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.

• If $A$, then $A$. What can you deduce if $\bot\lor B$? Nov 30 '19 at 16:31
• Then either B or a contradiction or both are true? Nov 30 '19 at 16:34
• Don't you mean $\bot$ (falsum) instead of $\top$ (true)? Nov 30 '19 at 16:34
• i dont know how to do that sign Nov 30 '19 at 16:35
• \bot (bottom) Nov 30 '19 at 16:36

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:

$$1. A \lor (\bot \lor B) \ Premise$$

$$2. \quad A \ Assumption$$

$$3. \quad A \lor B \ \lor \ Intro \ 2$$

$$4. \quad \bot \lor B \ Assumption$$

$$5. \quad \quad \bot \ Assumption$$

$$6. \quad \quad B \ \bot \ Elim \ 5$$

$$7. \quad \quad B \ Assumption$$

$$8. \quad B \ \lor \ Elim \ 4,5-6,7-7$$

$$9. \quad A \lor B \ \lor \ Intro \ 8$$

$$10. A \lor B \ \lor \ Elim \ 1,2-3,4-9$$