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1 L -Semantics as a Fibring over a Combination of Temporal and Kripke Frames

1 L -Semantics as a Fibring over a Combination of Temporal and Kripke Frames

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Multiple-Source Approximation Systems, Evolving Information Systems



231



– (σ, σ ) ∈ Rns6 if and only if σ = τ Rl and σ = τ t for some l ∈ {1, 2, . . . , N }

and t ∈ N, where char(σ) = char(τ ) = n;

– (σ, σ ) ∈ Rn7 if and only if σ = σRn t for some t ∈ N;

– (σ, σ ) ∈ Rns7 if and only if σ = σ Rn t for some t ∈ N.

Since the N -prefixes σ and σRl will be interpreted to represent the same

object, if we want the accessibility relation to be reflexive, then R1 will not be

sufficient for the purpose and we shall require R2 and R3 as well. For a similar

reason, we will require Rn5 , Rn6 , Rn7 in addition to Rn4 to give the accessibility

relation for the class of dynamic K spaces. Rns4 , Rns5 , Rns6 , Rns7 will be used to

give symmetry.

The following definition gives the accessibility relations for different classes

of dynamic spaces using the relations defined above.

Definition 53. Let I ∈ {K, K4, T, B, S4, KT B, KB4, S5} and n ≤ N . We

define the I-accessibility relation for characteristic n, RnI , on the set P (N ) \ {ε}

of N -prefixes as follows. Let us write R∗ to denote the transitive closure of the

relation R.



















RnK := Rn4 ∪ Rn5 ∪ Rn6 ∪ Rn7 .

RnT := R1 ∪ R2 ∪ R3 ∪ RnK .

RnK4 := (RnK )∗ .

RnB := Rn4 ∪ Rn5 ∪ Rn6 ∪ Rn7 ∪ Rns4 ∪ Rns5 ∪ Rns6 ∪ Rns7 .

RnKB4 := (RnB )∗ .

RnS4 := (RnT )∗ .

RnKT B := R1 ∪ R2 ∪ R3 ∪ RnB .

RnS5 := (RnKT B )∗ .



Now, we are in a position to give the tableau extension rules for each of the

classes of dynamic I spaces, I ∈ {K, K4, T, B, S4, KT B, KB4, S5}. Figures 3–4

give the rules for these classes corresponding to the N -prefixed wffs. We call

these (I, N )-tableau extension rules. Except one, the rules for the Since operator

are not listed, as these can be given in the lines of those for the Until operator.

The rules involving temporal operators, that is, ⊕, , U, S-rules, will be called

the temporal rules.

In order to understand the tableau extension rules, let us consider a wff ⊕α.

To evaluate ⊕α at the time point t at the object w, we need to evaluate the

wff α at the same object w, but at the time point t + 1. To capture this fact,

the ⊕ rule is designed such that if the branch contains σm ⊕ α, m ∈ N, then

using the ⊕(b)-rule, we introduce the wff σmRl+1 α (assuming char(σm) = l).

The N -prefix in the latter is used to indicate the same object referred to by σm

(recall the comments following Definition 49), along with the fact that we have

shifted focus to the time point l + 1. Similarly, one may explain the rule U(d).

Note that in a model M with number of time points N , for any w ∈ U and

n < N , we have M, n, w |= ¬(XUY ), if and only if either M, k, w |= ¬Y for all

k with n ≤ k ≤ N , or there exists s with n ≤ s < N such that M, s, w |= ¬X

and for all k such that n ≤ k ≤ s, M, k, w |= ¬Y . Observe that the wff XUY



232



Md.A. Khan

1st



σα

σα1

σα2



2nd



σβ



σβ1



α-rule



σRn ⊕, n < N



σn+1

⊕0



σβ2

β-rule



σRn t1 t2 · · · tr ⊕, n < N



σn+1

⊕0



⊕(a)-rule



σRn , n > 1



σn−1

0



σRn t1 t2 · · · tr , n > 1



σn−1

0



⊕(b)-rule



(a)-rule



(b)-rule



σν

σ ν0



σπ

σ π0



I

, σ is

where (σ, σ ) ∈ Rn

used on the branch and

char(σ) = char(σ ) = n.



σRn F p

σF p



σ is an unrestricted simple p is propositional variable.

extension of σ.



ν-rule



π-rule



σF (XUY ), char(σ) = N

σF Y

U end point-rule



σRn T p

σT p



Propositional-rule



σF (XSY ), char(σ) = 1

σF Y

S first point-rule



Fig. 3. (I, N )-tableau extension rules.



can be made false at (n, w) in N − n + 1 possible ways. Accordingly, the rule

U(d) for τ F (XUY ), n < N , where τ = σRn t1 t2 · · · tr , asks for the introduction



in the

of N − n + 1 branches. We again note that all N -prefixes of the kind τm

rule indicate the same object referred to by τ , along with the fact that we have

shifted focus to the time point m. The tableau extension rules for the other wffs

may be understood in a similar manner.

Given a N -prefix σ, we would like to investigate next what would be the form

of any N -prefix μ related to σ by the relation RnI . We would then be able to see

the possible forms of an accessible N -prefix – this is crucial for the application of

the ν rule. Moreover, the information about the form of such N -prefixes will also

be used for obtaining important results related with the soundness, completeness



Multiple-Source Approximation Systems, Evolving Information Systems



Fig. 4. (I, N )-tableau extension rules, continued.



233



234



Md.A. Khan



and termination of the proof procedures. The form of μ depends on I as well

as on the form of σ itself. Here we will only give the propositions determining

the form for I = S5, but one can obtain similar results for other Is. In fact, the

argument will be simpler for all the other choices of Is.

Let us first consider the N -prefix of the form τ := τ Rn t1 t2 · · · ts , s ≥ 1 and

the relation RnS5 . We investigate the form of μ for which we have (τ, μ) ∈ RnS5 .

Since RnS5 is reflexive, we should have (τ, τ ) ∈ RnS5 . Moreover, due to symmetry

and transitivity of RnS5 , we also expect to have (τ, τ Rn d1 d2 · · · dk ) ∈ RnS5 . Furthermore, as mentioned above, σ and σRl will be interpreted to represent the

same object, and thus we should also have (τ, τ Rl ) and (τ, τ Rn d1 d2 · · · dk Rl ) ∈

RnS5 . In fact, the following proposition shows that these are the only possibilities

for μ.

Proposition 67. Let τ = τ Rn t1 t2 · · · ts , s ≥ 1 and (τ, μ) ∈ RnS5 . Then μ must

be in either of the following forms.

(a)

(b)

(c)

(d)



μ

μ

μ

μ



is

is

is

is



τ

τ

τ

τ



.

Rl , where Rl ∈ {R1 , R2 , . . . , RN }.

Rn d1 d2 · · · dk , where di ∈ N.

Rn d1 d2 · · · dk Rl , where Rl ∈ {R1 , R2 , . . . , RN } and di ∈ N.



Proof. The proposition is proved by showing that if

τ RnKT B σ1 RnKT B σ2 · · · RnKT B σj ,

then σj = τ or σj is in either of the above-mentioned forms. We use induction

on j. Note that since τ Rn t1 t2 · · · ts ∈ P (N ), we have char(τ ) = n.

Basis case: Let j = 1.

Since (τ, σ1 ) ∈ RnKT B and RnKT B = R1 ∪ R2 ∪ R3 ∪ Rn4 ∪ Rn5 ∪ Rn6 ∪ Rn7 ∪ Rns4

∪Rns5 ∪Rns6 ∪Rns7 , we must have (τ, σ1 ) ∈ R for some R ∈ {R1, R2, R3, Rn4 , Rn5 ,

Rn6 , Rn7 , Rns4 , Rns5 , Rns6 , Rns7 }. But note that (τ, σ1 ) ∈ R for R ∈ {R2, Rn5 , Rn6 , Rn7 }.

Therefore, σ1 = τ (in the case when (τ, σ1 ) ∈ R1), or σ1 must be in one of the

following forms:

















τ Rl (when (τ, σ1 ) ∈ R3), or

τ Rn t1 t2 · · · ts ts+1 (when (τ, σ1 ) ∈ Rn4 ), or

τ Rn , where s = 1 (when (τ, σ1 ) ∈ Rns4 ), or

τ Rn t1 t2 · · · ts−1 , where s ≥ 2 (when (τ, σ1 ) ∈ Rns4 ), or

τ Rl , where s = 1 (when (τ, σ1 ) ∈ Rns5 ), or

τ Rn t1 t2 · · · ts−1 Rl , where s ≥ 2 (when (τ, σ1 ) ∈ Rns6 ), or

τ , where s = 1 (when (τ, σ1 ) ∈ Rns7 ).



Thus in each case, we obtain σj in the desired form.

Induction case: Suppose (a) σj = τ , or σj is in one of the following forms:

(b) τ Rl , or

(c) τ Rn d1 d2 · · · dk , or

(d) τ Rn d1 d2 · · · dk Rl



Multiple-Source Approximation Systems, Evolving Information Systems



235



Case (a): σj = τ

Since char(τ ) = n and (σj , σj+1 ) ∈ RnKT B , σj+1 = τ (when (σj , σj+1 ) ∈ R1)

or σj+1 must be in either of the following forms:

– τ Rl (when (σj , σj+1 ) ∈ R3), or

– τ Rn d1 (when (σj , σj+1 ) ∈ Rn7 ).

Case (b): σj is of the form τ Rl .

In this case, σj+1 = τ (when (σj , σj+1 ) ∈ R2), or σj+1 must be in either of the

following forms:

– τ Rl (when (σj , σj+1 ) ∈ R1), or

– τ Rl d1 , l = n (when (σj , σj+1 ) ∈ Rn4 ), or

– τ Rn d1 , (when (σj , σj+1 ) ∈ Rn5 ).

Case (c): σj is of the form τ Rn d1 d2 · · · dk .

Then, σj+1 = τ and k = 1 (when (σj , σj+1 ) ∈ Rns7 ), or σj+1 must be in one of

the following forms:

















τ

τ

τ

τ

τ

τ

τ



Rn d1 d2 · · · dk (when (σj , σj+1 ) ∈ R1), or

Rn d1 d2 · · · dk Rl (when (σj , σj+1 ) ∈ R3), or

Rn d1 d2 · · · dk dk+1 (when (σj , σj+1 ) ∈ Rn4 ), or

Rn d1 d2 · · · dk−1 and k ≥ 2 (when (σj , σj+1 ) ∈ Rns4 ), or

Rn and k = 1 (when (σj , σj+1 ) ∈ Rns4 ), or

Rl and k = 1 (when (σj , σj+1 ) ∈ Rns5 ), or

Rn d1 d2 · · · dk−1 Rl and k ≥ 2 (when (σj , σj+1 ) ∈ Rns6 ).



Case (d): σj is of the form τ Rn d1 d2 · · · dk Rl .

Then, σj+1 must be in either of the following forms:

– τ Rn d1 d2 · · · dk Rl (when (σj , σj+1 ) ∈ R1), or

– τ Rn d1 d2 · · · dk (when (σj , σj+1 ) ∈ R2), or

– τ Rn d1 d2 · · · dk dk+1 (when (σj , σj+1 ) ∈ Rn6 ).

Thus, in each case, we obtain σj+1 in the desired form. This completes the

proof.

As a direct consequence of Proposition 67, we obtain the following corollary.

Corollary 15. Let μ, τ ∈ P (N ) be such that (i) τ = τ Rn t1 t2 · · · ts , s ≥ 1, (ii)

(τ, μ) ∈ RnS5 and (iii) char(μ) = n. Then μ must be either τ Rn , or of the form

τ Rn d1 d2 · · · dk , di ∈ N.

We end this section with the two following propositions, similar to

Proposition 67.

Proposition 68. Let τ = τ Rh t1 t2 · · · ts Rn ∈ P (N ), (so, s ≥ 1, h = n) and

(τ, μ) ∈ RnS5 . Then μ = τ Rh t1 t2 · · · ts or μ must be in either of the form (a)

τ Rh t1 t2 · · · ts Rl , or (b) τ d1 d2 · · · dr , or (c) τ d1 d2 · · · dr Rl .

Proposition 69. Let τ = τ Rh t1 t2 · · · ts , s ≥ 1, h = n and (τ, μ) ∈ RnS5 . Then

μ = τ or μ must be in either of the form (a) τ Rl , or (b) τ Rn d1 d2 · · · dr , or (c)

τ R n d1 d2 · · · dr R l .



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Md.A. Khan



10.1



Soundness



In this section, we shall prove the soundness of the tableau-based proof procedures proposed in Sect. 10. We begin with the following definitions. Recall the

relations RnI given by Definition 53.

Definition 54.

(i) Let S be a set of N -prefixed wffs and F := F1 , F2 , . . . FN be a dynamic I

space, where Fi = (W, Pi ). By an I-interpretation of S in F, we mean a

mapping Int from the set of N -prefixes that occur in S to W such that the

following conditions are satisfied:

(a) If (σ, σ ) ∈ RnI then (Int(σ), Int(σ )) ∈ Pn , n = 1, 2, . . . , N .

(b) Int(σ) = Int(σRl ), for all l ∈ N.

(ii) A set S of N -prefixed wffs is said to be (I, N )-satisfiable if there exists some

dynamic I space F := F1 , F2 , . . . , FN of cardinality N , an I-interpretation

Int of S in F, and a valuation function V : P → 2W such that for each

σZ ∈ S,

M, n, Int(σ) |= Z

where char(σ) = n and M = (F, V ).

(iii) A branch of an (I, N )-tableau is said to be (I, N )-satisfiable if the set of

N -prefixed wffs on it is (I, N )-satisfiable.

(iv) An (I, N )-tableau is said to be (I, N )-satisfiable if some branch of it is

(I, N )-satisfiable.

Observe that the condition (b) in the definition of interpretation signifies that

the N -prefixes σ and σRl represent the same object.

Example 16. Let us consider a (S5, 2)-tableau for the wff F r → ♦p ∨ q,

p, q, r ∈ P V , given in Fig. 5.

Let S be the set of all 2-prefixed wffs occurring on the right branch of the

tableau of Fig. 5. Note that {R1 1, R1 11, R1 1R2 } is set of all 2-prefixes that occur

in S. Let us consider the dynamic S5 space F := F1 , F2 , where Fi := (W, Pi ),

W := {x, y}, W/P1 := W × W and W/P2 := {{x}, {y}}. Let Int be the function

from the set of 2-prefixes occurring in S to W defined as

Int(R1 1) = Int(R1 1R2 ) = x;

Int(R1 11) = y.

Note that Int is an interpretation of S in F as it satisfies both the defining

conditions (a), and (b) of an I-interpretation (cf. Definition 54). In fact, condition

/ R2S5 ,

(a) is a direct consequence of the fact that (R1 1, R1 11), (R1 1R2 , R1 11) ∈

which, in turn, follows from Proposition 67. Further, using Int and the valuation

V which maps p to ∅, q and r to {x}, one can show that the right branch of

the above tableau is (S5, 2)-satisfiable. For instance, for the wff R1 1F (F r →

♦p∨ q) lying on the right branch, we have M, 1, Int(R1 1) |= F (F r → ♦p∨ q)

as M, 2, x |= r, but M, 1, x |= ♦p and M, 1, x |= q, M := (F, V ).



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