CÁLCULO DEL COEFICIENTE DE TORRENCIALIDAD (Kb)

Now we extend the proof systemΣ_MLTL− by axioms that characterise the “keep”-

operators and call the new system ΣMLTL. The additional axioms are drawn to- gether in figure 4.6. We already have defined in the previous section the mappings τandκ for all pMLTL-formulas. All the lemmas proved there concerning these mappings hold also for pMLTL.

All axioms and rules ofΣ_MLTL−

(ax12)`keep_a⇒(a.α⇔ da.α) forα∈N∗

(ax13)`keep_a∧a.b⇒keep_b

(ax14)`a[false]∧ da[false]⇒keep_a

(ax15)`a[keep_b]∧a.b⇒keep_b∧ da.b

(ax16)`keepb∧a.b∧ da.b⇒a[keepb]

(ax17)`a.b[false]∧ da.b[false]⇒a[keep<sub>b</sub>] (ax18)`ahkeep_bi ∧a.b[false]⇒ da.b[false]

Figure 4.6: The systemΣMLTL

We want to show thatΣMLTL is a complete axiomatisation for pMLTL. Again, we assume a finite and consistent set of formulas and construct a model for it. First we extend the mappingθin the following way:

θ(

_F

):= 15

[

i=1

whereθ1, . . . ,θ8are defined as before andθ9, . . . ,θ15 as given in fig. 4.7.

θ9(

_F

) = {a.αhtruei|keep_a∈

_F

anda.αhtruei ∈

_F

} θ10(

_F

) = {a.α[false]|keep_a ∈

_F

anda.α[false]∈

_F

} θ11(

_F

) = {bhtruei|¬keep_b,b[false]∈

_F

}

θ12(

_F

) = {a.b.αhtruei|a[keep_b],ahtruei,a.b.αhtruei ∈

_F

} θ13(

_F

) = {a.b.α[false]|a[keep_b],ahtruei,a.b.α[false]∈

_F

} θ14(

_F

) = {a.bhtruei|¬a[keep_b],a.b[false]∈

_F

}

θ15(

_F

) = {a.b[false]|a.bhtruei,keep_b,¬a[keep_b]∈

_F

} Figure 4.7: Definition ofθ9, . . . ,θ15

In the next proposition we show that prop. 4.19 holds also for the extendedθ. Proposition 4.24 Let

_F

be a set of formulas. Then the following hold:

1. `

_F

⇒ eθ(

_F

).

2. If

_F

is consistent, then so isθ(

_F

).

Proof. The proof is the same as for proposition 4.19, we only need to consider the new cases in the definition ofθ.

- F ≡ a.αhtruei ∈ θ9(

_F

): by definition we have `

_F

⇒keep_a∧a.αhtruei, hence by (ax12) it follows that`

_F

⇒ ea.αhtruei.

- F ≡a.α[false]∈θ10(

_F

): using axiom (ax12),(T5) and the definition ofθ10 we conclude`

_F

⇒ ¬ea.αhtruei. The assertion follows with the aid of (ax6) and (T5).

- F ≡bhtruei ∈θ11(

_F

): by the definition ofθ11 holds¬keep_b,b[false]∈

_F

. By (ax14), (prop) and (ax6) it follows`

_F

⇒ e¬b[false], hence`

_F

⇒ ebhtruei.

76 4.2. Axiomatisation of propositional MLTL

- F ≡a.b.αhtruei ∈θ12(

_F

): Observe that a.b.αhtruei ∈

_F

and a[keep_b]∈

_F

by the definition ofθ12. We give a derivation of

_F

⇒ ea.b.α.

(1)

_F

⇒a.b.αhtruei def. ofθ12 (2)

_F

⇒a.bhtruei (1),(T7),(prop) (3)

_F

⇒b.αhtruei (1),(T7),(prop) (4)

_F

⇒a[keep_b] def. ofθ12 (5)

_F

⇒keep_b (ax15),(4),(2),(prop) (6)

_F

⇒ ea.bhtruei (ax15),(4),(2),(prop) (7)

_F

⇒ eb.αhtruei (ax12),(3),(5),(prop) (8)

_F

⇒ e(a.b∧b.α) (6),(7),(T10),(prop) (9) a.b∧b.α⇒a.b.α (T7),(prop) (10) e(a.b∧b.α)⇒ ea.b.α (nex),(9),(ax7),(prop) (11)

_F

⇒ ea.b.αhtruei (8),(10),(prop)

- F ≡a.b.α[false]∈θ13(

_F

): Note that a[keep_b],ahtruei,a.b.α[false]∈

_F

by the definition ofθ13. We give a derivation of

_F

⇒ ea.b.α[false]:

(1)

_F

⇒ahkeep_bi ∧a.b.α[false] def. ofθ12,(T6),(prop) (2) ahkeep_bi ∧a.b[false]⇒ ea.b[false] (ax18)

(3) a.b[false]⇒a.b.α[false] (T7),(T5),(prop) (4) ea.b[false]⇒ ea.b.α[false] (3),(nex),(ax7),(prop) (5)

_F

∧a.b[false]⇒ ea.b.α[false] (1),(2),(4),(prop) (6) ahkeep_bi ∧a.bhtruei ⇒keep_b (ax15),(prop) (7) a.bhtruei ∧a.b.α[false]⇒b.α[false] (T5),(T7),(prop) (8) keep_b∧b.α[false]⇒ eb.α[false] (ax12),(prop)

(9) eb.α[false]⇒ ea.b.α[false] (ax2),(nex),(ax7),(prop) (10)

_F

∧a.bhtruei ⇒ ea.b.α[false] (1),(6),(7),(8),(9),(prop) (11)

F

⇒ ea.b.α[false] (5),(10),(prop)

- F ≡a.bhtruei ∈θ14(

_F

): we have¬a[keep_b],a.b[false]∈

_F

by the definition ofθ14. By (ax17) it follows that`

_F

⇒ ¬ea.b[false], hence`

_F

⇒ ea.bhtruei by (ax6), (T5), (nex) and (ax7).

- F ≡ a.b[false]∈ θ15(

_F

): then it holds a.bhtruei,keep_b,¬a[keep_b]∈

_F

by the definition ofθ15. By (ax16) and by propositional reasoning it follows that `

_F

⇒ ¬ea.bhtruei, hence` ea.b[false]by (ax6), (T5), (nex) and (ax7).

The graph

_T

(

_F

)is redefined: the successors of a node are defined with respect to the newθ. The notion of a complete path is unchanged.

As our next step, we are going to construct a run based on a complete path

F

0,

F

1. . .in the graph

T

(

F

). Letσ0denote the sequence(t₀0,λ0₀)(t₁0,λ0₁). . .where (t_i0,λ0_i) is defined as in sec. 4.1, starting with the set

F

i. Note that this run is

in general no model for the set

F

, as formulas of the form¬keep_a are not ne- cessarily satisfied. In order to see this, consider for example the following set:

F

={¬keep_a,ahtruei, eahtruei}. Let

G

denote an arbitrary completion of

F

. A possible complete path in

T

(

_F

)is the following:

G

,{ahtruei},/0,/0. . .

The construction yields the following runσ:

q q a q q - - q - q a ε ε . . .

Obviously, forσdoes not hold the formula¬keep_a. The problem is that a formula of the form ¬keep_a possibly requires the existence of some new name that does not occur in

_F

whereas in the runσonly names occur that occur in some formula in

_F

.

To solve this problem, we have to modify the trees of the run based on a complete path slightly. For every name a for which ¬keep_a ∈S

i≥0

F

i, let na ∈N be a

name that does not occur in any of the formulas in S

i≥0

F

i, with na 6=nb for

a 6=b. Note that it is possible to choose such names, asS

i≥0

F

i is a finite set.

Intuitively, the name na will have to ensure that the formula ¬keepa holds: if

¬keep_a has to hold for a sub-runσ|i and the namena does not occur in ti, then

na is put below the namea in the treeti+1. Conversely, ifna is already inti, then

78 4.2. Axiomatisation of propositional MLTL

We define the run σ= (t0,λ0)(t1,λ1). . . by induction on i ∈ ω. We let again σ0_{= (t}0

0,λ00)(t10,λ01). . ., with ti0= (N0i, <0i), be the run based on a complete path

F

0,

F

1, . . .as described above and let(t0,λ0):= (t₀0,λ0₀). Leti ∈ω and assume that(t0,λ0), . . . ,(ti,λi)are already constructed.

Ni+1:=N0i+1∪ {na|¬keepa ∈

F

i,a∈N0_i,a∈N0_i+1andna ∈/Ni}

∪ {na|na ∈Ni andkeepb ∈

F

i for someb withna <i b}

and the binary relation<i+1onNi+1is defined as follows:

a<i+1b:⇔              a <0_i+1b ¯ a ≤0_i+1b a <0_i+1b¯ ¯ a <0_i+1b¯ ifa,b∈N0_i+1

ifb∈N0_i+1anda =na¯ for some ¯a∈N ifa ∈N0_i+1andb =n_b¯ for some ¯b∈N ifa =na¯,b=n_b¯ anda<0_i+1b

The assignmentsλi are defined by

λi(a):=    λ0 i(a) /0 ifa ∈N0_i ifa =na¯ for some ¯a ∈N0i

Informally, this definition indicates thatna is put immediately belowa. To illus-

trate the construction, we consider a sequence

F

∗₀,

_F

∗₁, . . .where

F

∗₀is some com- pletion of

F

0={2abhtruei,2chtruei,¬keepb, ekeepa,2¬keepc}and

F

∗i+1 is some completion of θ(

_F

∗_i). The prefix of one possible resulting run – first only applying the original definition from sec. 4.1, that is, without the described modi- fication – is the following:

q q q @ q @ a b c q q q @ q @ a b c q q q q q q q q @ @ - @ @ - - a c b a b c . . .

This is obviously not a model of

_F

0, as the formulas¬keepb and2¬keepc are not

satisfied. The modification according to our definition yields the following run:

q q q q @ @ - nc a b c q q q q q q q q q q q q q q q q @ @ @@ - - @ @ a c b nb a b nb c nc a b c . . .

The first trees are identical. Then the name nb is put below the nameb as well

as the namenc belowc in order to satisfy formulas¬keepb and2¬keepc. In the

next stepnc disappears as required by formula2¬keepc, but nodenb is still kept

as required bykeep_a ∈

_F

∗₁(this follows from ekeep_a ∈

_F

0) anda.bhtruei ∈

F

∗₁. In the last treenb does not appear (the auxiliary nodes are not kept if not explicitly

required by some keep operator) whereasnc reappears due to¬keepc ∈

F

∗2. We show thatti is a tree.

Lemma 4.25 Letti = (Ni, <i)be defined as described above. Thenti is a tree.

Proof. In order to prove the claim we have to show that

1. the relations<i are irreflexive.

2. the relations<i are transitive.

3. ifa,b,c∈Ni,a6=b,c <i a andc<i b then eithera<i b orb<i a.

To 1.: Assume thata <i a. We have to distinguish two cases.

Case: a∈N0_i. Thena <0_i a by definition in contradiction to the irreflexivity of the relation<0_i.

Case:a=na¯ for some ¯a∈N0_i. Then ¯a<0_i a¯ by definition, hence we obtain the same contradiction as above.

To 2.: Assume thata<i b andb <i c. We have to show that a <i c. Again, we

have to distinguish different cases.

Case: a,b,c∈N0_i. Then a<_i0 b andb<0_i c by definition, hencea<0_i c by the transitivity of<0_i. By the definition of<i this meansa <i c.

Case: a,b ∈N0_i andc =nc¯ for some ¯c ∈N0_i. Then a<0_i b andb <0_i c¯, hence

a<0_i c¯. Latter impliesa<i c.

Case: a,c ∈N0_i and b = n_b¯ for some ¯b ∈ N0_i. Then a <0_i b and b ≤0_i c by definition, hencea<0_i c and soa<i c by the definition of<i.

80 4.2. Axiomatisation of propositional MLTL

Case:b,c∈N0_i anda=na¯ for some ¯a∈N0i. Then ¯a≤0i bandb<0ic. It follows

¯

a<0_i c, hencena¯ <i c by definition.

Case: a ∈N0_i, b=n_b¯ andc =nc¯ for some ¯b,c¯ ∈N0i. Thena <i0 b¯ and ¯b <0i c¯

by definition. It followsa<0_i c¯, hencea<i c.

Case:b∈N0_i,a=na¯ andc=nc¯ for some ¯a,c¯∈N0_i. Then ¯a≤_i0bandb<0_i c¯ by definition. This implies ¯a<0_i c¯ as<0_i is transitive, hencea<i cby definition.

Case: c∈N0_i, a =na¯ andb =n_b¯ for some ¯a,b¯ ∈N0_i. Then ¯a <0_i b¯ and ¯b≤0_i c by definition. Hence ¯a<0_i cwhich impliesa<i c.

Case:a=na¯,b=n_b¯ andc=nc¯ for some ¯a,b¯,c¯∈N0_i. Then ¯a<0_ib¯ and ¯b<0_i c¯ by definition, hence ¯a<0_i c¯. From this followsa <i cby the definition of<i.

To 3.: Again, different cases have to be considered.

Case:a,b,c∈N0_i. Thenc<0_i aandc<0_i bby definition and as(t_i0, <0_i)is a tree it followsa<0_i b orb<0_i a. Hence,a<i borb<i a.

Case: a,b ∈N0_i and c=nc¯ for some ¯c ∈N0_i. Then ¯c ≤0_i a and ¯c ≤0_i b, which impliesa<0_i b orb<0_i a. Hence,a<i borb<i a by the definition of<i.

Case: c ∈N0_i, a=na¯ and b=n_b¯ for some ¯a,b¯ ∈N0_i with ¯a 6=b¯ (as na¯ 6=n_b¯ by assumption). Thenc<0_i a¯ andc<0_i b¯ by the definition of<i, hence ¯a <0_i b¯

or ¯b<0_i a¯. In the first case it followsa <i b, in the second caseb<i a by the

definition of<i.

Case: a,c ∈N0_i and b =n_b¯ for some ¯b ∈N0_i. Then c <0_i a and c <0_i b¯ by definition. It followsa<0_i b¯ or ¯b<0_i a, hencea≤i b orb <i a. Sincea6=b by

assumption, the assertion follows.

Case: b,c ∈N0_i and a=na¯ for some ¯a ∈N0i. This case is symmetrical to the

previous case.

Case:a∈N0_i,b=n_b¯ andc=nc¯ for some ¯b,c¯∈N0i. Then ¯c≤i0 aand ¯c<0i b¯. If

¯

c=a then we havea<0_i b by the latter. Otherwise we obtaina<0_i b¯ or ¯b<0_i a, hencea<i b orb<i a by definition.

Case: b ∈N0_i, a =na¯ andc =nc¯. for some ¯a,c¯ ∈N0i. This is symmetrical to

Case: a =na¯,b =n_b¯ andc=nc¯ for some ¯a,b¯,c¯ ∈N0i. Then we have ¯c<0i a¯

and ¯c<0_i b¯ by definition, hence ¯a<0_ib¯ or ¯b<0_i a¯ ast_i0is a tree. It followsa<i b

orb<i aby the definition of<i.

We prove the following lemma about the sequenceσ.

Lemma 4.26 Letσ= (t0,λ0)(t1,λ1). . .be defined as above and leti ∈ωbe ar-

bitrary. Then the following holds:

1. If keep_b ∈τκ(

_F

i), then keepb ∈

F

i iff ti↓b=ti+1↓b.

2. If a[keep_b]∈τκ(

_F

i), then a[keepb]∈

F

i iff a∈/Ni orti↓a.b=ti+1↓a.b. Proof.

1. Only if: Letkeep_b ∈

_F

i. First we show thatt_i0↓b =t_i0₊₁↓b. To this end, it is

enough to prove that for alla,c∈Nholds the following:

(a) a<0_i b iffa<0_i+1b

(b) Ifa<0_i b, thenc<_i0 a iffc<0_i+1a

To a): By the definition of the relations<0_j we have to showb.ahtruei ∈

_F

i

iff b.ahtruei ∈

_F

i+1. Since keepb ∈

F

i and since

F

i+1 is a completion of θ(

_F

i), this follows from the definition ofθ.

To b): Leta <0_i b, i.e.b.ahtruei ∈

_F

i. We have to show thata.chtruei ∈

F

i

iffa.chtruei ∈

_F

i+1.

Only if: Assumea.chtruei ∈

_F

i. Observe thata,b,c are pairwise distinct

and henceb.a.chtruei ∈τκ(

_F

i). By (T7) it follows`

F

i ⇒b.a.chtruei,

henceb.a.chtruei ∈

_F

i by proposition 4.11,1. By the definition ofθit fol-

lows thatb.a.chtruei ∈

_F

i+1. Since`b.a.chtruei ⇒a.chtrueiby (T0), it follows thata.chtruei ∈

_F

i+1.

82 4.2. Axiomatisation of propositional MLTL

If: Assume a.chtruei ∈

_F

i+1. Since a,b,c are pairwise distinct and as

b.ahtruei ∈

_F

i+1 bya), it follows as above thatb.a.chtruei ∈

F

i+1. As

a,b,c ∈nm(

_F

i), we have b.a.chtruei ∈τκ(

F

i). By the definition ofθ

it follows that b.a.c[false]∈/

_F

i, hence b.a.chtruei ∈

F

i. The assertion

follows now as in the previous case.

Now we consider the “auxiliary” namesnc. From the definition of the trees

ti it follows directly that if a nodenc occurs inti, then its position is imme-

diately below the nodec. Hence, it suffices to show for everync thatnc <i b

iffnc <i+1b.

Only if: Assume first thatnc <i b. Sincekeepb ∈

F

i by assumption, the

definition ofti+1implies thatnc ∈Ni+1. Furthermore, by the definition of <i+1it follows thatnc <i+1b, that is,nc <i+1b.

If: For the converse, assume thatnc <i+1b. Thenc ≤i+1b by definition, that is, eitherc=borb.chtruei ∈

_F

i+1.

Case: c =b. Since keep_b ∈

_F

i and nc ∈Ni+1, the definition of ti+1 impliesnb ∈Ni. By the definition of<i it followsnb <i b.

Case: c <i+1 b. By definition, this means b.chtruei ∈

F

i+1. Since keep_b ∈

_F

i, by the definition of θ it follows b.chtruei ∈

F

i. Hence,

`

_F

i ⇒keepc by (ax13), in particular¬keepc ∈/

F

i. Again, it follows

by the definition ofti+1thatnc ∈Ni. Asc<i b(sinceb.chtruei ∈

F

i),

we concludenc <i b by the definition of<i.

If: Assume thatkeep_b ∈/

_F

i, that is, as

F

i is complete,¬keepb ∈

F

i.

Case:n_b ∈/Ni. There are two possibilities. Eitherb∈/Ni orb∈Ni.

In the first case we haveb[false]∈

_F

i, hence bhtruei ∈

F

i+1 by the defi- nition ofθ. Hence,b∈/Ni butb∈Ni+1, in particular,ti↓b6=ti+1↓b. If b ∈Ni and b ∈/ Ni+1, then we already have ti↓b 6=ti+1↓b since latter is the empty tree whereas the first is not. If b ∈Ni+1, then nb ∈Ni+1 by the definition ofti+1, hencenb <i+1b. It followsti↓b 6=ti+1↓b since

n_b <i+1b butnb 6<i b.

Case: nb ∈Ni. Assume nb ∈Ni+1. By the definition of ti+1, this is only possible if there is a namecwithkeep_c∈

_F

i andnb<i c. By the definition

of<i it followsb≤i c, that is,b=c orc.bhtruei ∈

F

i. It follows in both

cases - in the latter by (ax13) and by lemma 4.13 - that`

_F

i ⇒keepb, in

contradiction to the consistency of

_F

i.

2. Only if: Assume thata[keep_b]∈

_F

i andahtruei ∈

F

i.

Case: b <i a, that is, a.bhtruei ∈

F

i. By (ax15) and (prop) it follows

that `

_F

i ⇒keepb holds. As keepb ∈τκ(

F

i), it holds keepb ∈

F

i by

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