4.3.xa. Exercise answers

1. a.
│A ∧ B 1
├─
1 Ext │A (2)
1 Ext │B
2 Wk │A ∨ B X,(3)
│●
├─
3 QED │A ∨ B
b.
│A ∧ B 1
├─
1 Ext │A
1 Ext │B (2)
2 Wk │B ∨ C X,(3)
│●
├─
3 QED │B ∨ C
c.
│A ∨ B 1
│¬ A (1)
├─
1 MTP │B (2)
│●
├─
2 QED │B
d. Although the following is a possible approach, the derivation in 4.2.xa is probably more natural:
│A ∨ (A ∧ B) 2
├─
││¬ A (2),(4)
│├─
2 MTP ││A ∧ B 3
3 Ext ││A (4)
3 Ext ││B
││●
│├─
4 Nc ││⊥ 1
├─
1 IP │A
e.
│A ∨ B 3
│¬ (A ∧ C) 2
│¬ (B ∧ C) 4
├─
││C (2),(5)
│├─
2 MPT ││¬ A (3)
3 MTP ││B (4)
4 MPT ││¬ C (5)
││●
│├─
5 Nc ││⊥ 1
├─
1 RAA │¬ C
f.
│A ∧ (B ∨ C) 1
├─
1 Ext │A (4)
1 Ext │B ∨ C 3
│
││¬ C (3)
│├─
3 MTP ││B (4)
4 Adj ││A ∧ B X,(5)
││●
│├─
5 QED ││A ∧ B 2
├─
2 PE │(A ∧ B) ∨ C
OR
│A ∧ (B ∨ C) 1
├─
1 Ext │A (3)
1 Ext │B ∨ C 2
│
││B (3)
│├─
3 Adj ││A ∧ B X,(4)
4 Wk ││(A ∧ B) ∨ C X,(5)
││●
│├─
5 QED ││(A ∧ B) ∨ C 2
│
││C (6)
│├─
6 Wk ││(A ∧ B) ∨ C X,(7)
││●
│├─
7 QED ││(A ∧ B) ∨ C 2
├─
2 PC │(A ∧ B) ∨ C
g.
│A ∨ B 1
│C (2),(5)
├─
││A (2)
│├─
2 Adj ││A ∧ C X,(3)
3 Wk ││(A ∧ C) ∨ (B ∧ C) X,(4)
││●
│├─
4 QED ││(A ∧ C) ∨ (B ∧ C) 1
│
││B (5)
│├─
5 Adj ││B ∧ C X,(6)
6 Wk ││(A ∧ C) ∨ (B ∧ C) X,(7)
││●
│├─
7 QED ││(A ∧ C) ∨ (B ∧ C) 1
├─
1 PC │(A ∧ C) ∨ (B ∧ C)
OR
│A ∨ B 1
│C (2),(4)
├─
││¬ (A ∧ C) 2
│├─
2 MPT ││¬ A (3)
3 MTP ││B (4)
4 Adj ││B ∧ C X,(5)
││●
│├─
5 QED ││B ∧ C 1
├─
1 PE │(A ∧ C) ∨ (B ∧ C)
h.
│A ∨ B 1
│¬ A ∨ C 2
├─
││A (2)
│├─
2 MTP ││C (3)
3 Wk ││B ∨ C X,(4)
││●
│├─
4 QED ││B ∨ C 1
│
││B (5)
│├─
5 Wk ││B ∨ C X,(6)
││●
│├─
6 QED ││B ∨ C 1
├─
1 PC │B ∨ C
OR
│A ∨ B 2
│¬ A ∨ C 3
├─
││¬ B (2)
│├─
2 MTP ││A (3)
3 MTP ││C (4)
││●
│├─
4 QED ││C 1
├─
1 PE │B ∨ C
i.
│A (2),(3)
├─
││¬ (A ∧ B) 2
│├─
2 MPT ││¬ B (3)
3 Adj ││A ∧ ¬ B X,(4)
││●
│├─
4 QED ││A ∧ ¬ B 1
├─
1 PE │(A ∧ B) ∨ (A ∧ ¬ B)
 
│(A ∧ B) ∨ (A ∧ ¬ B) 3
├─
││¬ A (2),(5)
│├─
2 Wk ││¬ (A ∧ B) X,(3)
3 MTP ││A ∧ ¬ B 4
4 Ext ││A (5)
4 Ext ││¬ B
││●
│├─
5 Nc ││⊥ 1
├─
1 IP │A
Although the derivation above for the second entailment is possible, the derivation for it in 4.2.xa is probably more natural
2. a.
│A ∨ A 2
├─
││¬ A (2),(3)
│├─
2 MTP ││A (3)
││●
│├─
3 Nc ││⊥ 1
├─
1 IP │A
Another somewhat artificial approach. 		 
│A (1)
├─
1 Wk │A ∨ A X,(2)
│●
├─
2 QED │A ∨ A
b.
│A ∨ B 1
├─
││A (2)
│├─
2 Wk ││B ∨ A X,(3)
││●
│├─
3 QED ││B ∨ A 1
│
││B (4)
│├─
4 Wk ││B ∨ A X,(5)
││●
│├─
5 QED ││B ∨ A 1
├─
1 PC │B ∨ A
 
│B ∨ A 2
├─
││¬ A (2)
│├─
2 MTP ││B (3)
││●
│├─
3 QED ││B 1
├─
1 PE │A ∨ B
As was the case with the derivations in 4.2.xa, each of the above approaches could have been used for both entailments.
  c.
│(A ∨ B) ∨ C 3
├─
││¬ A (4)
│├─
│││¬ C (3)
││├─
3 MTP │││A ∨ B 4
4 MTP │││B (5)
│││●
││├─
5 QED │││B 2
│├─
2 PE ││B ∨ C 1
├─
1 PE │A ∨ (B ∨ C)
The derivation at the right can be compared to the one in 4.2.3
│A ∨ (B ∨ C) 1
├─
││A (2)
│├─
2 Wk ││A ∨ B X,(3)
3 Wk ││(A ∨ B) ∨ C X,(4)
││●
│├─
4 QED ││(A ∨ B) ∨ C 1
│
││B ∨ C 5
│├─
│││B (6)
││├─
6 Wk │││A ∨ B X,(7)
7 Wk │││(A ∨ B) ∨ C X,(8)
│││●
││├─
8 QED │││(A ∨ B) ∨ C 5
││
│││C (9)
││├─
9 Wk │││(A ∨ B) ∨ C (10)
│││●
││├─
10 QED │││(A ∨ B) ∨ C 5
│├─
5 PC ││(A ∨ B) ∨ C 1
├─
1 PC │(A ∨ B) ∨ C
  d.
│A ∨ (B ∧ ¬ B) 2
├─
││¬ A (2)
│├─
2 MTP ││B ∧ ¬ B 3
3 Ext ││B (4)
3 Ext ││¬ B (4)
││●
│├─
4 Nc ││⊥ 4
├─
1 IP │A
 
│A (1)
├─
1 Wk │A ∨ (B ∧ ¬ B) X,(2)
│●
├─
2 QED │A ∨ (B ∧ ¬ B)
  e.
│¬ (A ∨ B) (4),(7)
├─
│││A (3)
││├─
3 Wk │││A ∨ B X,(4)
│││●
││├─
4 Nc │││⊥ 2
│├─
2 RAA ││¬ A 1
│
│││B (6)
││├─
6 Wk │││A ∨ B X,(7)
│││●
││├─
7 Nc │││⊥ 5
│├─
5 RAA ││¬ B 1
├─
1 Cnj │¬ A ∧ ¬ B
 
│¬ A ∧ ¬ B 1
├─
1 Ext │¬ A (3)
1 Ext │¬ B (4)
│
││A ∨ B 3
│├─
3 MTP ││B (4)
││●
│├─
4 Nc ││⊥ 2
├─
2 RAA │¬ (A ∨ B)
  f.
│¬ (A ∧ B) 2
├─
││A (2)
│├─
2 MPT ││¬ B (3)
││●
│├─
3 QED ││¬ B 1
├─
1 PE │¬ A ∨ ¬ B
 
│¬ A ∨ ¬ B 3
├─
││A ∧ B 2
│├─
2 Ext ││A (3)
2 Ext ││B (4)
3 MTP ││¬ B (4)
││●
│├─
4 Nc ││⊥ 1
├─
1 RAA │¬ (A ∧ B)
3. a. This derivation is unchanged from 4.2.xa
│A ∨ B 2
│A
├─
││B
│├─
│││A
││├─
│││○ A, B ⇏ ⊥
││├─
│││⊥ 2
││
│││B
││├─
│││○ A, B ⇏ ⊥
││├─
│││⊥ 2
│├─
2 PC ││⊥ 1
├─
1 RAA │¬ B


A B A ∨ B ,A /¬ B
T T Ⓣ Ⓣ Ⓕ
  b.
│A ∨ (B ∧ C) 3,8
├─
│││¬ A (3)
││├─
3 MTP │││B ∧ C 4
4 Ext │││B (5)
4 Ext │││C
│││●
││├─
5 QED │││B 2
│├─
2 PE ││A ∨ B 1
│
│││¬ C (7)
││├─
7 Wk │││¬ (B ∧ C) X,(8)
8 MTP │││A
│││○ A, ¬ C ⇏ ⊥
││├─
│││⊥ 9
│├─
6 IP ││C 1
├─
1 Cnj │(A ∨ B) ∧ C
 
│(A ∨ B) ∧ C 1
├─
1 Ext │A ∨ B 3
1 Ext │C (4)
│
││¬ A (3)
│├─
3 MTP ││B (4)
4 Adj ││B ∧ C X,(5)
││●
│├─
5 QED ││B ∧ C 2
├─
2 PE │A ∨ (B ∧ C)

Each of the following divides the one open gap:
A B C A ∨ (B ∧ C) / (A ∨ B) ∧ C
T T F Ⓣ F T Ⓕ
T F F Ⓣ F T Ⓕ
    Although the use of Wk and MTP shortens the whole first derivation, it actually delays the dead end, which would have been reached after stage 7 if the first premise had been exploited by PC in the second gap. As in 4.2.xa, the second derivation is unnecessary once a dead-end gap is found in the first.
  c.
│¬ (A ∨ B) (4)
├─
││A (3)
│├─
│││B
││├─
3 Wk │││A ∨ B X,(4)
│││●
││├─
4 Nc │││⊥ 2
│├─
2 RAA ││¬ B 1
├─
1 PE │¬ A ∨ ¬ B

The following divide the first and second open gap, respectively:
A B ¬ A ∨ ¬ B / ¬ (A ∨ B)
F T T Ⓣ F Ⓕ T
T F F Ⓣ T Ⓕ T
 
│¬ A ∨ ¬ B 2
├─
││A ∨ B 3,4
│├─
│││¬ A (3)
││├─
3 MTP │││B
│││○ ¬ A, B ⇏ ⊥
││├─
│││⊥ 2
││
│││¬ B (4)
││├─
4 MTP │││A
│││○ A, ¬ B ⇏ ⊥
││├─
│││⊥ 2
│├─
2 PC ││⊥ 1
├─
1 RAA │¬ (A ∨ B)
Glen Helman 25 Aug 2005