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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.

VakarianWrex
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  • What are you trying to prove? Are $M \to J, A \to J, \lnot M \to A, A \to \lnot J, \vdash M, J, \lnot A$ all that was given? –  Oct 15 '14 at 00:57
  • Yes, so the Premises are on the left and the conclusion is on the right. The question is that there is a party and you have to fulfill the guests wishes so: (i) If mary goes, so will jane. (ii) If amy goes, so will jane (iii) If mary doesn't go amy will go (iv) If amy goes, jane will not go. So prove M, J and $\lnot$ A from the previous constraints that I mentioned – VakarianWrex Oct 15 '14 at 01:07
  • I see, the only thing I hadn't seen before was $\vdash$ –  Oct 15 '14 at 01:18
  • ahh yes, that comma at the end may have been what caused the confusion. Question has been fixed – VakarianWrex Oct 15 '14 at 01:19

2 Answers2

1

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 $0$:

  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.
Kenny Lau
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0

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].

Mauro ALLEGRANZA
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