A dynamic knowledge base and its data compression with homomorphism

2.1 Knowledge bases

Recall that R is said to be a binary relation on U if R ∈ 2^U×U. Sometimes, if U = {x₁, x₂, . . . , x _m }, then a binary relation R is represented by the following matrix:

( R ( x 1 , x 1 ) R ( x 1 , x 2 ) . . . R ( x 1 , x m ) R ( x 2 , x 1 ) R ( x 2 , x 2 ) . . . R ( x 2 , x m ) . . . . . . . . . . . . R ( x m , x 1 ) R ( x m , x 2 ) . . . R ( x m , x m ) ) , where R ( x i , x j ) = { 1 , ( x i , x j ) ∈ R ; 0 , ( x i , x j ) ∉ R .

Suppose that R is a binary relation on U. Then R is said to be equivalence if R is reflexive, symmetric and transitive.

In this paper, R * ( U ) denotes the set of all equivalence relations on U.

For R ∈ R * ( U ) and x ∈ U, [x]  _R  = {y ∈ U : xRy} denotes the equivalence class of x. Additionally, U / R = { [ x ] R : x ∈ U } .

If R ∈ R * ( U ) , then for x ∈ U, R _s  (x) = [x]  _R ; if R ⊆ R * ( U ) , then ind ( R ) ∈ R * ( U ) .

Definition 2.1. ([33]) A pair (U, R) is called a knowledge base, if  R ⊆ R * ( U ) .

For each equivalence relation in a knowledge base, one can construct lower approximation and upper approximation of any subset on the universe in the following definition.

Definition 2.2. ([15]) Suppose that (U, R) is a knowledge base. For any X ∈ 2 ^U and R ∈ R, define R _ ( X ) = { x ∈ U : [ x ] R ⊆ X } , R ¯ ( X ) = { x ∈ U : [ x ] R ∩ X ≠ ∅ } . Then R _ ( X ) and R ¯ ( X ) are called R-lower approximation and R-upper approximation with respect to R, respectively.

Bnd R ( X ) = R ¯ ( X ) - R _ ( X ) is said to be the boundary region of X.

A set is rough if its boundary region is not empty; otherwise, it is crisp. Thus, X is rough if R _ ( X ) ≠ R ¯ ( X ) .

Definition 2.3. ([33]) Suppose that (U, R) is a knowledge base and P ⊆ R.

(1) P is said to be equivalent to R (or P is said to be a coordinate subfamily of R), if ind (P) = ind (R).

(2) R ∈ P is said to be independent in P, if ind (P - {R}) ≠ ind (P); P is said to be a independent subfamily of R, if ∀ R ∈ P, R is independent in P.

(3) P is said to be a knowledge reduction of R, if P is both coordinate and independent.

Suppose that (U, R) is a knowledge base. We denote the set of all coordinate subfamilies (resp., all knowledge reductions) of R by co (R) (resp., red (R)).

Obviously, P ∈ red ( R ) ⇔ P ∈ co ( R ) and ∀ Q ⊂ P , Q ∉ co ( R ) .

Theorem 2.4. Suppose that (U, R) is a knowledge base. Then

(1) P ∈ co (R) ⇔ U/P = U/R.

(2) P ∈ red (R) ⇔ U/P = U/R and ∀ R ∈ P,  U/P ⊂ U/(P - {R}). Denote [ x ] R = ⋂ R ∈ R [ x ] R , U / R = { [ x ] R : x ∈ U } .

2.2 Relation information systems

Definition 2.5. ([15]) An information system is a pair (U, A) of non-empty finite sets U and A, where U is a set of objects and A is a set of attributes; each attribute a ∈ A is a function a : U → V _a , where V _a is the set of values (called domain) of attribute a.

If (U, A) is an information system and B ⊆ A, then an equivalence relation (or indiscernibility relation) R _B can be defined by ( x , y ) ∈ R B ⇔ a ( x ) = a ( y ) , ∀ a ∈ B .

Definition 2.6. ([23]) A pair (U, R) is said to be a relation information system, if R ⊆ 2^U×U.

Obviously, a knowledge base is a special kind of relation information system.

Definition 2.7. Suppose that (U, A) is an information system. Put R = { R { a } : a ∈ A } . Then the pair (U, R) is said to be the relation information system induced by the information system (U, A).

In fact, (U, R) in Definition 2.7 is a knowledge base.

2.3 Consistent mappings

Definition 2.8. ([23, 24]) Suppose that  U and V are finite sets, f: U → V a mapping and R ∈ 2^U×U. Let [ x ] f = { u ∈ U : f ( u ) = f ( x ) } , [ x ] R = { u ∈ U : R s ( u ) = R s ( x ) } . Then {[x]  _f  : x ∈ U} and {(x)  _R  : x ∈ U} are two partitions on U. If [x]  _f  ⊆ R _s  (u) or [x]  _f ∩ R _s  (u) = ∅ for any x, u ∈ U, then f is said to be a type-1 consistent mapping with respect to R on U. If [x]  _f  ⊆ (x)  _R for any x ∈ U, then f is said to be a type-2 consistent mapping with respect to R on U.

Remark 2.9.

(1) For each x ∈ U, [x]  _f  = f^-1 (f (x)) .

(2) If R ∈ R * ( U ) , then for each x ∈ U, (x)  _R  = [x]  _R .

(3) f is type-1 ⇔ If [x]  _f ∩ R _s  (u) ≠ ∅, then [x]  _f  ⊆ R _s  (u)

                   ⇔  If [x]  _f  ⊈ R _s  (u), then [x]  _f ∩ R _s  (u) = ∅,

(4) f is type-2 ⇔  If f (u) = f (x), then R _s  (u) = R _s  (x).

Definition 2.10. ([23]) Suppose that f: U → V is a mapping and R ⊆ 2^U×U. If f is type-1 (resp. type-2) consistent with respect to R on U for every R ∈ R, then f is said to be type-1 (resp. type-2) consistent with respect to R on U.

2.4 Homomorphisms between relation information systems

In this subsection, we give definition of homomorphism between relation information systems.

Definition 2.11. ([23, 24]) Suppose that f : U → V is a mapping. Define f ˆ : 2 U × U → 2 V × V , R → f ˆ ( R ) = ⋃ x ∈ U ( { f ( x ) } × f ( R s ( x ) ) ) ; f ˆ - 1 : 2 V × V → 2 U × U , T → f ˆ - 1 ( T ) = ⋃ y ∈ V ( { f - 1 ( y ) } × f - 1 ( T s ( y ) ) ) . Then f ˆ and f ˆ - 1 are called the R-mapping and inverse R-mapping induced by f, respectively.

Obviously, y 1 f ˆ ( R ) y 2 ⇔ ∃ x 1 , x 2 ∈ U , y 1 = f ( x 1 ) , y 2 = f ( x 1 ) and x 1 Rx 2 , x 1 f ˆ - 1 ( T ) x 2 ⇔ ∃ y 1 , y 2 ∈ V , y 1 = f ( x 1 ) , y 2 = f ( x 1 ) and y 1 Ty 2 .

For R ⊆ 2^U×U, denote f ˆ ( R ) = { f ˆ ( R ) : R ∈ R } .

Proposition 2.12. ([23]) Suppose that f : U → V is a mapping and R ⊆ 2^U×U. If f is both type-1 and type-2 consistent with respect to R, then f ˆ ( ind ( R ) ) = ind ( f ˆ ( R ) ) .

Proposition 2.13. Suppose that f : U → V is a mapping and R ⊆ 2^U×U. If f is both type-1 and type-2 consistent with respect to R, then for P ⊆ R, f ˆ ( ind ( P ) ) = ind ( f ˆ ( P ) ) .

Proof. Note that f is both type-1 and type-2 consistent with respect to R. Then f is both type-1 and type-2 consistent with respect to P.

By Proposition 2.12, f ˆ ( ind ( P ) ) = ind ( f ˆ ( P ) ) .□

Definition 2.14. ([23]) Let (U, R) be a relation information system and f : U → V a mapping. The pair ( V , f ˆ ( R ) ) is called an f-induced relation information system of (U, R).

Definition 2.15. ([23]) Suppose that (U, R) is a relation information system and ( V , f ˆ ( R ) ) an f-induced relation information system of (U, R). If f is both type-1 and type-2 consistent with respect to R on U, then f is said to be a homomorphism from (U, R) to ( V , f ˆ ( R ) ) . We write ( U , R ) ∼ f ( V , f ˆ ( R ) ) .

Theorem 2.16. Suppose that f : U → V is both type-1 and type-2 consistent with respect to R on U. Then

(1) R ∈ R * ( U )  ⇔  f ˆ ( R ) ∈ R * ( V ) ;

(2) If R ∈ R * ( U ) , then for each x ∈ U, f ( [ x ] R ) = [ f ( x ) ] f ˆ ( R ) .

Proposition 2.17. If ( U , R ) ∼ f ( V , f ˆ ( R ) ) with R = {R₁, . . . , R _m }, then V / f ˆ ( R ) = { ⋂ i = 1 m f ( [ x ] R i ) : x ∈ U }

Theorem 2.18. If ( U , R ) ∼ f ( V , f ˆ ( R ) ) , then ( U , R ) is a knowledge base ⇔ ( V , f ˆ ( R ) ) is a knowledge base .

Proposition 2.19. Suppose that (U, R) is a knowledge base. If ( U , R ) ∼ f ( V , f ˆ ( R ) ) , then P ∈ co ( R ) ⇔ f ˆ ( P ) ∈ co ( f ˆ ( R ) ) .

Theorem 2.20. Suppose that (U, R) is a knowledge base. If ( U , R ) ∼ f ( V , f ˆ ( R ) ) , then P ∈ red ( R ) ⇔ f ˆ ( P ) ∈ red ( f ˆ ( R ) ) .

4.1 Adding knowledge

In this subsection, we illustrate how to compress a dynamic knowledge base when adding a knowledge to it by using homomorphism.

Theorem 4.1. If ( U , R ) ∼ f ( V , f ˆ ( R ) ) , then P ∈ co ( S ( t + 1 ) ) ⇔ f ˆ ( P ) ∈ co ( T ( t + 1 ) ) , P ∈ red ( S ( t + 1 ) ) ⇔ f ˆ ( P ) ∈ red ( T ( t + 1 ) ) .

Theorem 4.2. Suppose that a knowledge base S^(t) = (U, R) at time (t) be compressed into T^(t) = (V, f (R)) where R = {R₁, . . . , R _m } and ( U , R ) ∼ f ( V , f ˆ ( R ) ) . Suppose that that a new knowledge R⁺ is added to S^(t) at time (t + 1) such that ^R + = R ∪ {R⁺} and S^(t+1) = (U, ^R +). Then for the image knowledge base T ( t + 1 ) = ( V , f ˆ ( R + ) ) at time (t + 1), the following properties hold:

(1) If ∀ y ∈ V , [ y ] f ˆ ( R ) ⊆ [ y ] f ˆ ( R + ) , then T^(t+1) = T^(t),

(2) If ∃ y * ∈ V , [ y * ] f ˆ ( R ) ⊈ [ y * ] f ˆ ( R + ) , then T^(t+1) ≠ T^(t), V / f ˆ ( R + ) = V / f ˆ ( R ) ∧ V / f ˆ ( R + ) .

Proof. (1) Suppose ∀ y ∈ V, [ y ] f ˆ ( R ) ⊆ [ y ] f ˆ ( R + ) . Then

[ y ] f ˆ ( R + ) = ⋂ Q ∈ f ˆ ( R + ) [ y ] Q = ( ⋂ Q ∈ f ˆ ( R ) [ y ] Q ) ∩ [ y ] f ˆ ( R + ) = [ y ] f ˆ ( R ) ∩ [ y ] f ˆ ( R + ) = [ y ] f ˆ ( R ) .

Thus, [ y ] f ˆ ( R + ) = [ y ] f ˆ ( R ) . This implies V / f ˆ ( R + ) = V / f ˆ ( R ) .

Then f ˆ ( R + ) = f ˆ ( R ) .

Hence T^(t+1) = T^(t);

(2) Let ∃ y^* ∈ V, [ y * ] f ˆ ( R ) ⊈ [ y * ] f ˆ ( R + ) . By (1), [ y * ] f ˆ ( R + ) = [ y * ] f ˆ ( R ) ∩ [ y * ] f ˆ ( R + ) ⊂ [ y * ] f ˆ ( R ) . This implies [ y * ] f ˆ ( R + ) ≠ [ y * ] f ˆ ( R ) .

Then V / f ˆ ( R + ) ≠ V / f ˆ ( R ) and so f ˆ ( R + ) ≠ f ˆ ( R ) .

Let w ∈ V / f ˆ ( R + ) . Obviously, ∃ y ∈ V such that w = [ y ] f ˆ ( R + ) . By (1), [ y ] f ˆ ( R + ) = [ y ] f ˆ ( R ) ∩ [ y ] f ˆ ( R + ) . Since [ y ] f ˆ ( R ) ∈ V / f ˆ ( R ) and [ y ] f ˆ ( R + ) ∈ V / f ˆ ( R + ) , then w ∈ V / f ˆ ( R ) ∧ V / f ˆ ( R + ) .

On the other hand, let w ∈ V / f ˆ ( R ) ∧ V / f ˆ ( R + ) , then ∃ w 1 ∈ V / f ˆ ( R ) , w 2 ∈ V / f ˆ ( R + ) such that w = w₁ ∩ w₂. So w 1 = [ y 1 ] f ˆ ( R ) , w 2 = [ y 2 ] f ˆ ( R + ) for some y₁, y₂ ∈ V. This implies w = [ y 1 ] f ˆ ( R ) ∩ [ y 2 ] f ˆ ( R + ) . Pick y₃ ∈ w, that is y 3 ∈ [ y 1 ] f ˆ ( R ) ∩ [ y 2 ] f ˆ ( R + ) . It follows from y 3 ∈ [ y 1 ] f ˆ ( R ) , y 3 ∈ [ y 2 ] f ˆ ( R + ) . We have [ y 3 ] f ˆ ( R ) = [ y 1 ] f ˆ ( R ) , [ y 3 ] f ˆ ( R + ) = [ y 2 ] f ˆ ( R + ) . Then w = [ y 3 ] f ˆ ( R ) ∩ [ y 3 ] f ˆ ( R + ) . By (1), [ y 3 ] f ˆ ( R + ) = [ y 3 ] f ˆ ( R ) ∩ [ y 3 ] f ˆ ( R + ) . Then w ∈ V / f ˆ ( R + ) . Thus V / f ˆ ( R + ) = V / f ˆ ( R ) ∧ V / f ˆ ( R + ) . □

Theorem 4.3. Let a knowledge base S^(t) = (U, R) at time (t) be compressed into T^(t) = (V, f (R)) such that S^(t) ∼  _f T^(t). Suppose that k new knowledge R 1 + , ⋯ , R k + is added to S^(t) at time (t + 1) such that S^(t+1) ∼  _f T^(t+1) where R + = R ∪ { R 1 + , ⋯ , R k + } , S^(t+1) = (U, ^R +) and T ( t + 1 ) = ( V , f ˆ ( R + ) ) . Then for the image knowledge base T^(t+1) at time (t + 1), the following properties hold:

(1) If ∀ y ∈ V , [ y ] f ˆ ( R ) ⊆ ⋂ i = 1 k [ y ] f ˆ ( R i + ) , then T^(t+1) = T^(t),

(2) If ∃ y * ∈ V , [ y * ] f ˆ ( R ) ⊈ ⋂ i = 1 k [ y ] f ˆ ( R i + ) , then T^(t+1) ≠ T^(t), V / f ˆ ( R + ) = V / f ˆ ( R ) ∧ V / f ˆ ( R 1 + ) ∧ ⋯ ∧ V / f ˆ ( R k + ) .

Proof. The proof is similar to Theorem 4.3.□ Algorithm 2 is an incremental algorithm for compressing a knowledge base under homomorphism when adding an equivalence relation. Step 2 is to compute the knowledge U/R _i with respect to R _i and the time complexity is O (|R| × |U|²); Step 3 and Setp 4 are to construct the image knowledge base and compute the knowledge V / f ˆ ( R ) and the time complexity is O (|R| × |V|²); Step 5 is to judge whether V / f ˆ ( R + ) satisfies the first situation. The time complexity of this step is O (|V|); Step 9 is to compute the partition with respect to V / f ˆ ( R + ) . The time complexity of this step is O (|V|); Step 11 and Setp 12 are to compute the reduction of knowledge base ( V , f ˆ ( R + ) ) and the time complexity is O (|^R +| !). Then the total time complexity is O (|R| × (|U|² + |V|²) + |V| + |^R +| !).

Example 4.4. (Continued from Example 3.1) Let ^R + = R ∪ {R⁺} where R⁺ is an equivalence relation on U. Put

R + = ( 1 1 0 1 0 0 1 1 1 1 1 0 0 0 0 1 1 0 1 0 0 1 1 1 1 1 0 0 0 0 0 0 1 0 1 1 0 0 0 0 0 1 1 1 1 1 1 0 1 0 0 1 1 1 1 1 0 0 0 0 0 0 1 0 1 1 0 0 0 0 0 1 1 1 1 0 0 1 0 1 1 0 0 0 0 0 1 1 1 1 1 1 0 1 0 0 1 1 1 1 1 0 0 0 0 1 1 0 1 0 0 1 1 1 1 1 0 0 0 0 1 1 0 1 0 0 1 1 1 1 1 0 0 0 0 1 1 0 1 0 0 1 1 1 1 1 0 0 0 0 1 1 0 1 0 0 1 1 1 1 1 0 0 0 0 0 0 1 0 1 1 0 0 0 0 0 1 1 1 1 0 0 1 0 1 1 0 0 0 0 0 1 1 1 1 0 0 1 0 1 1 0 0 0 0 0 1 1 1 1 0 0 1 0 1 1 0 0 0 0 0 1 1 1 1 ) .

We have

U/R⁺  =   {{x₁, x₂, x₄, x₇, x₈, x₉, x₁₀, x₁₁} , {x₃, x₅, x₆, x₁₂, x₁₃, x₁₄, x₁₅}} .

It is easy to verify that f is both type-1 and type-2 consistent with respect to R⁺ on U.

Then f is both type-1 and type-2 consistent with respect to ^R + on U.

Thus ( U , R + ) ∼ f ( V , f ˆ ( R + ) ) .

We have

            V / f ˆ ( R + ) = { { y 1 , y 2 , y 5 } , { y 3 , y 4 , y 6 } } .

By Example 3.1, V / f ˆ ( R ) = { { y 1 } , { y 2 , y 5 } , { y 3 , y 4 } , { y 6 } } .

Then  V / f ˆ ( R + ) = V / f ˆ ( R ) ∧ V / f ˆ ( R + ) = { { y 1 } , { y 2 , y 5 } , { y 3 , y 4 } , { y 6 } } .

Thus

co ( f ˆ ( R + ) ) = { { f ˆ ( R 1 ) , f ˆ ( R 2 ) } , { f ˆ ( R 2 ) , f ˆ ( R + ) } , { f ˆ ( R 1 ) , f ˆ ( R 2 ) , f ˆ ( R 3 ) } ,

  { f ˆ ( R 1 ) , f ˆ ( R 2 ) , f ˆ ( R + ) } , { f ˆ ( R 2 ) , f ˆ ( R 3 ) , f ˆ ( R + ) } } and  red ( f ˆ ( R + ) ) = { { f ˆ ( R 1 ) , f ˆ ( R 2 ) } , { f ˆ ( R 2 ) , f ˆ ( R 4 ) } } .

By theorem 2.20, we get red ( R + ) = { { R 1 , R 2 } , { R 2 , R 4 } } .

4.2 Removing knowledge

In this subsection, we illustrate how to compress a dynamic knowledge base when removing a knowledge from it by using homomorphism.

Theorem 4.5. Let a knowledge base S^(t) = (U, R) at time (t) be compressed into T^(t) = (V, f (R)) where R = {R₁, . . . , R _m } and ( U , R ) ∼ f ( V , f ˆ ( R ) ) . Suppose that a knowledge R^- ∈ R is removed from S^(t) at time (t + 1) such that ^R - = R - {R^-} and S^(t+1) = (U, ^R -). Then for the image knowledge base T ( t + 1 ) = ( V , f ˆ ( R - ) ) at time (t + 1), the following properties hold:

(1) If ∀ y ∈ V , [ y ] f ˆ ( R - ) ⊆ [ y ] f ˆ ( R - ) , then T^(t+1) = T^(t),

(2) If ∃ y * ∈ V , [ y * ] f ˆ ( R ) ⊈ [ y * ] f ˆ ( R - ) , then T^(t+1) ≠ T^(t), V / f ˆ ( R ) = V / f ˆ ( R - ) ∧ V / f ˆ ( R - ) .

Proof. (1) Suppose that for each y ∈ V, [ y ] f ˆ ( R - ) ⊆ [ y ] f ˆ ( R - ) . Then

[ y ] f ˆ ( R ) = ⋂ Q ∈ f ˆ ( R ) [ y ] Q = ( ⋂ Q ∈ f ˆ ( R - ) [ y ] Q ) ∩ [ y ] f ˆ ( R - ) = [ y ] f ˆ ( R - ) ∩ [ y ] f ˆ ( R - ) = [ y ] f ˆ ( R - ) .

Thus, [ y ] f ˆ ( R - ) = [ y ] f ˆ ( R ) . This implies V / f ˆ ( R - ) = V / f ˆ ( R ) .

Then f ˆ ( R - ) = f ˆ ( R ) .

Hence T^(t+1) = T^(t);

(2) Let ∃ y^* ∈ V, [ y * ] f ˆ ( R - ) ⊈ [ y * ] f ˆ ( R - ) . By (1), [ y * ] f ˆ ( R ) = [ y * ] f ˆ ( R - ) ∩ [ y * ] f ˆ ( R - ) ⊂ [ y * ] f ˆ ( R - ) . This implies [ y * ] f ˆ ( R - ) ≠ [ y * ] f ˆ ( R ) .

Then V / f ˆ ( R - ) ≠ V / f ˆ ( R ) and so f ˆ ( R - ) ≠ f ˆ ( R ) .

Let w ∈ V / f ˆ ( R ) . Obviously, ∃ y ∈ V such that w = [ y ] f ˆ ( R ) . By (1), [ y ] f ˆ ( R ) = [ y ] f ˆ ( R - ) ∩ [ y ] f ˆ ( R - ) . Since [ y ] f ˆ ( R - ) ∈ V / f ˆ ( R - ) and [ y ] f ˆ ( R - ) ∈ V / f ˆ ( R - ) , then w ∈ V / f ˆ ( R - ) ∧ V / f ˆ ( R - ) .

On the other hand, let w ∈ V / f ˆ ( R - ) ∧ V / f ˆ ( R - ) , then ∃ w 1 ∈ V / f ˆ ( R - ) , w 2 ∈ V / f ˆ ( R - ) such that w = w₁ ∩ w₂. So w 1 = [ y 1 ] f ˆ ( R - ) , w 2 = [ y 2 ] f ˆ ( R - ) for some y₁, y₂ ∈ V. This implies w = [ y 1 ] f ˆ ( R - ) ∩ [ y 2 ] f ˆ ( R - ) . Pick y₃ ∈ w, that is y 3 ∈ [ y 1 ] f ˆ ( R - ) ∩ [ y 2 ] f ˆ ( R - ) . It follows from y 3 ∈ [ y 1 ] f ˆ ( R - ) , y 3 ∈ [ y 2 ] f ˆ ( R - ) . We have [ y 3 ] f ˆ ( R - ) = [ y 1 ] f ˆ ( R - ) , [ y 3 ] f ˆ ( R - ) = [ y 2 ] f ˆ ( R - ) . Then w = [ y 3 ] f ˆ ( R - ) ∩ [ y 3 ] f ˆ ( R - ) . By (1), [ y 3 ] f ˆ ( R ) = [ y 3 ] f ˆ ( R - ) ∩ [ y 3 ] f ˆ ( R - ) . Then w ∈ V / f ˆ ( R ) . Thus V / f ˆ ( R ) = V / f ˆ ( R - ) ∧ V / f ˆ ( R - ) . □

Theorem 4.6. Let a knowledge base S^(t) = (U, R) at time (t) be compressed into T^(t) = (V, f (R)) such that S^(t) ∼  _f T^(t). Suppose that k knowledge R 1 - , ⋯ , R k - ∈ R is removed from S^(t) at time (t + 1) such that R - = R - { R 1 - , ⋯ , R k - } and S^(t+1) = (U, ^R -). Then for the image knowledge base T ( t + 1 ) = ( V , f ˆ ( R - ) ) at time (t + 1), the following properties hold:

(1) If ∀ y ∈ V , [ y ] f ˆ ( R - ) ⊆ ⋂ i = 1 k [ y ] f ˆ ( R i - ) , then T^(t+1) = T^(t),

(2) If ∃ y * ∈ V , [ y * ] f ˆ ( R ) ⊈ ⋂ i = 1 k [ y ] f ˆ ( R i - ) , then T^(t+1) ≠ T^(t), V / f ˆ ( R ) = V / f ˆ ( R - ) ∧ V / f ˆ ( R 1 - ) ∧ ⋯ ∧ V / f ˆ ( R k - ) .