The contradiction sign in logic I got a logic question.
The premise is $(A \lor ( \bot \lor B))$ 
The goal is $A \lor B$. 
I can only use intro elim and reit as rules. I know that the contradiction sign can be used stand alone to show that there is a contradiction obviously but how does that work in this case. What does it do here?
Also I am guessing that for the proof I have to proof that the contradiction sign is Always false so I can say that B is Always true so I can use if an only if to equal the right side to B. But because I don't fully understand how the contradiction sign works in this case I don't know how to do that.
 A: Given that $\bot$ is always false, the only way for $\bot \lor B$ to be true is for $B$ to be true. 
Indeed, the expression $\bot \lor B$ is equivalent to just $B$, and thus $A \lor (\bot \lor B)$ is equivalent to $A \lor B$.
Ok, but how do you prove that using your rules of inference? You say that you have to prove that $\bot$ is always false, but that is typically a given. In fact, you must have some inference rule dealing with the $\bot$, and most likely that is:
$\bot$
$\therefore P \ \bot \ Elim$
where $P$ is any expression you want ... which is valid, because anything follows from a contradiction.
Other than that, the proof is really just a proof by cases, i.e. Use $\lor $ Elim:
$1. A \lor (\bot \lor B) \ Premise$
$2. \quad A \ Assumption$
$3. \quad A \lor B \ \lor \ Intro \ 2$
$4. \quad \bot \lor B \ Assumption$
$5. \quad \quad \bot \ Assumption$
$6. \quad \quad B \ \bot \ Elim \ 5$
$7. \quad \quad B \ Assumption$
$8. \quad B \ \lor \ Elim \ 4,5-6,7-7$
$9. \quad A \lor B \ \lor \ Intro \ 8$
$10. A \lor B \ \lor \ Elim \ 1,2-3,4-9$
