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It seems to be quite simple but I've yet to solve it. There is this, Given p ⇒ q and m ⇒ p ∨ q, how would I prove m ⇒ q? but it does not satisfy the Fitch System I am using since the 7th and 8th step are not possible as assuming q after line 6 would form another subproof.

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Alexander Michalak
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  • So what rules are available to you? The proof provided in the post you link to uses fairly standard Fitch rules, so if that doesn't work for you you will really need to explain what rules you have, otherwise we can't help. – Bram28 Mar 28 '17 at 00:43
  • http://imgur.com/qQmIjFT I get to the point where, if I followed the directions in the other post, I would add another assumption on line 7 and then proceed to use an Or Elimination on lines 4,5-6,7-7. For some reason, this is not possible. – Alexander Michalak Mar 28 '17 at 02:10
  • Ah. It looks like a bug in the machine implementation. Perhaps you are required to discharge the assumption, with implication introduction to obtain $p\to q$, before proceeding to a second sub-proof? Then you would use disjunction elimination as $(p\vee q), p\to q, q\to q\vdash q$. – Graham Kemp Mar 28 '17 at 02:19
  • @GrahamKemp That's an idea! Alexander, try that ... Also, what program are you using here? is this available for free online somewhere? – Bram28 Mar 28 '17 at 02:20
  • http://intrologic.stanford.edu/exercises/exercise_04_04.html Knock yourselves out. – Alexander Michalak Mar 28 '17 at 02:23
  • @AlexanderMichalak Yup, just found it myself and just completed the proof. it is exactly what GrahamKemp said: you just need to do a conditional introduction on line 7 to get p=>q, then assume q on line 8, do a conditional introduction to get q =>q, and then do a Or elimination on lines 4,7,9, and conclusde with the conditional inteoduction to get m=>q – Bram28 Mar 28 '17 at 02:28
  • @Bram28 I was able to complete the proof by using the methods discussed previously in the comment tree. http://i.imgur.com/tlxS0WV.png – Alexander Michalak Mar 28 '17 at 02:33
  • @AlexanderMichalak Great! Yeah, that's exactly what I got. – Bram28 Mar 28 '17 at 02:36

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as assuming q after line 6 would form another subproof.

It is another subproof.   Trivially: assuming $q$, infers $q$.   It is the "first and last line of the second subproof." as stated.

However, I would use (with reiterations to make it clearer): $$\begin{array}{r|l:l} 1. & p\to q & \text{Premise 1}\\ 2. & m\to (p\vee q) & \text{Premise 2} \\ \hdashline 3. & \quad q & \text{Assumption} \\ \hline 4. & q\to q & 3,3, \text{Implication Introduction} \\ \hdashline 5. & \quad m & \text{Assumption} \\ 6. & \quad m\to (p\vee q) &2,\text{Reiteration} \\ 7. & \quad (p\vee q) & 5,6,\text{Implication Elimination} \\ 8. & \quad p\to q & 1, \text{Reiteration}\\ 9. & \quad q\to q & 4,\text{Reiteration} \\ 10. & \quad q & 7,8,9, \text{Disjunction Elimination} \\ \hline \Box & m\to q & 5,10,\text{Implication Introduction} \end{array}$$

Graham Kemp
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