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I'm trying to attempt a proof which I think is quite similar to disjunctive syllogism, if not equivalent: $a \, \lor \, b \vdash \neg a \rightarrow b$

I've started with the first case of the disjunction by assuming $a$ and $\neg a$ to get a contradiction and therefore you can get $b$, which leads to the implication $\neg a \rightarrow b$. My issue is with the second case of just assuming $b$.

I tried just assuming $\neg a$ at that point but you can't build an implication out of that. Am I on the right track?

werner_b
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    Hint: you can do disjunction elimination inside of conditional introduction – Ten O'Four Sep 11 '21 at 05:17
  • From assumption $b$, by $(\to \text I)$, we have $\lnot a \to b$, and it is done. – Mauro ALLEGRANZA Sep 11 '21 at 08:10
  • See this post – Mauro ALLEGRANZA Sep 11 '21 at 08:11
  • @MauroALLEGRANZA but $b$ does not rely on $\neg a$, and an implication introduction is invalid. This is the case in the system I'm using which checks whether each step is valid – werner_b Sep 12 '21 at 03:51
  • @TenO'Four thanks for the hint, but I'm still confused. By eliminating the disjunction how does that lead to a conditional? Not sure how I can eliminate it at all. Is this separate to the first case which I mentioned in the question? – werner_b Sep 12 '21 at 03:52
  • If you have $\lnot a$ as a premise what can you use disjunction elimination to derive? As an aside, what you've said to Mauro isn't the full story. You can use a technique called augmentation. $\begin{array}{l}1.&Q&\text{Premise}\2.&P&\text{A for CP}\3.&P\land Q&\text{$\land$I 1, 2}\4.&Q&\text{$\land$E 3}\5.&P\to Q&\text{$\to$I 2, 4}\\end{array}$ – Ten O'Four Sep 12 '21 at 04:07
  • To be extra clear, some proof systems will allow you to use reiteration to bring Q into the scope of the cp, others don't have it as a rule and you have to use augmentation. It amounts to the same thing, though. – Ten O'Four Sep 12 '21 at 04:21
  • @TenO'Four hm okay, I think my understanding of disjunction elimination is flawed then because I'm still not entirely sure. I completely understand the intuition behind it and why it's true for both disjuncts, just struggling to find the exact steps to do it. I'll keep working on it with your hints in mind, thank you so much for helping – werner_b Sep 12 '21 at 04:51
  • I can give you an example of the disjunction elimination rule in action if you like? – Ten O'Four Sep 12 '21 at 05:01
  • You can simply assume $\lnot a$, re-iterate $b$ and then derive $\lnot a \to b$, discharging the assumption. This is a case of $(\to \text I)$ that allows the discharge of "unused" assumptions; the reason is that $b \to (\lnot a \to b)$ is a tautology. – Mauro ALLEGRANZA Sep 12 '21 at 09:20

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Hint: disjunction elimination rule in action. We'll show $P\lor Q\vdash Q\lor P$. The numbers in curly braces, $\{\}$, are the set of dependencies and the numbers in brackets, $(\;)$, are showing which dependency gets discharged.

$\begin{array}{} \{1\}&1.&P\lor Q&\text{P}\\ \{2\}&2.&P&\text{A for $\lor$E}\\ \{2\}&3.&Q\lor P&\text{2 $\lor$I}\\ \{4\}&4.&Q&\text{A for $\lor$E}\\ \{4\}&5.&Q\lor P&\text{4 $\lor$I}\\ \{1\}&6.&Q\lor P&\text{1, (2), 3, (4), 5 $\lor$E}\\ \end{array}$

Ten O'Four
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    thank you, I finally just figured it out with a combination of this and your description of augmentation. – werner_b Sep 12 '21 at 10:56
  • Well done If you're after an extra challenge, $\lnot a \to b\vdash a \lor b$ as well. If you get stuck just ask another question – Ten O'Four Sep 12 '21 at 11:51