7.5.xa. Exercise answers

1.       instance for a   instance for b   instance for c
  a. ∀x Fx   Fa   Fb   Fc
  b. ∀y Fy   Fa   Fb   Fc
  c. ∀x Rxa   Raa   Rba   Rca
  d. ∀x Saxb   Saab   Sabb   Sacb
  e. ∀x ∀y Rxy   ∀y Ray   ∀y Rby   ∀y Rcy
  f. ∀x (Fx → Gx)   Fa → Ga   Fb → Gb   Fc → Gc
  g. ∀x (Fx → Gd)   Fa → Gd   Fb → Gd   Fc → Gd
  h. ∀x (Fx → ∀y Rxy)   Fa → ∀y Ray   Fb → ∀y Rby   Fc → ∀y Rcy
  i. ∀x (Fx → ∀x Rxx)   Fa → ∀x Rxx   Fb → ∀x Rxx   Fc → ∀x Rxx
2. a.
│∀x Fx a:1
│∀x (Fx → Gx) a:2
├─
1 UI │Fa (3)
2 UI │Fa → Ga 3
3 MPP │Ga (4)
│●
├─
4 QED │Ga
  b.
│∀x (Fx ∧ Gx) a:1, b:3
├─
1 UI │Fa ∧ Ga 2
2 Ext │Fa (5)
2 Ext │Ga
3 UI │Fb ∧ Gb 4
4 Ext │Fb
4 Ext │Gb (5)
5 Adj │Fa ∧ Gb (6)
│●
├─
6 QED │Fa ∧ Gb
  c.
│∀x Rxa b:1
│∀x (Rbx → Gx) a:2
├─
1 UI │Rba (3)
2 UI │Rba → Ga 3
3 MPP │Ga (4)
│●
├─
4 QED │Ga
  d.
│∀x Fx a:2
│∀x (Fx → Gx) a:3
├─
│ⓐ
2 UI ││Fa (4)
3 UI ││Fa → Ga 4
4 MPP ││Ga (5)
││●
│├─
5 QED ││Ga 1
├─
1 UG │∀x Gx
  e.
│∀x (Fx ∧ Gx) a:3,b:7
├─
││ⓐ
3 UI │││Fa ∧ Ga 4
4 Ext │││Fa
4 Ext │││Ga (5)
│││●
││├─
5 QED │││Fa 2
│├─
2 UG ││∀x Fx 1
│
││ⓑ
7 UI │││Fb ∧ Gb 8
8 Ext │││Fb
8 Ext │││Gb (9)
│││●
││├─
9 QED │││Gb 6
│├─
6 UG ││∀x Gx 1
├─
1 Cnj │∀x Fx ∧ ∀x Gx
 
│∀x Fx ∧ ∀x Gx 1
├─
1 Ext │∀x Fx a:3
1 Ext │∀x Gx a:4
│ⓐ
3 UI ││Fa (5)
4 UI ││Ga (5)
5 Adj ││Fa ∧ Ga X, (6)
││●
│├─
6 QED ││Fa ∧ Ga 2
├─
2 UG │∀x (Fx ∧ Gx)
    The term a could have been used again as the parameter of the second general argument of the derivation on the left since we require only that a parameter not appear outside its scope line in the gap which is developed by introducing the general argument, and the first general argument is boxed off from the gap that is developed by setting up the second one. But we will not be short of letters to use as parameters, so it will be easier to see that the requirement is satisfied if we use a new parameter for each general argument in a derivation.
  f.
│∀x ∀y Rxy a:1, b:3
├─
1 UI │∀y Ray b:2
2 UI │Rab (5)
3 UI │∀y Rby b:4, a:6
4 UI │Rbb (5)
5 Adj │Rab ∧ Rbb X, (7)
6 UI │Rba (7)
7 Adj │(Rab ∧ Rbb) ∧ Rba X, (8)
│●
├─
8 QED │(Rab ∧ Rbb) ∧ Rba
  g.
│∀x ∀y Rxy b:2
├─
│ⓑ
2 UI ││∀y Rby a:3
3 UI ││Rba (4)
││●
│├─
4 QED ││Rba 1
├─
1 UG │∀y Rya
Notice that the term a cannot be used as the parameter of the general argument in this derivation because it already appears in the gap (specifically, in its goal) when the general argument is introduced.
  h.
│∀x ∀y (Rxy → ¬ Ryx) a:3
├─
│ⓐ
│││Raa (5), (6)
││├─
3 UI │││∀y (Ray → ¬ Rya) a:4
4 UI │││Raa → ¬ Raa 5
5 MPP │││¬ Raa (6)
│││●
││├─
6 Nc │││⊥ 2
│├─
2 RAA ││¬ Raa 1
├─
1 UG │∀x ¬ Rxx
  i.
│∀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)
Glen Helman 25 Aug 2005