OBJETIVOS ESPECÍFICOS
5. DISEÑO Y CONSTRUCCIÓN DE LA UNIDAD PILOTO DE FLOCULACIÓN LASTRADA
To overcome the issues pointed out during the previous examples, we need to restrict the considered knowledge base. We do this as follows:
Definition 3.4.5 (Minimal inconsistency property). An inconsistent infinite knowledge base K satisfies the minimal inconsistency property if
• given a set H ∈ SI (K), there is a set H0 ⊆ H with H0 ∈ SImin(K),
• given a hitting set S of SImin(K), there is a minimal hitting set S0⊆ S of SImin(K).
Given this property, the collection of maximal consistent subsets of an infinite knowledge base is as well-behaving as in the finite case.
Proposition 3.4.6. Let K be an inconsistent infinite knowledge base satisfying the minimal inconsistency property. If H ⊆ K is a consistent subset of K, then there is a maximal consistent subsetH0 ∈ Cmax(K) with H ⊆ H0. In particular,H0 = K \ S for a minimal
hitting setS of SImin(K).
Proof. Let H ⊆ K be consistent. We show that S = K \ H is a hitting set of SImin(K).
Assume this is not the case, i. e., there is a set I ∈ SImin(K) with S ∩ I 6= ∅. Hence,
I ⊆ K \ S and in particular I ⊆ K \ S = H. Since I is strongly inconsistent, the set H with I ⊆ H is inconsistent, which contradicts our assumption. We thus infer that S = K\H is a hitting set of SImin(K). Due to the minimal inconsistency property, there is a minimal
hitting set S0 of SImin(K) with S0 ⊆ S.
It is left to show that H0 = K \ S0 is maximal consistent. This can be done similarly to the proof of Theorem 3.1.12: If there is a consistent set H00 with H0 ( H00, it is easy to show that K \ H00 is a hitting set of SImin(K) as well, which contradicts minimality
of S0. Thus, each proper superset of H0 is inconsistent. Furthermore, assume H0 is not consistent. Since each superset of H0 is inconsistent (as seen before), H0 is in particular strongly inconsistent. We utilize the minimal inconsistency property of K again to infer that there is a minimal strongly inconsistent set H00with H00⊆ H0. Since H00 ⊆ H0 = K \ S0,
we obtain H00 ∩ S0 = ∅, so S0 is not a hitting set of SI
min(K). This is a contradiction.
Therefore, H0∈ Cmax(K). Since H was an arbitrary consistent set, the claim follows.
The minimal inconsistency property ensures that consistent subsets can always be turned into maximal consistent ones. This leads to the question whether the converse holds as well, i. e., can we guarantee the minimal inconsistency property if each consistent subset H of K can be turned into a maximal consistent one? The answer to this particular question is negative, but this will not disturb us in any way. For this, recall Theorem 3.1.19, i. e., S is a minimal hitting set of coCmax(K) if and only if S ∈ SImin(K). Hence, to ensure that a
given set H ∈ SI (K) contains a minimal strongly inconsistent set, we need to ensure that a given hitting set S of coCmax(K) contains a minimal hitting set S0 ⊆ S.
Although this cannot be guaranteed in general as seen in Example 3.4.4, the minimal inconsistency property ensures the required structure of hitting sets of coCmax(K):
Proposition 3.4.7. Let K be an inconsistent infinite knowledge base satisfying the minimal inconsistency property. IfS is a hitting set of coCmax(K), then there is a minimal hitting setS0 ⊆ S of coCmax(K). In particular, S0∈ SImin(K).
Proof. To see this, we prove that S ∈ SI (K) if and only if S is a hitting set of coCmax(K).
“⇐”: Let S be a hitting set of coCmax(K). Assume for the sake of contradiction that S is
not in SI (K). Then, there is a consistent set H with S ⊆ H. According to Proposition 3.4.6, there is a maximal consistent set H0 ∈ Cmax(K) with H ⊆ H0. In particular,
S ∩ (K \ H0) ⊆ S ∩ (K \ H) = ∅. Hence, S is not a hitting set of coCmax(K), which is a contradiction.
“⇒”: Now let S ∈ SI (K). Similarly, assume S is not a hitting set of coCmax(K). Then,
there is a set H ∈ coCmax(K) with H ∩ S = ∅. By definition, K \ H is consistent and since
H ∩ S = ∅, we have S ⊆ K \ H which contradicts strong inconsistency of S.
The previous result motivates considering a dual counterpart to the minimal inconsistency property, defined as follows:
Definition 3.4.8 (Maximal consistency property). An inconsistent infinite knowledge base K satisfies the maximal consistency property if
• given a consistent set H ⊆ K, there is a set H0∈ C
max(K) with H ⊆ H0,
• given a hitting set S of coCmax(K), there is a minimal hitting set S0 of coCmax(K)
with S0 ⊆ S .
Combining Proposition 3.4.6 and Proposition 3.4.7, we get: If K satisfies the minimal hit- ting set property, then it satisfies the maximal consistency property as well. As already mentioned, the converse is also true:
Proposition 3.4.9. Let K be an inconsistent infinite knowledge base. If K satisfies the maximal consistency property, then it satisfies the minimal hitting set property as well. Proof. Let H ∈ SI (K). As seen in the proof of Proposition 3.4.7, this is the case if and only if H is a hitting set of coCmax(K). By assumption, there is a minimal hitting set H0of
coCmax(K) with H0 ⊆ H. Hence, H0 ∈ SImin(K).
Now assume we are given a hitting set S of SImin(K). Then, H = K \ S is consistent.
Again by assumption, there is a set H0 ∈ Cmax(K) with H ⊆ H0 and as usual we see that now, S0= K \ H0must be a minimal hitting set of SImin(K).
So, the results of this section can be summarized as follows.
Theorem 3.4.10. Let K be an inconsistent infinite knowledge base. Then, K satisfies the minimal inconsistency property if and only if it satisfies the maximal consistency property. Moreover,S is a minimal hitting set of SImin(K) if and only if K \ S ∈ Cmax(K) and S is
a minimal hitting set ofcoCmax(K) if and only if S ∈ SImin(K).
This result emphasizes the fact that infinite knowledge bases are similar in their spirit to finite ones, except for the lack of maximal resp. minimal sets with certain properties. When- ever we make appropriate assumptions about our knowledge base (for example the minimal inconsistency property), we can work with our hitting set duality as usual. The definition of the minimal inconsistency property is tailored ensures the existence of maximal consistent sets. Interestingly, it also yields an analogous result for the hitting sets of coCmax(K). The
converse Proposition 3.4.9 emphasizes the close link between the notions we investigate.
3.4.3 Compact Logics
Having established Theorem 3.4.10, a naturally arising questions is which significant ex- amples of logics satisfy the minimal inconsistency property. A rather popular result about inconsistency of propositional knowledge bases is the compactness theorem.
Theorem 3.4.11 (Compactness Theorem, see [94]). If K is a propositional knowledge base, thenK is consistent iff each finite subsets H ⊆ K is.
One would expect this property to ensure a pleasant behavior of inconsistent sets as well as their hitting sets. Indeed, it is easy to see that propositional knowledge bases satisfy the first item of the minimal inconsistency property: K is inconsistent iff there is a finite inconsistent subset H. Hence, given an inconsistent subset H ⊆ K, there is a finite minimal inconsistent subset contained in H.
For the second item of the minimal inconsistency property, we need to take the hitting sets into account. The compactness theorem ensures that each H ∈ SImin(K) is finite and
hence, S must be a hitting set of a collection of finite sets. In fact, this ensures that any hitting set can be turned into a minimal one. This result does not seem to be obvious and has, to the best of our knowledge, not been stated explicitly before.
Lemma 3.4.12. Let X be a set of finite subsets of a countable set X. If S is a hitting set of X , then there is a minimal hitting set S0 ⊆ S of X .
Proof. Note that X is a countable collection of sets, so we may assume X is of the form X = {Xi| i ∈ N} with Xi ⊆ X for each i ∈ N.
Let S be a hitting set of X . We construct a minimal hitting set S0 of X inductively as follows. Let us first consider X1. Since X1is finite, the set S ∩X1is finite as well. Consider
a maximal set Y1of removable elements in X1∩ S, i. e., let
Y1∈ max{Y ⊆ S ∩ X1 | S \ Y is a hitting set of X }
where at least one such maximal set exists since S ∩ X1is finite. Moreover, if we are given
Y1, . . . , Yn−1, then we set Yn∈ max ( Y ⊆ S \ n−1 [ i=1 Yi ! ∩ Xn| S \ n−1 [ i=1 Yi∪ Y ! is a hitting set of X ) (3.4) where again at least one such maximal set exists since
S \ n−1 [ i=1 Yi ! ∩ Xn
is finite. Note that the Yiare pairwise disjoint by definition. We claim that
S0 := S \ [
i∈N
Yi
!
is a minimal hitting set of X . First observe that S0 is indeed a hitting set, since in (3.4) the set Ynwe remove is constrained accordingly.
Now assume S0is not minimal. Hence, there is an element xm ∈ X such that S0\ {xm}
is a hitting set of X as well. Let Xnbe such that xm∈ Xn. We see that this contradicts the
construction of Yn. More precisely, since xm∈ S0we have xm ∈/ Sn−1i=1 Yiand hence,
Yn∪ {xm} ⊆ S \ n−1 [ i=1 Yi ! ∩ Xn holds. Clearly, S \ n−1 [ i=1 Yi∪ Yn∪ {xm} !
is a hitting set of X and thus, Ym ∈ max/ ( Y ⊆ S \ n−1 [ i=1 Yi ! ∩ Xn| S \ n−1 [ i=1 Yi∪ Y ! is a hitting set of X ) , which is a contradiction.
Equipped with this lemma, we see that propositional logic satisfies the minimal inconsis- tency property.
Theorem 3.4.13. Let K be an inconsistent infinite knowledge base of the propositional logic LP. ThenK satisfies both the minimal inconsistency as well as the maximal consistency
property.
Proof. Let H ⊆ K be strongly inconsistent. Since propositional logic is monotonic, this simply means that H is inconsistent. Due to the compactness theorem, H possesses a fi- nite inconsistent subset H0 ⊆ H. This implies that there is a minimal inconsistent subset H00 ⊆ H0. This is the first item of the minimal inconsistency property. Now consider SImin(K) = Imin(K). Since this a collection of finite sets of a countable set, we apply
Lemma 3.4.12 in order to obtain the second item of the minimal inconsistency property. Due to Theorem 3.4.10, the maximal consistency property is satisfied as well.
The reader may observe that the proof of Theorem 3.4.13 does not rely on specific properties of propositional logic, but only on the structural properties that can be inferred from the compactness theorem. So the following is a sufficient condition for a knowledge base to satisfy both the minimal inconsistency as well as the maximal consistency property. Definition 3.4.14 (The compactness property). An inconsistent infinite knowledge base K satisfies the compactness property if
• a subset H ⊆ K is strongly inconsistent iff it possesses a finite subset H0 ⊆ H with H0 ∈ SI (K).
The name “compactness property” is motivated by the fact that logics where every knowl- edge base satisfies the above property are called compact. We now have:
Theorem 3.4.15. Let K be an inconsistent infinite knowledge base which satisfies the com- pactness property. ThenK satisfies both the minimal inconsistency as well as the maximal consistency property.
The proof is similar to the proof of Theorem 3.4.13. Instead of applying the compactness theorem, we require K to possess the compactness property and hence, we infer the result analogously.
In a similar fashion we see that the following dual property also implies both the mini- mal inconsistency as well as the maximal consistency property.
Definition 3.4.16 (The co-compactness property). An inconsistent infinite knowledge base K satisfies the co-compactness property if
• for each consistent H ⊆ K there is a superset H ⊆ H0 with H0∈ Cmax(K) such that
K \ H0is finite.
Now we may proceed as in the above theorem. Given a consistent H ⊆ K, we can move to a maximal one H0 (first item of the maximal consistency property). Moreover, K \ H0 is finite for each H0 ∈ Cmax(K), and hence we can apply Lemma 3.4.12 which yields the
second item of the maximal consistency property. Thus:
Theorem 3.4.17. Let K be an inconsistent infinite knowledge base which satisfies the co- compactness property. Then, K satisfies both the minimal inconsistency as well as the maximal consistency property.
We want to mention that in general, properties like the minimal inconsistency property or the compactness property are rare, especially when considering non-monotonic logics. This becomes apparent in view of Example 3.4.2: First, this example works for nearly all relevant semantics for AFs in the sense that there is no non-empty extension (see Chapter 5 below). Second, it is straightforward to construct similar examples for ASP or other non- monotonic frameworks like Reiter’s default logic [95] via infinite chains of default-negated atoms. However, besides being interesting from a theoretical point of view, the results of this section can be applied to certain classes of knowledge bases. After all, the compactness propertyis inspired by the behavior of propositional logic and so does not appear unnatural. The compactness theorem is also true for e.g. first order or modal logic. Moreover, in Chapter 5 we will see how to apply our results to so-called finitary AFs.