasoning looks fine to me.  For comparison, here is a calculational approach which may shed some additional light.
$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \unicode{x201c}\text{#2}\unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\then}{\mathrel{\Rightarrow}}
\newcommand{\ref}[1]{\text{(#1)}}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
$First, let's simplify things by abbreviating the "target expressions" $\;(\forall x)(P(x))\;$, $\;(\forall x)(Q(x))\;$, $\;(\exists x)(\lnot R(x))\;$, and $\;(\exists x)(S(x))\;$ as $\;p\;$, $\;q\;$, $\;r\;$, and $\;s\;$, respectively.  By several appeals to DeMorgan, that allows us to write the original proposition as
$$
\tag 0
p \land \lnot q \;\then\; (p \;\equiv\; r \lor s)
$$
which is a lot easier on the eyes.
What does it mean for $\ref 0$ to be false?  Let's calculate:
$$\calc
\lnot(p \land \lnot q \;\then\; (p \;\equiv\; r \lor s))
\calcop\equiv{use antecedent $\;p\;$ in the consequent of $\;\then\;$; simplify $\;\true \equiv X\;$ to $\;X\;$}
\lnot(p \land \lnot q \;\then\; r \lor s)
\calcop\equiv{write $\;X \then Y\;$ as $\;\lnot X \lor Y\;$; DeMorgan}
\lnot(\lnot p \lor q \lor r \lor s)
\calcop\equiv{DeMorgan}
p \land \lnot q \land \lnot r \land \lnot s
\endcalc$$
Therefore $\ref 0$ being false implies, and is even equivalent to, $\;p\;$ being true and $\;q\;$, $\;r\;$ and $\;s\;$ all being false.
