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.

Here is yet another proof, using the fact that $\;P \Rightarrow Q\;$ can be rewritten as $\;P \lor Q \equiv Q\;$.  Introducing that equivalence makes possible the following proof which uses only equivalences:
\begin{align}
& A' \;\lor\; B' \\
\equiv & \qquad \text{"using $\;A \Rightarrow A'\;$ and $\;B \Rightarrow B'\;$, in the above form"} \\
& A \lor A' \;\lor\; B \lor B' \\
\equiv & \qquad \text{"symmetry of $\;\lor\;$; using $\;A \lor B\;$"} \\
& \text{true} \;\lor\; A' \lor B' \\
\equiv & \qquad \text{"simplify using $\;P \lor \text{true} \;\equiv\; \text{true}\;$"} \\
& \text{true} \\
\end{align}
