I need to prove that the following argument is valid using Natural Deduction:

1.  $[\lnot (B \lor \lnot I) \rightarrow (\lnot L \land J)]$

2.  $[\lnot L \rightarrow (M \land B)]$

3.  $\lnot (B \lor \lnot I)$ 

$\therefore \quad    (M \lor E)$

I'm very new to this, so I'd appreciate very much if someone can help me though the process. What's listed above are the premises, from which I need to determine if they validly lead to the conclusion at the bottom.  


closed as off-topic by Hayden, Mauro ALLEGRANZA, Carl Mummert, user147263, user642796 Oct 1 '14 at 4:32

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    $\begingroup$ Hi! Simply posting homework-type questions is not encouraged on this site. You should explain what you have attempted so far and where you are stuck. $\endgroup$ – Eike Schulte Sep 30 '14 at 11:14

From $(1)$ and $(3)$, by modus ponens (aka $\rightarrow$-elimination), infer

$(4)\quad \lnot L \land J$.

From $(4)$ infer

$(5) \quad \lnot L,\;$ by $\land$-elimination.

From $(2)$ and $(5)$, by modus ponens (aka $\rightarrow$-elimination) infer

$(6)\quad M\land B$.

$(7)\quad M,\;$ using $(6)$ and $\land$-elimination.

$\therefore\quad M \lor E,\;$ from $(7)$ and $\lor$-Introduction.


What does $E$ mean? Other then that we have:

$\neg L \wedge J$ (using $\rightarrow$E on 3. and 1.)

$\neg L$ (using $\wedge E$)

$M \wedge B$ (using $\rightarrow$E on 2.)

$M$ (using $\wedge$E)

$M \vee E$ (using $\vee$I)

  • $\begingroup$ Huh. My answer is a duplicate. Should I delete it? $\endgroup$ – Stefan Perko Sep 30 '14 at 11:19
  • $\begingroup$ Not necessarily. The timestamps make it clear that your answer is independent. Sometimes it is a good idea to remove duplicates, but I doubt anyone thinks less of you, if you let this answer stay. $\endgroup$ – Jyrki Lahtonen Dec 30 '14 at 21:02

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