I have to prove the conclusion $(A \to B) \vee (B \to C)$ using nothing but formal proof rules for propositional logic (so introduction and elimination rules). Here's what I've done so far.
Basically, I've assumed the negation of the conclusion and my goal of the sub-proof is to prove a contradiction. After assuming the negation, I was able to prove that $\neg(A \to B)$ and $\neg(B \to C)$ which means I can say that $\neg(A \to B) \wedge \neg(B \to C)$ is true under this assumption. The problem now is that I have no idea where to go from here. All I need to do is show a contradiction but I have no idea how I would get to that contradiction.
P.S. This proof doesn't have any premises.
You would not have a contradiction in your negated statement if both expressions $A \to B$ and $B\to C$ are false. $A\to B$ can only be false if $B$ is false. But if $B$ is false then $B\to C$ is always true. Therefore at least one of the expressions will be true, which makes the overall expression always false, thus leading to a contradiction. (When in doubt, just do truth tables).
You can also use the fact that the negation of $P\to Q$ is $P\wedge\neg Q$. Therefore we have $$A\wedge\neg B\wedge B\wedge\neg C$$
Since $\neg B\wedge B$ is always false, you have a contradiction (the overall negated expression is false no matter what the truth values of $A$, $B$, and $C$ are).
I'll let you add in all of the formal labelings of each step.