1. |
Here's the way to do this one with DI. (You might instead have
assumed 'T' at line 2. This leads to a contradiction within the subderivation
and the conclusion of '~T' outside the subderivation. The rest is
easy.)
Premise |
|
1 |
|
[(T>L)&(~T>L)]&~L |
1 &E |
|
2 |
|
(T>L)&(~T>L) |
2 IM |
|
3 |
|
(~TvL)&(~T>L) |
3 IM |
|
4 |
|
(~TvL)&(~~TvL) |
4 DN |
|
5 |
|
(~TvL)&(TvL) |
5 CM |
|
6 |
|
(Lv~T)&(TvL) |
6 CM |
|
7 |
|
(Lv~T)&(LvT) |
7 DI |
|
8 |
|
Lv(~T&T) |
1 &E |
|
9 |
|
~L |
8,9 DS |
|
10 |
|
~T&T |
10 &E |
|
11 |
|
T |
10 &E |
|
12 |
|
~T |
|
2. |
Premise |
|
1 |
|
[A>(C&D)]&~(A>C) |
1 &E |
|
2 |
|
A>(C&D) |
2 IM |
|
3 |
|
~Av(C&D) |
3 DI |
|
4 |
|
(~AvC)&(~AvD) |
4 &E |
|
5 |
|
~AvC |
5 IM |
|
6 |
|
A>C |
1 &E |
|
7 |
|
~(A>C) |
|
3. |
Premise |
|
1 |
|
A>(G>L) |
Premise |
|
2 |
|
~(G>L)vK |
Premise |
|
3 |
|
~(~K>~A) |
2 IM |
|
4 |
|
(G>L)>K |
1,4 HS |
|
5 |
|
A>K |
3 TR |
|
6 |
|
~(A>K) |
|
4. |
Premise |
|
1 |
|
~(A>~B)&T |
|
Premise |
|
2 |
|
(A=~B)v(S&~T) |
|
Assumption |
|
3 |
|
....what if |
A=~B |
3 EQ |
|
4 |
|
....then... |
(A>~B)&(~B>A) |
4 &E |
|
5 |
|
....then... |
A>~B |
1 &E |
|
6 |
|
....then... |
~(A>~B) |
3-6 ~I |
|
7 |
|
~(A=~B) |
|
2,7 DS |
|
8 |
|
S&~T |
|
1 &E |
|
9 |
|
T |
|
8 &E |
|
10 |
|
~T |
|
|