3.4.xa. Exercise answers

1. a.
│¬ B
├─
││A ∧ ¬ B 2
│├─
2 Ext ││A
2 Ext ││¬ B
││○ A, ¬ B ⊭ ⊥
│├─
││⊥ 1
├─
1 RAA │¬ (A ∧ ¬ B)
AB¬B/¬(A∧¬B)
TFⓉⒻTT
  b.
│¬ (A ∧ B) 3,8
├─
│││A (5)
││├─
│││││●
││││├─
5 QED │││││A 4
││││
││││││¬ B
│││││├─
││││││○ A, ¬ B ⊭ ⊥
│││││├─
││││││⊥ 6
││││├─
6 IP │││││B 4
│││├─
4 Cnj ││││A ∧ B 3
││├─
3 CR │││⊥ 2
│├─
2 RAA ││¬ A 1
│
│││B (11)
││├─
││││││¬ A
│││││├─
││││││○ ¬ A, B ⊭ ⊥
│││││├─
││││││⊥ 10
││││├─
10 IP │││││A 9
││││
│││││●
││││├─
11 QED │││││B 9
│││├─
9 Cnj ││││A ∧ B 8
││├─
8 CR │││⊥ 7
│├─
7 RAA ││¬ B 1
├─
1 Cnj │¬ A ∧ ¬ B
AB¬(A∧B)/¬A∧¬B
TFⓉFFⒻT(lurks n the first gap)
FTⓉFTⒻF(lurks in the second gap)

 

  c.
│¬ (A ∧ B) 3
│¬ (B ∧ C) 7
├─
││A ∧ C 2
│├─
2 Ext ││A
2 Ext ││C (10)
││
││││●
│││├─
5 QED ││││A 4
│││
│││││¬ B
││││├─
││││││││¬ B
│││││││├─
││││││││○ A, ¬ B, C ⊭ ⊥
│││││││├─
││││││││⊥ 9
││││││├─
9 IP │││││││B 8
││││││
│││││││●
││││││├─
10 QED │││││││C 8
│││││├─
8 Cnj ││││││B ∧ C 7
││││├─
7 CR │││││⊥ 6
│││├─
6 IP ││││B 4
││├─
4 Cnj │││A ∧ B 3
│├─
3 CR ││⊥ 1
├─
1 RAA │¬ (A ∧ C)
ABC¬(A∧B),¬(B∧C)/¬(A∧C)
TFTⓉFⓉFⒻT
2. a.
│¬ (A ∧ ¬ B) 2
├─
││¬ B (5)
│├─
│││││¬ A
││││├─
│││││○ ¬ A, ¬ B ⊭ ⊥
││││├─
│││││⊥ 4
│││├─
4 IP ││││A 3
│││
││││●
│││├─
5 QED ││││¬ B 3
││├─
3 Cnj │││A ∧ ¬ B 2
│├─
2 CR ││⊥ 1
├─
1 RAA │B
AB¬(A∧¬B)/B
FFⓉFTⒻ
b.
│¬ (A ∧ B) 3
├─
││B ∧ A 2
│├─
2 Ext ││B (6)
2 Ext ││A (5)
││
││││●
│││├─
5 QED ││││A 4
│││
││││●
│││├─
6 QED ││││B 4
││├─
4 Cnj │││A ∧ B 3
│├─
3 CR ││⊥ 1
├─
1 RAA │¬ (B ∧ A)
c.
│¬ (A ∧ ¬ B) 3
├─
││B ∧ ¬ A 2
│├─
2 Ext ││B
2 Ext ││¬ A
││
│││││¬ A
││││├─
│││││○ ¬ A, B ⊭ ⊥
││││├─
│││││⊥ 5
│││├─
5 IP ││││A 4
│││
│││││B
││││├─
│││││○ ¬ A, B ⊭ ⊥
││││├─
│││││⊥ 6
│││├─
6 IP ││││¬ B 4
││├─
4 Cnj │││A ∧ ¬ B 3
│├─
3 CR ││⊥ 1
├─
1 RAA │¬ (B ∧ ¬ A)
AB¬(A∧¬B)/¬(B∧¬A)
FTⓉFFⒻTT
  d.
│¬ (A ∧ B) 3,8
│¬ (B ∧ C) 12
│B (6),(10),(14)
├─
│││A (5)
││├─
│││││●
││││├─
5 QED │││││A 4
││││
│││││●
││││├─
6 QED │││││B 4
│││├─
4 Cnj ││││A ∧ B 3
││├─
3 CR │││⊥ 2
│├─
2 RAA ││¬ A 1
│
│││C (15)
││├─
││││││¬ A
│││││├─
││││││││●
│││││││├─
14 QED ││││││││B 13
│││││││
││││││││●
│││││││├─
15 QED ││││││││C 13
││││││├─
13 Cnj │││││││B ∧ C 12
│││││├─
12 CR ││││││⊥ 11
││││├─
11 IP │││││A 9
││││
│││││●
││││├─
10 QED │││││B 9
│││├─
9 Cnj ││││A ∧ B 8
││├─
8 CR │││⊥ 7
│├─
7 RAA ││¬ C 1
├─
1 Cnj │¬ A ∧ ¬ C
│¬ (A ∧ B) 3
│¬ (B ∧ C) 8
│B (6),(10)
├─
│││A (5)
││├─
│││││●
││││├─
5 QED │││││A 4
││││
│││││●
││││├─
6 QED │││││B 4
│││├─
4 Cnj ││││A ∧ B 3
││├─
3 CR │││⊥ 2
│├─
2 RAA ││¬ A 1
│
│││C (11)
││├─
│││││●
││││├─
10 QED │││││B 9
││││
│││││●
││││├─
11 QED │││││C 9
│││├─
9 Cnj ││││B ∧ C 8
││├─
8 CR │││⊥ 7
│├─
7 RAA ││¬ C 1
├─
1 Cnj │¬ A ∧ ¬ C

The derivation at the left above exploits resources in their order of appearance; while the alternative derivation at the right chooses, at stage 8, the resource that is most closely connected with other resources of the gap in which it is exploited. Notice that derivation at the left is eventually led, at stage 12, to exploit the same resource to the same effect.

e.
│¬ (A ∧ ¬ (B ∧ ¬ (C ∧ ¬ D))) 3
├─
││A ∧ ¬ (B ∧ D) 2
│├─
2 Ext ││A (5)
2 Ext ││¬ (B ∧ D) 8
││
││││●
│││├─
5 QED ││││A 4
│││
│││││B ∧ ¬ (C ∧ ¬ D) 7
││││├─
7 Ext │││││B (10)
7 Ext │││││¬ (C ∧ ¬ D) 12
│││││
│││││││●
││││││├─
10 QED │││││││B 9
││││││
││││││││¬ D (15)
│││││││├─
│││││││││││¬ C
││││││││││├─
│││││││││││○ A, B, ¬ C, ¬ D ⊭ ⊥
││││││││││├─
│││││││││││⊥ 14
│││││││││├─
14 IP ││││││││││C 13
│││││││││
││││││││││●
│││││││││├─
15 QED ││││││││││¬ D 13
││││││││├─
13 │││││││││C ∧ ¬ D 12
│││││││├─
12 CR ││││││││⊥ 11
││││││├─
11 IP │││││││D 9
│││││├─
9 Cnj ││││││B ∧ D 8
││││├─
8 CR │││││⊥ 6
│││├─
6 RAA ││││¬ (B ∧ ¬ (C ∧ ¬ D)) 4
││├─
4 Cnj │││A ∧ ¬ (B ∧ ¬ (C ∧ ¬ D)) 3
│├─
3 CR ││⊥ 1
├─
1 RAA │¬ (A ∧ ¬ (B ∧ D))
ABCD¬(A∧¬(B∧¬(C∧¬D)))/¬(A∧¬(B∧D))
TTFFⓉFFTTFTⒻTTF
Glen Helman 16 Jul 2012