Are the converses of the following mentioned theorems also true? Let $X$ be any non-empty set
Let $Y$ be any non-empty set
Let $P(m)$ be any propositional statement containing object $m \in X$
Let $Q(n)$ be any propositional statement containing object $n \in Y$
Let $R(m,n)=P(m)\ \wedge\ Q(n)$
Then we know that the following two Theorems are correct:

First Theorem is the following:
$\Bigg\{[\forall\ m \in X,\ P(m)]\ \wedge\ [\forall\ n \in Y,\ Q(n)]\Bigg\}$
implies
$\Bigg\{[\forall\ m \in X,\forall\ n \in Y] [P(m)\ \wedge\ Q(n) ]\Bigg\}$
QUESTION 1: Is the converse of first theorem also true?

Second Theorem is the following:
$\Bigg\{[\exists\ m \in X,\ P(m)]\ \wedge\ [\exists\ n \in Y,\ Q(n)]\Bigg\}$
implies
$\Bigg\{[\exists\ m \in X,\exists\ n \in Y] [P(m)\ \wedge\ Q(n) ]\Bigg\}$
QUESTION 2: Is the converse of second theorem also true?

Showing converse of first theorem is not necessarily true
$\Bigg\{[\forall\ m \in X,\forall\ n \in Y] [P(m)\ \wedge\ Q(n) ]\Bigg\} \tag 1$
$\Rightarrow$
$\Bigg\{[m \in X\ \wedge\ n \in Y] \Rightarrow  [P(m)\ \wedge\ Q(n)]\Bigg\} \tag 2$
does not necessarily imply
$\Bigg\{[m \in X\ \Rightarrow\ P(m)] \wedge\  [n \in Y\ \Rightarrow\ Q(n)]\Bigg\} \tag 3$
$\Rightarrow$
$\Bigg\{[\forall\ m \in X,\ P(m)]\ \wedge\ [\forall\ n \in Y,\ Q(n)]\Bigg\} \tag 4$
By making a truth table, we can see that the statement in second line of proof does not necessarily imply the statement in third line of proof.
Therefore, the statement in first line of proof does not necessarily imply the statement in fourth line of proof.
Thus we have shown that the converse of first theorem is not necessarily true.

Am I correct? If no, please point out the mistakes in the above proof.
 A: The first converse is true if both sets are nonempty or both sets are empty.
The second one is always true.

The statement you say is not necessarily an implication actually is. Let me prove you the first converse provided both sets are nonempty (I don't go through your rewriting with $\implies$)
Assume $\Bigg\{[\forall\ m \in X,\forall\ n \in Y] [P(m)\ \wedge\ Q(n) ]\Bigg\}$.
Let $m\in X$. $Y$ is nonempty so there exists some $n\in Y$. Consider it. Then by assumption, $P(m) \wedge Q(n)$. In particular, $P(m)$. We have shown $\forall m\in X, P(m)$.
Let $n\in Y$. $X$ is nonempty so there exists some $m\in X$. Consider it. Then by assumption, $P(m) \wedge Q(n)$. In particular, $Q(n)$. We have shown $\forall n\in Y, P(n)$.
Finally we have shown $(\forall m\in X, P(m))\wedge (\forall n\in Y, P(n))$.
A: Your $(2)\not\Rightarrow(3)$ is correct and your truth table must have shown you that the implication fails exactly when
$$(n\notin Y\land m\in X\land\neg P(m))\lor(m\notin X\land n\in Y\land\neg Q(n)).$$
This indicates that the counterexamples to the converse of theorem 1 are exactly  when either $Y=\varnothing$ and $\exists m\in X\;\neg P(m)$, or $X=\varnothing$ and $\exists n\in Y\;\neg Q(n)$.
The same resoning (with truth tables) certainly showed you that theorem 2 and its converse are "true", as well as theorem 1.
