Finding a logical expression (under some constrains) s.t. it is equivalent to another one In this question, it was made clear, when
$\bullet$ some statement $A$ is stronger  than another statement $B$, namely if $A\Rightarrow B$ holds; and when the statement $A$ is weaker than another statement $B$, namely if $B\Rightarrow A$ holds.
and
$\bullet$ when a theorem $A\Rightarrow B$ (every mathematical theorem is  from the point of view of propositional logic an implication) is stronger another theorem $A'\Rightarrow B'$, namely if $[(A'\Rightarrow A) \ \land (B=B')]$ or $[(B\Rightarrow B') \ \land (A=A')]$ holds (i.e. the hypothesis of the first theorem is weaker (viewed as a statement) than the hypothesis if the second and the conclusions are equivalent or if the conclusion of the first theorem is stronger than the conclusion of the second theorem and the hypothesis' are equal).
(One could also define analogously the notation if "weaker" for theorems - for statements we had to define this notion - but for the sake of simplicity let's treat only the case of "stronger" in this question.)
Since a theorem is also a statement, I'm interested in the connection between the use of the words stronger for statements and theorems:
Let $A$,$A'$,$B$ and $B'$ be some statements and $$ \boldsymbol{X}:=[(A' \Rightarrow A) \land (B=B')] \lor [(B \Rightarrow B') \land (A=A')] $$ and $$\boldsymbol{Y}:=(A \Rightarrow B) \Rightarrow ( A' \Rightarrow B').$$ Then $\boldsymbol{X}$ models the statement, that "the theorem "$A\Rightarrow B$ is stronger than the theorem $A' \Rightarrow B'$", and $\boldsymbol{Y}$ models the statement that "the statement $A \Rightarrow B$ is stronger than the statement $A' \Rightarrow B'$".
Now it isn't very hard to prove that $$ \boldsymbol{X} \Rightarrow \boldsymbol{Y},$$
but the converse doesn't hold.
My question is, how would I have to modify the definition of one theorem being stronger than another, i.e. of $\boldsymbol{X}$, in a meaningful way that actually reflects how we think about one theorem being stronger than another ?
(So letting the definition of one theorem being stronger than another be something tautological, like $\boldsymbol{X}=\boldsymbol{Y}$, or something meaningless for our way to think of theorems, like $\boldsymbol{X}=\top$, is not allowed.)
For example, suppose we let $\boldsymbol{X}:=A' \Rightarrow A )\lor (B \Rightarrow B')$, meaning the hypothesis of the first theorem has to be weaker than that of the second theorem (but we don't say anything about how the conclusion of the theorems have to relate to each other) or the conclusion of the first theorem is stronger than the one of then second theorem but we don't say how the hypothesis' have to relate). The we get that $\boldsymbol{X}\Leftarrow \boldsymbol{Y}$, but this time the other direction fails.
BTW, we can also convince ourselves that modifying only $\boldsymbol{X}$ makes sense, since tampering with the definition of $\boldsymbol{Y}$ would be pointless because the meaning of the words "stronger" and "weaker" for statements is clear and since the correct way to model some mathematical theorem with hypothesis $A$ and conclusion $B$ via propositional logic is via the implication $A \Rightarrow B$.
I'm also aware that this might be a soft question, since it what one understands to be a  "meaningful way" that reflects how we think about one theorem being stronger than another.
 A: I'm afraid there is no "good" answer to that. As you said, $X$ is a sufficient condition for $Y$, but not vice versa. For example: If $B\Rightarrow B'$ and $A'\Rightarrow A$ (the assumption is weaker and the conclusion stronger) then $Y$ holds but $X$ doesn't. So we could generalize $X$ to
$$X_1\,:=\,(B\Rightarrow B') \land (A'\Rightarrow A)$$
This is probably the best where we can get, but still $X_1$ isn't equivalent to $Y$. In fact, it cannot be. We cannot really compare two implicative theorems just by looking at their assumptions and conclusions separately. The reason is that we can have one theorem stronger than another without such a direct relationship between their assumptions and conclusions. Consider these formulas:
\begin{align}
U &\,:=\, A \Rightarrow B \\
V &\,:=\, ((A\lor C)\land D) \Rightarrow ((B\lor C)\land D)
\end{align}
$U$ is stronger than $V$ because $\vdash U\Rightarrow V$ (i.e. in the sense of your $Y$). But there is no reasonable correlation between $A$ and $(A\lor C)\land D$ or $B$ and $(B\lor C)\land D$. Depending on other circumstances ($C$ and $D$), $A$ can be stronger or weaker than $(A\lor C)\land D$.
So $Y$ (or something equivalent to it) is the only reasonable way to say that one theorem is stronger than another.
