Take the adaptive logic thus obtained and consider the example where ourpremise set is:
Before we get to our adaptive proof, observe three facts:



In the proof we first write down the premises:
1.  p ∨ q  Prem  ∅  
2.  ¬p  Prem  ∅  
3.  r  Prem  ∅  
4.  ¬r  Prem  ∅  
5.  r ∨ s  Prem  ∅  
Since p∧¬p is an abnormality and q ∨ (p∧¬p) LLLfollows from lines 1 and 2, we can apply Rc to lines 1 and 2 and conclude q, relying on the normal behavior of p ∧¬p:
6.  q  Rc: 1, 2  {p ∧¬p} 
7.  s  Rc: 4, 5  {r ∧¬r} 
However, we are faced with the following difficulty: Prem_{3} is clearly problematic. Our input data inform us that r doesn’t behave normally. In fact, according to our premises, both r and ¬r are true. But then, should we really conclude that s is true just because r ∨ s is true? The usual rationale behind the disjunctive syllogism (ϕ ∨ ψ,¬ϕ ⊢ ψ) is this: ϕ ∨ ψ tells us that at least one of the involved sentences is true; ¬ϕ, however, tells us that ϕ is false and hence it cannot be true. Therefore, the only option is that it is ψ that is true and its truth accounts for the truth of ϕ ∨ ψ. On the paraconsistent approach, however, a formula and its negation can be both true: the claim that ¬ϕ is true doesn’t entail the claim that ϕ cannot be true. For this reason, when both r and ¬r are true, the disjunctive syllogism applied to ¬r, r ∨ s might fail. For even if s is still (only) false, r ∨ s will still be true in virtue of r being true. Hence, our premise set LLLproves an abnormality: r ∧¬r.
8.  r ∧¬r  RU: 3, 4  ∅  
This also shows that r ∧¬r is not an abnormality on whose falsehood we can rely: it is actually derivable from our premises. But this, according to our marking regulations requires us to cancel any conclusion that relied on the normal behavior of r ∧¬r, which (in our case) means that we have to withdraw line 8 (which I mark by putting ‘⋎’ besides a line):
7.  s  Rc: 4, 5  {r ∧¬r}  ⋎ 
8.  r ∧¬r  RU: 3, 4  ∅  
So, as things stand, the only line with s as its formula has been canceled. It might be the case, however, that a formula actually is ALderivable from a set of premises even though it is a formula of a marked line. The fact that ϕ is a formula in a marked line only means that the way it is introduced in that line is unreliable. This doesn’t have to mean that ϕ is not ALderivable from the premises in general. Is it the case with s? A moment of consideration should convince us that it isn’t. From these premises, there is no other sensible way we could introduce s without relying on the normal behavior of r ∧¬r.
If we were simply reasoning in CLuN we also wouldn’t be allowed to introduce line 7 either, because we wouldn’t be allowed to use disjunctive syllogism in general. In our AL, however, we can still keep line 7 as long as the abnormality on which it depends is not CLuNderived from our premises. That is, q is finally derivable if no possible extension of our proof derives on ∅ a minimal Dabformula Dab(Δ) with (p ∧¬p) in Δ such that for no Δ′⊆ Δ, where (p ∧¬p) is not in Δ′, Dab(Δ′) is LLLderivable from the premises. Luckily, the question whether these formulas are CLuNderivable from Prem_{3} is decidable (as long as we stay on the level of propositional logic). Hence we can consider q to be finally derived, even though s is not derivable.
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