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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.

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

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

$2. \quad A \ Assumption$

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

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

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

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

$7. \quad \quad B \ Assumption$

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

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

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

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