Krdytkyu

I'm struggling to proof this both if I use or introduction rule $lor_{I_1}$ (to work on $D$) or or introduction rule $lor_{I_2}$ (to work on $C lor E$). Could you help me?

We will work by contradicition, starting assuming :

1) $lnot [D lor (C lor E)]$ --- assumed [a]

2) $lnot D$ --- assumed [b]

3) $lnot E$ --- assumed [c]

4) $A$ --- from 3) and premise-3

5) $lnot D land A$ --- from 2) and 4)

6) $B$ --- from 5) and premise-2

7) $lnot B lor C$ --- from 4) and premise-1

Now we need $lor$-elim on 7)

8) $lnot B$ --- assumed [d1] from 7)

9) $bot$ --- contradiction ! with 6) and 8)

10) $C$ --- assumed [d2] from 7)

11) $C lor E$ --- from 10)

12) $D lor (C lor E)$ --- from 11)

13) $bot$ --- contradiction ! with 1) and 12)

We have derived $bot$ in both cases of the $lor$-elim; thus we have :

14) $bot$ --- from 8)-9) and 10)-13) and 7) by $lor$-elim, discharging assumptions [d1] and [d2]

15) $E$ --- from 3) and 14) by RAA and DN, discharging [c]

16) $C lor E$ --- from 15)

17) $D lor (C lor E)$ --- from 16)

18) $bot$ --- contradiction ! with 1) and 17)

19) $D$ --- from 2) and 18) by RAA and DN, discharging [b]

20) $D lor (C lor E)$ --- from 19)

21) $bot$ --- contradiction ! with 1) and 20)

22) $D lor (C lor E)$ --- from 1) and 21) by RAA and DN, discharging [a].

Natural deduction proof of $(A to lnot B lor C), ((lnot D land A) to B), (lnot E to A) vdash D lor (C lor E)$

Here is a skeleton; just flesh it out.   The subproofs are mostly proofs by reduction to absurdity, and a proof by cases.

$$deffitch#1#2{~~begin{array}{|l}#1\hline#2end{array}}fitch{(A to lnot B lor C)\ ((lnot D land A) to B)\ (lnot E to A) }{fitch{lnot(Dlor (Clor E))}{fitch{~}{~\~\fitch{~}{fitch{~}{~\~\bot}\~\~\Dlor(Blor E)\bot}\~\fitch{~}{~\Dlor(Clor E)\bot}\~\bot}\~\~\~\Dlor(Clor E)\bot}\~\Dlor (Clor E)}$$

$A implies lnot B lor C$ is equivalent to $lnot Blor C$ or $lnot A$

$lnot D land A implies B $ is equivalent to $B$ or $Dlor lnot A$

$lnot E implies A$ is equivalent to $A$ or $E$

So you want to prove that $lnot A lor lnot Blor C$, $lnot Alor Blor D$, $Alor E$ gives you $Clor Dlor E$. Can you show this last step?

Here is a proof using the Law of Excluded Middle (LEM). Links to the text explaining the terms and the proof checker are at the bottom.

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/

P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/