Am I assuming too much in this Natural Deduction proof?

Question

So I need to prove the following using natural deduction:

$M \to J, A \to J, \lnot M \to A, A \to \lnot J \vdash M, J, \lnot A$

This is my proof so far:

1.) $M \to J$

2.) $A \to J$

3.) $\lnot M \to A$

4.) $A \to \lnot J$

5.) $(M \to J) \lor (A \to J) ----(\lor I 1,2)$

6.) $M ---- (\lor E 1,2,5)$ <- M is proven

7.) $J ---- (\lor E 1,2,5)$ <- J is proven

....(not sure how to prove $\lnot A$ yet)

So my question is, am I assuming to much? Am I doing this completely wrong? If so, where exactly am I assuming to much and do you have any hints or tips to lead me in the right direction? It seemed way to easy to prove M and J so it makes me think I'm jumping to conclusions.

Answer 1

From :

2) $A→J$

and

4) $A→¬J$

assuming : [a] $A$

we get $J$ and $\lnot J$ and thus, by $\land$-I, a contradiction :

$J \land \lnot J$.

Thus, by $\lnot$-E followed by $\lnot$-I we derive :

$\lnot A$

discharging [a].

Answer 2

In 6 and 7 you have wrongly used ∨E to obtain a proof of $M$ and a proof of $J$. It simply isn't how it works.

01. M→J    premise
02. A→J    premise
03. ¬M→A   premise
04. A→¬J   premise
  05. A    assumption
  06. J    MP 02 05
  07. ¬J   MP 04 05
  08. ⊥    contradiction 06 07
09. ¬A     ¬intro 05-08
  10. ¬M   assumption
  11. A    MP 03 10
  12. ⊥    contradiction 11 09
13. ¬¬M    ¬intro 10-12
14. M      ¬¬elim 13            *non-intuitionistic!
15. J      MP 01 14

The required results are 14, 15, 09.

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This does not work in intuitionistic logic.

Build a Kripke frame of $W=\{0\longrightarrow1\}$, where only $J$ is known at $0$ and only $M,J$ is known at $1$.

Then, in world $0$:

1. $M \to J$: in all worlds reachable, whenever $M$ is known, $J$ is also known.
2. $A \to J$: in all worlds reachable, whenever $A$ is known, $J$ is also known. (vacuously true)
3. $\neg M \to A$ in all worlds reachable, whenever $M$ is not reachable, then $A$ is known (vacuously true, as $M$ is reachable in all worlds).
4. $A \to \neg J$: in all worlds reachable, whenever $A$ is known, $\neg J$ is also known. (vacuously true)
5. $M$ is not known, contradicting the proposed theorem.