Phi 270
Fall 2013
site index
(
Site navigation is not working.
)
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
01 Aug 2013