Given $p$ ⇒ $q$, use the Fitch System to prove ¬p ∨ q.

Question

Disclaimer: I'm a complete newbie to the site, and I haven't fully figured out how to format properly. I do not have enough reputation to use the meta sandbox thread, so in addition to asking my question, I may use some space below to test some stuff out...

edit: I now have enough reputation to use the meta forums

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Here's what I've done so far, with regards to the question:

1. $p ⇒ q$ (premise)
   
2. (tab)|$¬q$ (assumption)
   
3. (tab)| (tab) |$p$ (assumption)
   
4. (tab)| (tab) |$¬q$ (reiteration: 2)
   
5. (tab)|$p ⇒ ¬q$ (implication introduction: 3, 4)
   
6. (tab)|$¬p$ (negation introduction: 1,5)

I thought of "Or-ing" $¬p$ with $q$ (Or introduction: 6) but that would not be a complete proof, because I would still be within a subproof. What steps am I missing? Or, perhaps, was I going in the wrong direction in the first place?

Also, please do not edit the format of my question. Instead, I would appreciate formatting suggestions, especially including the following:

• how do I format a tab?
• how do I format logic symbols like "and"/"or", without using copy+paste.
• how do I format Fitch-style logic problems. I don't want to use "|" as a substitute for the solid line going down.

Answer 1

A bit long winded, due to the logic software's inability to do a contradiction intro:

1.  p => q            Premise
2.    ~(~p | q)       Assumption
3.      ~p            Assumption
4.      ~p | q        Or Introduction: 3
5.    ~p => ~p | q    Implication Introduction: 3, 4
6.     ~p             Assumption
7.     ~(~p | q)      Reiteration: 2
8.    ~p => ~(~p | q) Implication Introduction: 6, 7
9.    ~~p             Negation Introduction: 5, 8
10.   p               Negation Elimination: 9
11.   q               Implication Elimination: 1, 10
12.   ~p | q          Or Introduction: 11
13. ~(~p | q) => ~p | q        Implication Introduction: 2, 12
14.  ~(~p | q)                 Assumption
15. ~(~p | q) => ~(~p | q)     Implication Introduction: 14, 14
16. ~~(~p | q)                 Negation Introduction: 13, 15
17. ~p | q                     Negation Elimination: 16

Bam, done.

Answer 2

I am not sure why you start by assuming $\neg q$ ...

In stead, try a proof by contradiction: assume $\neg (\neg p \lor q)$, and derive a contradiction between that and your premise.

In general, if your goal is ever a disjunction $A \lor B$, there are basically 3 strategies:

1. If you're lucky, you already have (or can get to) $A$ or $B$ by itself ... so then you can just do $\lor \: Intro$

2. You have some other disjunction (e.g $C \lor D$) to work with. In that case, a fruitful approach may be to do a proof by case of the $C \lor D$: So do one subproof assuming $C$, and a second subproof assuming $D$: chances are that one of them leads to $A$, and that the other one leads to $B$. And then in either case you can do $\lor \: Intro$ at the end of the subproof to get $A \lor B$, which you can then pull out using $\lor \: Elim$

3. Do a proof by Contradiction, i.e. assume $\neg (A \lor B)$ and try to get a contradiction. The nice thing about this strategy is that after assuming $\neg (A \lor B)$, you can derive $\neg A$ as well as $\neg B$, so you get some nice 'small stuff' that you can use to combine with other premises you have on your way to a contradiction.