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I've been stuck on this one problem that should be really simple. If somebody could help me prove m ⇒ q from the premises p ⇒ q and m ⇒ p ∨ q, preferably using the Fitch system, I would greatly appreciate it.

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  1. $p \rightarrow q \qquad \qquad$ premise

  2. $m \rightarrow (p \lor q) \qquad$ premise

  3. $\qquad m \qquad \qquad$ Assumption

  4. $\qquad p \lor q \qquad $ $\rightarrow Elim \: 2,3$

  5. $\qquad \qquad p \qquad$ Assumption

  6. $\qquad \qquad q \qquad $$\rightarrow Elim \: 1,5$ (end of subproof)

  7. $\qquad \qquad q \qquad $Assumption (first and last line of second suproof)

  8. $\qquad q \qquad \qquad $$\lor Elim \: 4,5-6,7-7$

  9. $m \rightarrow q \qquad \qquad $$\rightarrow Intro \: 3-8$

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    I am a little confused about line 7. If we assume q, shouldn't it be in an additional subproof? The rest makes sense, although I haven't figured out how to make q => q in the software I'm using. – 2017-01-26
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    @AndrewGuo yes, exactly. I tried to indicate that: on line 5 you start a subproof, and the last lone of that subproof is line 6. Then on line 7 I start a second subproof, and line 7 is at the same time the last line of that subproof (yes, that is allowed .... If you don't like how that looks, you can always just reiterate p on line 8 from line 7, and make that the last line of the subproof. Oh. and you don't have to get $q\rightarrow q$: all you need is that subproof that starts with q and ends with q. i think i know the software you are using so i will add a pic later today – 2017-01-26
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    @AndrewGuo Added picture – 2017-01-26
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If $m$, then $p$ or $q$ by $m\Rightarrow p\lor q$. If not $p$, then $q$. If $p$ then $q$ by $p\Rightarrow q$. Hence $q$ in either case,