Am I assuming too much in this Natural Deduction proof? 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.
 A: 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.

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 $0$:


*

*$M \to J$: in all worlds reachable, whenever $M$ is known, $J$ is also known.

*$A \to J$: in all worlds reachable, whenever $A$ is known, $J$ is also known. (vacuously true)

*$\neg M \to A$ in all worlds reachable, whenever $M$ is not reachable, then $A$ is known (vacuously true, as $M$ is reachable in all worlds).

*$A \to \neg J$: in all worlds reachable, whenever $A$ is known, $\neg J$ is also known. (vacuously true)

*$M$ is not known, contradicting the proposed theorem.

A: 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].
