6.3.xa. Exercise answers

Some of the derivations below are given twice, once using only the basic identity rules EC, DC, QED=, and Nc= and a second time using MPP= and similar extensions for equations of other rules (see 6.3.3); either approach is entirely acceptable.

1.
│Fa → Ga 1
│Fa (1)
│a = b a—b
├─
1 MPP │Ga (2)
│●
├─
2 QED= │Gb
2.
│Fa → Ga 2
│Fb (3)
│a = b a—b
├─
││¬ Ga (2)
│├─
2 MTT ││¬ Fa (3)
││●
│├─
3 Nc= ││⊥ 1
├─
1 IP │Ga
 
│Fa → Ga 1
│Fb (1)
│a = b a—b
├─
1 MPP= │Ga (2)
│●
├─
2 QED │Ga
3.
│Fa ∧ a = gb 1
├─
1 Ext │Fa (4)
1 Ext │a = gb a—gb, b, c, gc
│
││¬ F(gc) (4)
│├─
│││b = c a—gb—gc, b—c
││├─
│││●
││├─
4 Nc= │││⊥ 3
│├─
3 RAA ││¬ b = c 2
├─
2 CP │¬ F(gc) → ¬ b = c
4.
│Fa → G(fa) 3
│G(fb) → Hb 5
│a = b a—b, fa—fb
├─
││Fb (4)
│├─
│││¬ Ha (7)
││├─
││││●
│││├─
4 QED= ││││Fa 3
│││
││││G(fa) 6
│││├─
│││││●
││││├─
6 QED= │││││G(fb) 5
││││
│││││Hb (7)
││││├─
│││││●
││││├─
7 Nc= │││││⊥ 5
│││├─
5 RC ││││⊥ 3
││├─
3 RC │││⊥ 2
│├─
2 IP ││Ha 1
├─
1 CP │Fb → Ha
 
│Fa → G(fa) 2
│G(fb) → Hb 3
│a = b a—b, fa—fb
├─
││Fb (2)
│├─
2 MPP= ││G(fa) (3)
3 MPP= ││Hb (4)
││●
│├─
4 QED= ││Ha 1
├─
1 CP │Fb → Ha
5.
│fa = b
│fc = d a, b—fa, c, d—fc
├─
││a = c ∨ b = d 2
│├─
│││a = c a—c, b—fa—fc—d
││├─
│││●
││├─
3 EC │││fa = d 2
││
│││b = d a, fa—b—d—fc, c
││├─
│││●
││├─
4 EC │││fa = d 2
│├─
2 CP ││fa = d 1
├─
1 CP │(a = c ∨ b = d) → fa = d
6.
│v = b
│o = p o—p, b—v
│¬ Fvi (3)
├─
││Foi (3)
│├─
│││b = p o—p—b—v
││├─
│││●
││├─
3 Nc= │││⊥ 2
│├─
2 RAA ││¬ b = p 1
├─
1 CP │Foi → ¬ b = p
F: [ _ is from _ ]; v: the vice president; b: Barack Obama; c: Joe Biden; p: the president; t: Illinois
Glen Helman 17 Jul 2012