phi_270_text_4_3

4.3.xa. Exercise answers

	│A ∧ B	1
	├─
1 Ext	│A	(2)
1 Ext	│B
2 Wk	│A ∨ B	X,(3)
	│●
	├─
3 QED	│A ∨ B

	│A ∧ B	1
	├─
1 Ext	│A
1 Ext	│B	(2)
2 Wk	│B ∨ C	X,(3)
	│●
	├─
3 QED	│B ∨ C

	│A ∨ B	1
	│¬ A	(1)
	├─
1 MTP	│B	(2)
	│●
	├─
2 QED	│B

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

	│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

	│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

	│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

	│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)

	│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)

	│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

	│A ∨ B	2
	│¬ A ∨ C	3
	├─
	││¬ B	(2)
	│├─
2 MTP	││A	(3)
3 MTP	││C	(4)
	││●
	│├─
4 QED	││C	1
	├─
1 PE	│B ∨ C

	│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

	│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

	│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.

	│(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

	│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)

	│¬ (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)

	│¬ (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)

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		Ⓣ			Ⓣ		Ⓕ

	│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.

	│¬ (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 10 Sep 2011