If one of the hypotheses holds, then one of the conclusions holds. (looking for a proof) Using a huge truth table, I proved the theorem below.
I cannot find a more elegant proof. I tried to rewrite expressions; e.g. using the distributive laws and the laws of absorption - to no avail. Is there another proof - or any hint? I know that the issue is very basic which makes me feel quite stupid.

Theorem
If we have


*

*$A \Rightarrow A'$ and

*$B \Rightarrow B'$ and

*$A \lor B$,


then we have $A' \lor B'$.

P.S.: I know that the theorem is simple and might be accepted without a proof. Still, I am looking for a rigorous proof.

So you want to prove $$((A \Rightarrow A') \wedge (B \Rightarrow B')) \Rightarrow (A \vee B \Rightarrow A' \vee B')$$
You can rewrite it as:
$$\begin{align}
& \equiv (\neg (A \Rightarrow A') \vee \neg (B \Rightarrow B')) \vee (A \vee B \Rightarrow A' \vee B') \\
& \equiv ((A \wedge \neg  A') \vee (B \wedge \neg B')) \vee ((\neg A \wedge \neg B) \vee (A' \vee B')) \\
& \equiv (A \wedge \neg A') \vee (B \wedge \neg B') \vee (\neg A \wedge \neg B) \vee A' \vee B'
\end{align}$$
Now it's just a matter of using distributivity law to regroup terms and eliminate all $(p \vee \neg p) \equiv \top$ to get $\top$ in the end.
