Phi 270
Fall 2013
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3.4.2. Some examples of consistency

The aim of this subsection is to consider a few examples, but its title makes a further general point. A counterexample that lurks in a dead-end open gap will make its active resources all true. That is, it will show that they form a consistent set. Counterexamples to arguments in chapter 2 did that, too, since they made all active resources of any gap they lurked in true, but now that is the full significance of a counterexample lurking in a dead-end gap since the goal of the gap is ⊥ and is therefore automatically false.

Here is a simple example that exhibits a common pattern.

│¬ (A ∧ B) 2
│A (4)
├─
││¬ B
│├─
││││●
│││├─
4 QED ││││A 3
│││
│││││¬ B
││││├─
│││││○ A, ¬ B ⊭ ⊥
││││├─
│││││⊥ 5
│││├─
5 IP ││││B 3
││├─
3 Cnj │││A ∧ B 2
│├─
2 CR ││⊥ 1
├─
1 IP │B
AB¬(AB),A/B
TFF

It may seem odd to continue to stage 5 since, before IP is applied, the resources of the second gap are fully exploited and its goal is not among them. So, in this case, it is clear before stage 5 that the gap will not close. But, with enough thought, it would have been clear before stage 1 that some gap would not close so the simple fact that a dead-end gap can be foreseen is not grounds for declaring one. A dead-end gap is an indication of failure made fully explicit. What we count as fully explicit is a conventional matter, and we will treat as fully explicit only what cannot be made more explicit by the system of derivations. In this case, that requires the final use of IP (though the closure of the first gap at stage 4 might have been ignored).

Here is a somewhat longer example. It is developed following the most straightforward approach, in which resources are exploited in the order in which they appear (when there is a choice).

│¬ (A ∧ ¬ B) 3
│¬ (A ∧ ¬ C) 6, 11
├─
││B ∧ ¬ C 2
│├─
2 Ext ││B
2 Ext ││¬ C (9),(14)
││
│││││¬ A
││││├─
││││││││¬ A
│││││││├─
││││││││○ ¬ A, B, ¬ C ⊭ ⊥
│││││││├─
││││││││⊥ 8
││││││├─
8 IP │││││││A 7
││││││
│││││││●
││││││├─
9 QED │││││││¬ C 7
│││││├─
7 Cnj ││││││A ∧ ¬ C 6
││││├─
6 CR │││││⊥ 5
│││├─
5 IP ││││A 4
│││
│││││B
││││├─
││││││││¬ A
│││││││├─
││││││││○ ¬ A, B, ¬ C ⊭ ⊥
│││││││├─
││││││││⊥ 13
││││││├─
13 IP │││││││A 12
││││││
│││││││●
││││││├─
14 QED │││││││¬ C 12
│││││├─
12 Cnj ││││││A ∧ ¬ C 11
││││├─
11 RC │││││⊥ 10
│││├─
10 RAA ││││¬ B 4
││├─
4 Cnj │││A ∧ ¬ B 3
│├─
3 CR ││⊥ 1
├─
1 RAA │¬ (B ∧ ¬ C)
ABC¬(A¬B),¬(A¬C)/¬(B¬C)
FTFFFFTTT

The derivation could have been shortened significantly by reversing the order in which the first two resources were exploited, but it would have been shorter still (no matter what order these resources were exploited in) if we stopped after reaching a dead-end gap at stage 8. Stopping then would be perfectly legitimate, and the derivation is continued here only for the sake of the example. One reason for continuing a dervation after an open gap has been reached would be that we wanted, for some reason, to discover all counterexamples to the ultimate argument. In fact, in this case, there is only one such counterexample, so both open gaps lead us to the same thing.

Glen Helman 01 Aug 2013