phi_270_text_6_4_xa

6.4.xa. Exercise answers

1.

a.

τ	Fτ
0	T
1	T
2	F

τ	Gτ
0	F
1	T
2	T

R	0	1	2
0	F	F	F
1	F	F	T
2	T	T	T

b.

c.

2.

a.

(F	a	∨	G	b)	→	R	a	b
F	3	T	T	4	Ⓕ	F	3	4

b.

R	(f	c	a)	(f	a	c)
Ⓣ	1	0	3	4	3	0

c.

f	a	b	=	f	b	a
0	3	4	Ⓕ	2	4	3

3.

a.

Without attachment rules:

	│a = a → Fa	2
	├─
	││¬ Fa	(2)
	│├─
2 MTT	││¬ a = a	(3)
	││●
	│├─
3 DC	││⊥	1
	├─
1 IP	│Fa

Using attachment rules:

	│a = a → Fa	2
	├─
1 CE	│a = a	X,(2)
2 MPP	│Fa	(3)
	│●
	├─
3 QED	│Fa

b.

	│¬ (Fa ∧ Fb)	3
	├─
	││¬ Fa
	│├─
	│││Fb	(3)
	││├─
3 MPT	│││¬ Fa
	│││○	¬ Fa,Fb ⊭ ⊥
	││├─
	│││⊥	2
	│├─
2 RAA	││¬ Fb	1
	├─
1 CP	│¬ Fa → ¬ Fb

range: 1, 2

a	b
1	2

τ	Fτ
1	F
2	T

¬	(F	a	∧	F	b)	/	¬	F	a	→	¬	F	b
Ⓣ	F	1	F	T	2		T	F	1	Ⓕ	F	T	2

c.

	│a = b ∨ b = a	1
	├─
	││a = b	a—b
	│├─
	│││●
	││├─
3 EC	│││a = b	2
	││
	│││●
	││├─
4 EC	│││b = a	2
	│├─
2 Cnj	││a = b ∧ b = a	1
	│
	││b = a	a—b
	│├─
	│││●
	││├─
6 EC	│││a = b	5
	││
	│││●
	││├─
7 EC	│││b = a	5
	│├─
5 Cnj	││a = b ∧ b = a	1
	├─
1 PC	│a = b ∧ b = a

d.

	│Fa → a = b	3
	│ga = b	a, b—ga
	│Ra(ga) → Fa	5
	│F(ga)
	├─
	││Raa	(6)
	│├─
	│││¬ R(ga)(ga)	(6)
	││├─
	│││││¬ Fa	(5)
	││││├─
5 MTT	│││││¬ Ra(ga)
	│││││○	b=ga,F(ga),Raa,¬ R(ga)(ga),
	│││││	b=ga,F(ga),¬ Fa,¬ Ra(ga) ⊭ ⊥
	││││├─
	│││││⊥	4
	│││├─
4 IP	││││Fa	3
	│││
	││││a = b	a—b—ga
	│││├─
	││││●
	│││├─
6 Nc=	││││⊥	3
	││├─
3 RC	│││⊥	2
	│├─
2 IP	││R(ga)(ga)	1
	├─
1 CP	│Raa → R(ga)(ga)

range: 1, 2

a	b
1	2

τ	gτ
1	2
2	1

τ	Fτ
1	F
2	T

R	1	2
1	T	F
2	F	F

F	a	→	a	=	b	,	g	a	=	b	,	R	a	(g	a)	→	F	a	,	F	(g	a)	/	R	a	a	→	R	(g	a)	(g	a)
F	1	Ⓣ	1	F	2		2	1	Ⓣ	2		F	1	2	1	Ⓣ	F	1		Ⓣ	2	1		T	1	1	Ⓕ	F	2	1	2	1

e.

	│a = b → Rac	3
	│¬ a = b → Rbc	2
	├─
	││¬ Rbc	(2),(4)
	│├─
2 MTT	││a = b	a—b, c; (3)
3 MPP	││Rac	(4)
	││●
	│├─
4 Nc=	││⊥	1
	├─
1 IP	│Rbc

Glen Helman 18 Jul 2012