Phi 270
Fall 2013
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7.6.xa. Exercise answers

  a.
│∀x ∀y (Rxy → ¬ Ryx) a:6
├─
│ⓐ
││ⓑ
││││¬ a = b
│││├─
│││││Rab ∧ Rba 5
││││├─
5 Ext │││││Rab (8)
5 Ext │││││Rba (9)
6 UI │││││∀y (Ray → ¬ Rya) b:7
7 UI │││││Rab → ¬ Rba 8
8 MPP │││││¬ Rba (9)
│││││●
││││├─
9 Nc │││││⊥ 4
│││├─
4 RAA ││││¬ (Rab ∧ Rba) 3
││├─
3 CP │││¬ a = b → ¬ (Rab ∧ Rba) 2
│├─
2 UG ││∀y (¬ a = y → ¬ (Ray ∧ Rya)) 1
├─
1 UG │∀x ∀y (¬ x = y → ¬ (Rxy ∧ Ryx))
  b.
│∀x ∀y (¬ x = y → ¬ (Rxy ∧ Ryx)) a:5
│∀x ¬ Rxx a:9
├─
│ⓐ
││ⓑ
││││Rab (7),(10)
│││├─
│││││Rba (7)
││││├─
5 UI │││││∀y (¬ a = y → ¬ (Ray ∧ Rya)) b:6
6 UI │││││¬ a = b → ¬ (Rab ∧ Rba) 8
7 Adj │││││Rab ∧ Rba X,(8)
8 MTT │││││a = b a—b
9 UI │││││¬ Raa (10)
│││││●
││││├─
10 Nc= │││││⊥ 4
│││├─
4 RAA ││││¬ Rba 3
││├─
3 CP │││Rab → ¬ Rba 2
│├─
2 UG ││∀y (Ray → ¬ Rya) 1
├─
1 UG │∀x ∀y (Rxy → ¬ Ryx)
  c.
│∀x ∀y ∀z ((Rxy ∧ Ryz) → Rxz) a:5
│∀x ¬ Rxx a:10
├─
│ⓐ
││ⓑ
││││Rab (8)
│││├─
│││││Rba (8)
││││├─
5 UI │││││∀y ∀z ((Ray ∧ Ryz) → Raz) b:6
6 UI │││││∀z ((Rab ∧ Rbz) → Raz) a:7
7 UI │││││(Rab ∧ Rba) → Raa 9
8 Adj │││││Rab ∧ Rba X,(9)
9 MPP │││││Raa (11)
10 UI │││││¬ Raa (11)
│││││●
││││├─
11 Nc │││││⊥ 4
│││├─
4 RAA ││││¬ Rba 3
││├─
3 CP │││Rab → ¬ Rba 2
│├─
2 UG ││∀y (Ray → ¬ Rya) 1
├─
1 UG │∀x ∀y (Rxy → ¬ Ryx)
  d.
(∀x: Px) (∀y: Py) (∀z: Pz ∧ Lzx) Lyz
(∀x:Px) (∀y:Py) (Lxy → (∀z:Pz) (∀w:Pw) Lzw)
│∀x (Px → ∀y (Py → ∀z ((Pz ∧ Lzx) → Lyz))) b:10, a:17
├─
│ⓐ
│││Pa (15), (18)
││├─
│││ⓑ
│││││Pb (11)
││││├─
││││││Lab (15)
│││││├─
││││││ⓒ
││││││││Pc (20)
│││││││├─
││││││││ⓓ
││││││││││Pd (13), (22)
│││││││││├─
10 UI ││││││││││Pb → ∀y (Py → ∀z ((Pz ∧ Lzb) → Lyz)) 10
11 MPP ││││││││││∀y (Py → ∀z ((Pz ∧ Lzb) → Lyz)) d:12
12 UI ││││││││││Pd → ∀z ((Pz ∧ Lzb) → Ldz) 13
13 MPP ││││││││││∀z ((Pz ∧ Lzb) → Ldz) a:14
14 UI ││││││││││(Pa ∧ Lab) → Lda 16
15 Adj ││││││││││Pa ∧ Lab X, (16)
16 MPP ││││││││││Lda (22)
17 UI ││││││││││Pa → ∀y (Py → ∀z ((Pz ∧ Lza) → Lyz)) 18
18 MPP ││││││││││∀y (Py → ∀z ((Pz ∧ Lza) → Lyz)) c:19
19 UI ││││││││││Pc → ∀z ((Pz ∧ Lza) → Lcz) 20
20 MPP ││││││││││∀z ((Pz ∧ Lza) → Lcz) d:21
21 UI ││││││││││(Pd ∧ Lda) → Lcd) 23
22 Adj ││││││││││Pd ∧ Lda X, (23)
23 MPP ││││││││││Lcd (24)
││││││││││●
│││││││││├─
24 QED ││││││││││Lcd 9
││││││││├─
9 CP │││││││││Pd → Lcd 8
│││││││├─
8 UG ││││││││∀w (Pw → Lcw) 7
││││││├─
7 CP │││││││Pc → ∀w (Pw → Lcw)) 6
│││││├─
6 UG ││││││∀z (Pz → ∀w (Pw → Lzw)) 5
││││├─
5 CP │││││Lab → ∀z (Pz → ∀w (Pw → Lzw)) 4
│││├─
4 CP ││││Pb → (Lab → ∀z (Pz → ∀w (Pw → Lzw))) 3
││├─
3 UG │││∀y (Py → (Lay → ∀z (Pz → ∀w (Pw → Lzw)))) 2
│├─
2 CP ││Pa → ∀y (Py → (Lay → ∀z (Pz → ∀w (Pw → Lzw)))) 1
├─
1 UG │∀x (Px → ∀y (Py → (Lxy → ∀z (Pz → ∀w (Pw → Lzw)))))

It would be easy to get lost in this argument, but the basic structure has just three parts: planning what must be shown (stages 1-9) and then applying the premise twice (stages 10-16 and 17-23) to take us first from Lab to Lda and then from Lda to Lcd. After stage 9, we have Lcd as the goal and Lab among the resources, and we also know that a, b, c, and d are all people. The premise tells us that anyone who loves is loved by everyone. It will then follow from Lab that the predicate [L _ a] is true of everyone, and it will follow from any predication of [Ld _ ] of a person that Lcd. Since Lda is both [L _ a]d and [Ld _ ]a, it can link the two applications of the premise.

  e.
│∀x ∀y Rxy b:7
│∀x (∀y Ryx → (Fx → Gx)) a:3
├─
│ⓐ
│││Fa (10)
││├─
3 UI │││∀y Rya → (Fa → Ga) 5
│││
││││¬ Ga (11)
│││├─
│││││ⓑ
7 UI ││││││∀y Rby a:8
8 UI ││││││Rba (9)
││││││●
│││││├─
9 QED ││││││Rba 4
││││├─
6 UG │││││∀y Rya 5
││││
│││││Fa → Ga 10
││││├─
10 MPP │││││Ga (11)
│││││●
││││├─
11 Nc │││││⊥ 5
│││├─
5 RC ││││⊥ 4
││├─
4 IP │││Ga 2
│├─
2 CP ││Fa → Ga 1
├─
1 UG │∀x (Fx → Gx)

There is not much alternative to the using RC to exploit ∀y Rya → (Fa → Ga). Although ∀y Rya follows from the premise, it is not an instance of it and thus does not come by UI; and, although the resources Fa and ¬ Ga together entail ¬ (Fa → Ga), we have no attachment rule implementing this entailment. So we do not have an opportunity to apply either MPP or MTT.

  f.
(∀x: Rax) Sax
Pa ∧ ∀x ¬ Sax
(∀x: Px ∧ ∀y ¬ Rxy) ∀z Fxz
∀x Fax
│∀x (Rax → Sax) c:9
│Pa ∧ ∀x ¬ Sax 1
│∀x ((Px ∧ ∀y ¬ Rxy) → ∀z Fxz) a:3
├─
1 Ext │Pa (7)
1 Ext │∀x ¬ Sax c:10
│ⓑ
3 UI ││(Pa ∧ ∀y ¬ Ray) → ∀z Faz 5
││
│││¬ Fab (14)
││├─
│││││●
││││├─
7 QED │││││Pa 6
││││
│││││ⓒ
9 UI ││││││Rac → Sac 11
10 UI ││││││¬ Sac (11)
11 MTT ││││││¬ Rac (12)
││││││●
│││││├─
12 QED ││││││¬ Rac 8
││││├─
8 UG │││││∀y ¬ Ray 6
│││├─
6 Cnj ││││Pa ∧ ∀y ¬ Ray 5
│││
││││∀z Faz b:13
│││├─
13 UI ││││Fab (14)
││││●
│││├─
14 Nc ││││⊥ 5
││├─
5 RC │││⊥ 4
│├─
4 IP ││Fab 2
├─
2 UG │∀x Fax

There were many other approaches that might have been attempted at stage 3. The key to seeing the approach that was taken is thinking through the content of the resources at that point. Since we have Al is a person and Al said nothing (the resources added at stage 1), the first premise should allow us to conclude that Al is a person who remembered nothing. The third premise should thus allow us to reach the goal of showing that Al forgot b. Stage 3 is a first step along these lines but we are not be able to add the resource needed to apply MPP to this conditional, so stages 4 and 5 set out to exploit it to complete a reductio.

2. a.
Everyone watched every snake
Every cobra is a snake
Everyone watched every cobra
(∀x: Px) (∀y: Sy) Wxy
(∀x: Cx) Sx
(∀x: Px) (∀y: Cy) Wxy
│∀x (Px → ∀y (Sy → Wxy)) a:5
│∀x (Cx → Sx) b:7
├─
│ⓐ
│││Pa (6)
││├─
│││ⓑ
│││││Cb (8)
││││├─
5 UI │││││Pa → ∀y (Sy → Way) 6
6 MPP │││││∀y (Sy → Way) b:9
7 UI │││││Cb → Sb 8
8 MPP │││││Sb (10)
9 UI │││││Sb → Wab 10
10 MPP │││││Wab (11)
│││││●
││││├─
11 QED │││││Wab 4
│││├─
4 CP ││││Cb → Wab 3
││├─
3 UG │││∀y (Cy → Way) 2
│├─
2 CP ││Pa → ∀y (Cy → Way) 1
├─
1 UG │∀x (Px → ∀y (Cy → Wxy))
  b.
No one watched every snake
Every snake is a reptile
No one watched every reptile
(∀x: Px) ¬ (∀y: Sy) Wxy
(∀x: Sx) Rx
(∀x: Px) ¬ (∀y: Ry) Wxy
│∀x (Px → ¬ ∀y (Sy → Wxy)) a:3
│∀x (Sx → Rx) b:9
├─
│ⓐ
│││Pa (4)
││├─
3 UI │││Pa → ¬ ∀y (Sy → Way) 4
4 MPP │││¬ ∀y (Sy → Way) 6
│││
││││∀y (Ry → Way) b:11
│││├─
│││││ⓑ
│││││││Sb (10)
││││││├─
9 UI │││││││Sb → Rb 10
10 MPP │││││││Rb (12)
11 UI │││││││Rb → Wab 12
12 MPP │││││││Wab (13)
│││││││●
││││││├─
13 QED │││││││Wab 8
│││││├─
8 CP ││││││Sb → Wab 7
││││├─
7 UG │││││∀y (Sy → Way) 6
│││├─
6 CR ││││⊥ 5
││├─
5 RAA │││¬ ∀y (Ry → Way) 2
│├─
2 CP ││Pa → ¬ ∀y (Ry → Way) 1
├─
1 UG │∀x (Px → ¬ ∀y (Ry → Wxy))
  c.
No one watched any snake
Every cobra is a snake
No one watched any cobra
(∀x: Sx) (∀y: Py) ¬ Wyx
(∀x: Cx) Sx
(∀x: Cx) (∀y: Py) ¬ Wyx
│∀x (Sx → ∀y (Py → ¬ Wyx)) a:5
│∀x (Cx → Sx) a:3
├─
│ⓐ
│││Ca (4)
││├─
3 UI │││Ca → Sa 4
4 MPP │││Sa (6)
5 UI │││Sa → ∀x (Px → ¬ Wxa) 6
6 MPP │││∀x (Px → ¬ Wxa) (7)
│││●
││├─
7 QED │││∀x (Px → ¬ Wxa) 2
│├─
2 CP ││Ca → ∀x (Px → ¬ Wxa) 1
├─
1 UG │∀x (Cx → ∀y (Py → ¬ Wyx))

The relative simplicity of this derivation is due to the fact that the difference between the first premise and the conclusion is not deeply embedded in their structures.

  d.
Everyone who likes every snake was pleased
Every snake is a reptile
Everyone who likes every reptile was pleased
(∀x: Px ∧ (∀y: Sy) Lxy) Dx
(∀x: Sx) Rx
(∀x: Px ∧ (∀y: Ry) Lxy) Dx
│∀x ((Px ∧ ∀y (Sy → Lxy)) → Dx) a:4
│∀x (Sx → Rx) b:11
├─
│ⓐ
│││Pa ∧ ∀y (Ry → Lay) 3
││├─
3 Ext │││Pa (7)
3 Ext │││∀y (Ry → Lay) b:13
4 UI │││(Pa ∧ ∀y (Sy → Lay)) → Da 6
│││
││││¬ Da (6)
│││├─
6 MTT ││││¬ (Pa ∧ ∀y (Sy → Lay)) 7
7 MPT ││││¬ ∀y (Sy → Lay) 8
││││
│││││ⓑ
│││││││Sb (12)
││││││├─
11 UI │││││││Sb → Rb 12
12 MPP │││││││Rb (14)
13 UI │││││││Rb → Lab 14
14 MPP │││││││Lab (15)
│││││││●
││││││├─
15 QED │││││││Lab 10
│││││├─
10 CP ││││││Sb → Lab 9
││││├─
9 UG │││││∀y (Sy → Lay) 8
│││├─
8 CR ││││⊥ 5
││├─
5 IP │││Da 2
│├─
2 CP ││(Pa ∧ ∀y (Ry → Lay)) → Da 1
├─
1 UG │∀x ((Px ∧ (∀y: Ry) Lxy) → Dx)
  e.
Everyone who likes any snake was pleased
Every cobra is a snake
Everyone who likes any cobra was pleased
(∀x: Sx) (∀y: Py ∧ Lyx) Dy
(∀x: Cx) Sx
(∀x: Cx) (∀y: Py ∧ Lyx) Dy
│∀x (Sx → ∀y ((Py ∧ Lyx) → Dy)) a:5
│∀x (Cx → Sx) a:3
├─
│ⓐ
│││Ca (4)
││├─
3 UI │││Ca → Sa 4
4 MPP │││Sa (6)
5 UI │││Sa → ∀y ((Py ∧ Lya) → Dy) 6
6 MPP │││∀y ((Py ∧ Lya) → Dy) (7)
│││●
││├─
7 QED │││∀y ((Py ∧ Lya) → Dy) 2
│├─
2 CP ││Ca → ∀y ((Py ∧ Lya) → Dy) 1
├─
1 UG │∀x (Cx → ∀y ((Py ∧ Lyx) → Dy))
Glen Helman 01 Aug 2013