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.

One of the ways is to get rid of disjunctions using implication. $A \lor B$ is the same as $\lnot B \Rightarrow A$, and $A' \lor B'$ is the same as $\lnot B' \Rightarrow A'$. Also, $B \Rightarrow B'$ is the same as its contrapositive $\lnot B' \Rightarrow \lnot B$.
So, now you are given that $\lnot B' \Rightarrow \lnot B$, $\lnot B \Rightarrow A$, and $A \Rightarrow A'$. You need to show that $\lnot B' \Rightarrow A'$. But now it follows from the "transitivity" of implication.
PS: this isn't very rigorous, but the way to make it more accurate depends on how things are defined, what can be considered as already known, and what standard of rigour you require.
