4. Operations with Nested Named Sets
Here, we study operations with set-theoretical completely amply nested named sets. In this case, reflections in named sets are binary relations, and all elements from the support and reflectors of these named sets are also set-theoretical named sets [
1].
The intricate structure of nested named sets implies that there are three types of operations with nested named sets:
Outer operations are operations with named sets without taking into account their nested structure, i.e., without changing named sets that are nested.
Inner operations are operations with named sets that are applied to the named sets nested within them, i.e., operations that cause changes (transformations) of named sets are induced by changes (transformations) of named sets nested within them.
Combined operations are operations with named sets based on their nested structure, i.e., operations that cause direct changes (transformations) of named sets and changes (transformations)induced by changes (transformations) of named sets nested within them.
Example 8.
Let us consider such an operation as the unification υ of a set-theoretical named set. In it, all elements from the set of names are changed for one element, i.e., the set of names becomes a one-element set. The result of unification is a singlenamed set, which is, in essence, an ordinary set [
1].
By its definition, unification υ is an outer operation with named sets. Example 9.
As an example of an inner operation, it is possible to take the inner unification ν of a named set. In it, ifXis a nested named set, then the first-level named sets nested inXare unified.
Example 10.
Let us consider the following operation. Taking a nested named setX, we delete all elements from the support ofXthat are not singlenamed sets and connections of these elements. This is a combined operation with named sets.
There are two classes of outer operations with nested named sets.
Free outer operations with nested named sets do not depend on the structure of nesting.
Conditional outer operations with nested named sets are dependent on the structure of nesting.
First-level conditional outer operations with nested named sets are dependent on the first-level nested named sets of the operated named sets.
n-level conditional outer operations with nested named sets are dependent on the nested named sets of level n of the operated named sets.
Flat first-level conditional outer operations with nested named sets are dependent only on the first-level nested named sets of the operated named sets.
Flat n-level conditional outer operations with nested named sets are dependent only on the nested named sets of level n of the operated named sets.
Free outer operations with nested named sets are operations with plain named sets, which are studied in [
1].
Here, we consider some inner operations with amply nested set-theoretical named sets, i.e., we assume that all elements of the supports and sets of names in given named sets are also set-theoretical named sets. For simplicity here, we define and study only first-level inner and combined operations, i.e., operations that involve only the top named sets and named sets that are elements of the support and the set of names of the top named set, i.e., first-level named sets. First, we introduce and study set-theoretical operations. The basic operations with sets are union and intersection. Analyzing the counterparts of these operations for named sets, we find that there are six such operations [
1]. For nested named sets, there are even more unions and intersections, which are introduced and studied below.
Definition 8. The first-level disjunctive unionX∪dYof two nested named setsX= (X, f, N) andY = (Y, g, M) is defined as the named setsZ= (Z, h, Q), in which:
The constructed relation h has the following property.
Condition D.
The named set Xi∪ Yj is connected to the named set Nk∪Mr by h if and only if Xi is connected to Nk by f or Yj is connected to Mr by g.
Note that all elements,
Xi from
X;
Yj from
Y;
Nk from
N; and
Mr from
M, are named sets. Here and in what follows, ∪ is the union of (flat) named sets studied in [
1].
Let us study properties of the first-level disjunctive union of nested named sets. It has many properties similar to but not identical to those of union of sets.
By definition, the empty named set Λ = (Ø, Ø, Ø) is a nested named set, and we have the following result.
Theorem 1 (Identity Law). X ∪d Λ = Λ ∪d X = X for any nested named set X = (X, f, N).
Indeed, Λ = (Ø, Ø, Ø), whereas Xi ∪ Ø = Xi, r ∪ Ø = r, and Nk ∪ Ø = Nk.
This result shows that the empty named set Λ is the identity element with respect to first-level disjunctive union.
Theorem 2 (Commutative Law). The first-level disjunctive union of named sets is commutative, i.e., X ∪d Y = Y ∪d X for any nested named sets X and Y.
Proof.
Let us take two named sets X = (X, r, N) and Y = (Y, q, M), building their first-level disjunctive unions X ∪d Y= (Z, h, Q) and Y ∪d X= (V, k, R). Then, according to Definition 8, we have:
The relation h satisfies Condition D.
The named set Xi ∪ Yj is connected to the named set Nk ∪ Mr by h if and only if Xi is connected to Nk by f or Yj is connected to Mr by g.
The relation k also satisfies Condition D.
The named set Xi ∪ Yj is connected to the named set Nk ∪ Mr by h if and only if Yj is connected to Mr by g or Xi is connected to Nk by f.
The union of (plain) named sets is a commutative operation [
1]. This means that for all named sets
Xi ∈
X and
Yj ∈
Y, we have
Zij =
Vij, and for all named sets
Mi ∈
M and
Nj ∈
N, we have
Qij =
Rij. Consequently, we have
Z =
V and
Q =
RIn addition, relations
h and
k are constructed in the same way and therefore coincide. Then, by definition, we have:
Theorem is proved. □
Theorem 3 (Associative Law). The first-level disjunctive union is associative, i.e., X ∪d (Y ∪d Z) = (X ∪d Y) ∪d Z for any nested named sets X, Y and Z.
Proof is similar to the proof of Theorem 3. However, instead of the commutativity of the union of (plain) named sets, we use the associativity of the union of (plain) named sets [
1].
Theorem 4 (Normalization Law). For any non-empty nested named sets X and Y, their first-level disjunctive union Y ∪d X is a normalized named set if and only if at least one of the named sets X and Y is a normalized named set.
Proof.
Sufficiency. Let us take two non-empty named sets X = (X, r, N) and Y = (Y, q, M), and suppose that the named set X is normalized. Building their first-level disjunctive union X ∪d Y= (Z, h, Q), we have:
The relation h satisfies Condition D.
The named set Xi ∪ Yj is connected to the named set Nk ∪ Mr by h if and only if Xi is connected to Nk by r or Yj is connected to Mr by q.
As the named set X is normalized, for any element b = Nk from N, there is an element a = Xi from X connected to b by r. Consequently, for any element Nk ∪ Mr from Q, the element Xi ∪ Yj from Z is connected to Nk ∪ Mr by h. As b is an arbitrary element from N, the named set Nk ∪ Mr is an arbitrary element from Q, and it has an element from Z connected to it by h. This means that Y ∪d X is a normalized named set.
When Y is a normalized named set, it is treated in the same way.
Sufficiency is proved.
Necessity is proved by contradiction. To do this, we take two non-empty named sets X = (X, r, N) and Y = (Y, q, M), and suppose that neither X nor Y is normalized. By definition, this means that there is an element b = Nk from N for which there is no element a = Xi from X connected to b. Besides, there is an element d = Mr from N for which there is no element c = Yj from Y connected to d. Consequently, there is no element Xi ∪ Yj from Z connected to Nk ∪ Mr. This means that Y ∪d X is not a normalized named set.
Thus, if the first-level disjunctive union Y ∪d X is a normalized named set, then at least one of the named sets X and Y must be normalized.
Theorem is proved. □
This theorem is complemented by the following result, which is implied by Theorem 1.
Proposition 3. The named set X∪d Λ = Λ ∪d X is normalized if and only if X is a normalized named set.
It is possible to characterize conormalization in a similar way to normalization.
Theorem 5 (Conormalization Law). For any non-empty nested named sets X and Y, their first-level disjunctive union X ∪d Y is a conormalized named set if and only if at least one of the named sets X and Y is a conormalized named set.
Proof is similar to the proof of Theorem 4.
Remark 6. It is also possible to prove Theorem 5 using the duality relation between a named set and its inverse [
1].
Theorem 5 is complemented by the following result, which is directly implied by Theorem 1.
Proposition 4. The named set X∪dΛ =Λ ∪d X is conormalized if and only if X is a conormalized named set.
Theorems 4 and 5 imply the following result.
Corollary 1 (Binormalization Law). For any non-empty nested named sets X and Y, their first-level disjunctive union Y ∪d X is a binormalized named set if and only if at least one of the named sets X and Y is a binormalized named set.
Propositions 3 and 4 imply the following result.
Corollary 2. The named set X∪d Λ =Λ ∪d X is binormalized if and only if X is a binormalized named set.
Remark 7. The analogues of Theorems 4 and 5 are not true for functional named sets because in a general case, the first-level disjunctive union of two nested named sets can be not functional (cofunctional) even when both named sets are functional (cofunctional), as the following example demonstrates.
Example 11. Let us consider nested named sets X = (X, r, N) and Y = (Y, q, M), where X = {X0}, N = {N0}, Y = {Y0}, M = {M0}, r connects X0 with N0, q connects Y0 with M0, X0 = (T, p, P), Y0 = (V, t, U), T = {a}, V = {b}, and N0 = M0 = ({1}, e, {1}), where e is the identity mapping.
Both named sets X and Y are functional. However, taking the first-level disjunctive union Y ∪d X = (Z, h, Q), we see that it is not a functional named set because by construction, the element X0 ∪ Y0 from its support Z is connected by the relation h to three elements:N0 ∪ N0 = N0, M0 ∪ M0 = M0, and N0 ∪ M0 from Q.
Theorem 6 (Idempotent Law). X ∪d X = X for any nested named set X if and only if the support S(X) and the set of names N(X) are closed with respect to the unions of their elements while the naming relation r(X) satisfies Condition D.
Proof. Sufficiency. Let us take a nested named set X = (X, r, N) and suppose that it satisfies the conditions of the theorem. Building the first-level disjunctive union X ∪d X= (Z, h, Q), we have Z = {X0 ∪ X0 = X0} and Q = {N0 ∪ N0 = N0}. Then the support S(X) is closed with respect to the unions of its elements. Consequently, by Definition 8, we have:
Besides, the set of names N(
X) is also closed with respect to the unions of its elements. Consequently, by Definition 8, we have:
As the relation h satisfies Condition D, by construction, the relation r satisfies this condition and thus, the relation h coincides with the relation r. As a result, we obtain the necessary equality X ∪d X = X.
Sufficiency is proved.
Necessity. If X = (X, r, N), X ∪d X= (Z, h, Q), and X ∪d X =X, then X = Z, i.e., the support, S(X) = X is closed with respect to the unions of its elements, N = Q, i.e., the set of names N(X) = N is closed with respect to the unions of its elements, and relations h and r coincide, and thus, r satisfies Condition C
Theorem is proved. □
Corollary 3. X ∪dX =X if the support S(X) consists of one non-empty element and the set of names N(X) also consists of only one non-empty element.
Remark 8. In a general case, the first-level disjunctive union of two nested named sets is not an idempotent operation, i.e., Theorem 6 is not true for all nested named sets, as the following example demonstrates.
Example 12.
Let us take a nested named set X = (X, r, N), in which the set X contains exactly two non-empty named sets, X1 and X2, and the set of names N contains only one named set. Building the first-level disjunctive union X ∪dX= (Z, h, Q), we see that the support Z contains three named sets, X1, X2 and X1∪ X2, which are equal neither to X1 nor to X2. Thus, Z is not equal to X, and X∪dXis not equal to X.
One more binary operation with nested named sets is first-level strict disjunctive union.
Definition 9. The first-level strict disjunctive union of two nested named sets X = (X, f, N) and Y = (Y, g, M), is defined as the named sets Z = (Z, h, Q) = X ∪sd Y, in which:
The relation h is constructed in the following way.
If Xi ∩ Yj ≠ Ø, Xi is connected to Nk or Yj is connected to Mr and Nk ∩ Mr ≠ Ø, then Xi ∪ Yj is connected to Nk ∪ Mr by h.
If Xi∩Yj ≠ Ø, Xi is connected to Nk, Yj is connected to Mr, and Nk∩Mr= Ø, then Xi ∪ Yj is connected to Nk and to Mr by h.
If for any Yj∈Y (Xi∩Yj = Ø), Xi is connected to Nk by f and ∀Mj∈M (Ni∩Mj=Ø), then Xi is connected to Nk by h.
If for any Xj∈X (Yi∩Xj=Ø),Yi is connected to Mk by f and∀Nj∈N (Mi∩Nj=Ø), then Yi is connected to Mk by h.
If for any Yj∈Y (Xi∩Yj=Ø), Xi is connected to Nk by f and Nk∩Mr≠Ø, then Xi is connected to Nk ∪ Mr by h.
If for any Xj∈X (Yi∩Xj=Ø), Yi is connected to Mk by f and Nk∩Mr ≠ Ø, then Yi is connected to Nk ∪ Mr by h.
Some properties of first-level strict disjunctive union are similar to properties of first-level disjunctive union, whereas others are dissimilar. For instance, Definition 9 implies the following results.
Theorem 7 (Identity Law). X ∪sd Λ = Λ ∪sd X =X for any nested named set X= (X, f, N).
This means that the empty named set Λ is the identity element with respect to strict first-level disjunctive union.
Theorem 8 (Commutative Law). X ∪sd Y= Y ∪sd X for any nested named sets X and Y.
First-level strict disjunctive union is intrinsically connected to the union of plain named sets. To show this, we use the following concept.
Definition 10. Two named sets X and Y are disjunctive on the first level if any two named sets Xi from X and Yj from Y do not intersect.
Let us consider two nested named sets X = (X, f, N) and Y = (Y, g, M).
Theorem 9. The first-level strict disjunctive union X ∪sdY = Y ∪sd X of two nested named sets X and Y coincides with their union as plain named sets if X and Y, as well as N and M, are disjunctive on the first level.
One more binary operation with nested named sets is first-level conjunctive union.
Definition 11. The first-level conjunctive union of two nested named sets X = (X, f, N) and Y = (Y, g, M), is defined as the named sets Z = (Z, h, Q) = X ∪c Y, in which:
The relation h is constructed in such a way that it satisfies the following condition.
Condition C.
The named set Xi ∪ Yj is connected to the named set Nk ∪ Mr by h if and only if Xi is connected to Nk by f and Yj is connected to Mr by g.
Let us study the properties of the first-level conjunctive union of nested named sets.
Definition 12. A disconnected named set ΛX,N has the following form:
Theorem 10. X ∪c Λ = Λ ∪c X = ΛX,N for any nested named set X = (X, f, N).
Proof is similar to the proof of Theorem 1.
Theorem 11 (Commutative Law). X ∪c Y= Y ∪c X for any nested named sets X and Y.
Proof is similar to the proof of Theorem 2.
Theorem 12 (Associative Law). The first-level conjunctive union is associative, i.e., X ∪c (Y ∪c Z) = (X ∪c Y) ∪c Z for any nested named sets X, Y, and Z.
Proof is similar to the proof of Theorem 3.
Theorem 13 (Normalization Law). For any non-empty nested named sets X and Y, their first-level conjunctive union Y ∪c X is a normalized named set if and only if both named sets X and Y are normalized named sets.
Proof.
Sufficiency. Let us take two named sets X = (X, r, N) and Y = (Y, q, M), and suppose that both of them are normalized. Building their first-level conjunctive union X ∪c Y= (Z, h, Q), we have:
The relation h satisfies Condition D.
The named set Xi ∪ Yj is connected to the named set Nk ∪ Mr by h if and only if Xi is connected to Nk by r or Yj is connected to Mr by q.
As the named set X is normalized, for any element b = Nk from N, there is an element a = Xi from X connected to b by r. As the named set Y is normalized, for any element d = Mr from M, there is an element c = Yj from Y connected to d by q. Consequently, for any element Nk ∪ Mr from Q, the element Xi ∪ Yj from Z is connected to Nk ∪ Mr. As b is an arbitrary element from N, the named set Nk ∪ Mr is an arbitrary element from Q, and it has an element from Z connected to it by h. This means that Y ∪c X is a normalized named set.
Sufficiency is proved.
We prove necessity by contradiction. Let us take two named sets X = (X, r, N) and Y = (Y, q, M), and suppose that either X or Y is not normalized. For convenience, we assume that X is not normalized. By definition, this means that there is an element b = Nk from N, for which there is no element a = Xi from X connected to b. Then, by the construction of the relation h for any element d = Mr from M, there is no element Xi ∪ Yj from Z connected to Nk ∪ Mr. This means that Y ∪c X is not a normalized named set.
Thus, if the first-level conjunctive union Y ∪c X is a normalized named set, then both named sets X and Y must be normalized.
Theorem is proved. □
Theorem 14 (Conormalization Law). For any nested named sets X and Y, their first-level conjunctive union Y ∪d X is a conormalized named set if and only if both named sets X and Y are conormalized.
Proof is similar to the proof of Theorem 14.
Theorem 14 and 15 imply the following result.
Corollary 4 (Binormalization Law). For any nested named sets X and Y, their first-level conjunctive union Y ∪d X is a binormalized named set if and only if both named sets X and Y, are binormalized.
To continue with properties of first-level conjunctive union, we remind that if
X = (
X,
f,
N) is a named set,
a ∈
X, and
f connects
a with
b, then
b is called a name of
a [
1].
Theorem 15 (Functionality Law). For any non-empty nested named sets X and Y, their first-level conjunctive union Y ∪c X is a functional named set if and only if both X and Y are functional named sets.
Proof. Necessity. Let us take two named sets X = (X, r, N) and Y = (Y, q, M), and suppose that X is not functional and both named sets are not empty. By definition, this means that there is an element a = Xi from X that has two names b and c, i.e., a is connected to elements b and c by the relation f. Then, by the construction of the conjunctive union Y ∪c X, any element Xi ∪ Yj from Z is connected to two elements from the set of names, N(Y ∪d X). This means that Y ∪d X is not a functional named set.
The case when Y is not a functional named set is treated in the same way.
Necessity is proved.
Sufficiency. Let us take two named sets X = (X, r, N) and Y = (Y, q, M), and suppose that both of them are functional. First, we also suppose that these named sets are not empty. Building their first-level conjunctive union X ∪d Y = (Z, h, Q), we have:
Z = {Zij = Xi ∪ Yj; Xi∈X and Yj∈Y}
Q = {Qkr = Nk ∪ Mr; Nk∈N and Mr∈M}
The relation h satisfies Condition D.
The named set Xi ∪ Yj is connected to the named set Nk ∪ Mr by h if and only if Xi is connected to Nk by f or Yj is connected to Mr by g.
As the named set X is functional, then any element a = Xi from X is connected to not more than one element b = Nk from N by the relation r. As the named set Y is functional, then any element d = Yj from Y is connected to not more than one element c = Mr from M by the relation q.
Consequently, the element Xi ∪ Yj is connected to not more than one element, Nk ∪ Mr from Q by the relation h. This means that the first-level conjunctive union X ∪c Y is a functional named set.
Theorem is proved. □
Note that the empty named set Λ is functional. Therefore, if we have the first-level conjunctive union X ∪c Λ, and X is functional, then X ∪c Λ is functional because according to Theorem 1, it is equal to Λ.
Theorem 16 (Cofunctionality Law). For any nested named sets X and Y, their first-level conjunctive union Y ∪c X is a cofunctional named set if and only if both X and Y are cofunctional named sets.
Proof is similar to the proof of Theorem 15.
Remark 9. It is also possible to prove Theorem 16 using the duality relation between a named set and its inverse [
1].
Theorems 15 and 16 imply the following result.
Corollary 5 (Individualization Law). For any nested named sets X and Y, their first-level conjunctive union Y ∪c X is an individually named set if and only if both X and Y are individually named sets.
We see that some properties of first-level conjunctive union are similar to properties of first-level disjunctive union, whereas others are dissimilar.
One more binary operation with nested named sets is strict first-level conjunctive union.
Definition 13. The first-level strict conjunctive union of two nested named sets X = (X, f, N) and Y = (Y, g, M) is defined as the named sets Z = (Z, h, Q) = X ∪scY, in which:
The relation,h, is constructed in the following way.
If Xi ∩ Yj ≠ Ø, Xi is connected to Nk, Yj is connected to Mr, and Nk ∩ Mr ≠ Ø, then X i∪ Yj is connected to Nk ∪ Mr by h.
If Xi ∩ Yj ≠ Ø, Xi is connected to Nk,Yj is connected to Mr, and Nk ∩ Mr = Ø, then Xi ∪ Yj is connected to Nk and to Mr by h.
If for any Yj∈Y (Xi ∩ Yj = Ø), Xi is connected to Nk by f and ∀Mj∈M (Ni ∩ Mj = Ø), then Xi is connected to Nk by h.
If for any Xj∈X (Yi ∩ Xj = Ø), Yi is connected to Mk by f and ∀Nj∈N (Mi ∩ Nj = Ø), then Yi is connected to Mk by h.
If for any Yj∈Y (Xi ∩ Yj = Ø), Xi is connected to Nk by f and Nk ∩ Mr ≠ Ø, then Xi is connected to Nk ∪ Mr by h.
If for any Xj∈X (Yi ∩ Xj = Ø), Yi is connected to Mk by f and Nk ∩ Mr ≠ Ø, then Yi is connected to Nk ∪ Mr by h.
Some properties of first-level strict conjunctive union are similar to properties of first-level conjunctive union, whereas others are dissimilar. In particular, Definition 13 directly implies the following result.
Theorem 17 (Identity Law). X ∪sc Λ = Λ ∪sc X = X for any nested named set X = (X, f, N).
This means that the empty named set Λ is the identity element with respect to strict first-level conjunctive union.
Theorem 18 (Commutative Law). X ∪sc Y = Y ∪sc X for any nested named sets X and Y.
Proof is similar to the proof of Theorem 2.
One more binary operation with nested named sets is first-level disjunctive intersection.
Definition 14. The first-level disjunctive intersection of two nested named sets X = (X, f, N) and Y = (Y, g, M) is defined as the named sets Z = (Z, h, Q), in which:
The relation h is constructed in the following way.
If Xi is connected to Nk by f or Yj is connected to Mr by g, then when Xi ∩ Yj is not empty, it is connected to Nk ∩ Mr by h.
Here and in what follows, ∩ is the intersection of (flat) named sets studied in [
1].
Theorem 19 (Identity Law). X ∩d Λ =Λ ∩d X =Λ for any nested named set X= (X, f, N).
Indeed, Λ = (Ø, Ø, Ø), while as Xi ∩ Ø = Ø, r ∩ Ø = Ø, and Nk ∩ Ø = Ø.
This result shows that the empty named set Λ is the null element with respect to first-level disjunctive intersection.
Theorem 20 (Commutative Law). X ∩d Y= Y ∩d X for any nested named sets X and Y.
Proof. Let us take two named sets X = (X, r, N) and Y = (Y, q, M), building their first-level disjunctive intersections X ∩d Y = (Z, h, Q) and Y ∩d X= (V, k, R). Then, according to Definition 8, we have:
The relation h is constructed in the following way.
If Xi is connected to Nk by r or Yj is connected to Mr by q, then when Xi ∩ Yj is not empty, it is connected to Nk ∩ Mr by h.
The relation k is constructed in the following way.
If Xi is connected to Nk by r or Yj is connected to Mr by q, then Xi ∩ Yj is connected to Nk ∩ Mr by k.
The intersection of (plain) named sets is a commutative operation [
1]. This means that for all named sets
Xi∈
X and
Yj∈
Y, we have
Zij =
Vij, and for all named sets
Mi∈
M and
Nj∈
N, we have
Qij =
Rij. Consequently, we obtain the equalities
Z =
V and
Q =
RIn addition, relations
h and
k are constructed in the same way and therefore coincide. Then by definition, we have:
Theorem is proved. □
Theorem 21 (Associative Law). The first-level disjunctive intersection is associative, i.e., X ∩d (Y ∩d Z) = (X ∩d Y) ∩d Z for any nested named sets X, Y, and Z.
Proof is similar to the proof of Theorem 3. However, instead of the associativity of the union of (plain) named sets, we use the associativity of the intersection of (plain) named sets [
1].
Theorem 22 (Normalization Law). For any non-empty nested named sets X and Y, their first-level disjunctive intersection Y ∩d X is a normalized named set if at least one of the named sets X and Y is a normalized named set and each of the elements from its support has a non-empty intersection with some elements from the support of the other named set.
Proof. Let us take two non-empty named sets X = (X, r, N) and Y = (Y, q, M), supposing that the named set X is normalized. Building their first-level disjunctive intersection X ∩d Y = (Z, h, Q), we have:
The relation h is constructed in the following way.
If Xi is connected to Nk by r or Yj is connected to Mr by q, then when Xi∩Yj is not empty, it is connected to Nk ∩ Mr by h.
As the named set X is normalized, for any element b = Nk from N, there is an element a = Xi, from X connected to b. By the initial conditions, there is an element Yj from Y such that the element Xi ∩ Yj from Z is not empty. Consequently, for any element Nk ∩ Mr from Q, the element Xi ∩ Yj from Z is connected to Nk ∩ Mr. As b is an arbitrary element from N, Nk ∩ Mr is an arbitrary element from Q. This means that Y ∩d X is a normalized named set.
The case when Y is a normalized named set is treated in the same way.
Theorem is proved. □
Remark 10. In a general case, the first-level disjunctive intersection Y∩d X is not always a normalized named set, even if both named sets Xand Y are normalized, as the following example demonstrates.
Example 13. Let us consider nested named setsX = (X, r, N) and Y = (Y, q, M), where X = {X0}; N = {N0}; Y = {Y0}; M = {M0}; r connects X0 with N0; q connects Y0 with M0; X0 = (T, p, P); Y0 = (V, t, U); T = {a}, V = {b}; and N0 = M0 = ({1}, e, {1}), where e is the identity mapping. Bothnamed sets X and Y are normalized. However, taking the first-level disjunctive intersection Y ∩dX = (Z, h, Q), we see that it is not a normalized named set because its support is empty, as X0 ∩ Y0 = Λ, whereas its set of names, Q, contains the named set N0 and is therefore not empty.
It is possible to characterize conormalization in a similar way to normalization.
Theorem 23 (Conormalization Law). For any non-empty nested named sets X and Y, their first-level disjunctive intersection Y ∩d X is a conormalized named set if at least one of the named sets X and Y is a conormalized named set and each of the elements from its set of names has a non-empty intersection with some elements from the set of names of the other named set.
Proof is similar to the proof of Theorem 22.
Remark 11. It is also possible to prove Theorem 23 using the duality relation between a named set and its inverse [
1].
Remark 12. In a general case, the first-level disjunctive intersection Y∩d X is not always a conormalized named set, even if both named sets Xand Y are conormalized.
Theorems 22 and 23 imply the following result.
Corollary 6 (Binormalization Law). For any non-empty nested named sets X and Y, their first-level disjunctive intersection Y ∩d X is a binormalized named set if and only if at least one of the named sets X and Y is a binormalized named set, each of the elements from its support has a non-empty intersection with some elements from the support of the other named set, and each of the elements from its set of names has a non-empty intersection with some elements from the set of names of the other named set.
Remark 13. The first-level disjunctive intersection of two nested named sets can be non-cofunctional, even if both nested named sets are cofunctional.
Remark 14. In a general case, the first-level disjunctive intersection of two nested named sets is not an idempotent operation, i.e., an analogue of Theorem 6 is not true for first-level disjunctive intersection and all nested named sets, as the following example demonstrates.
Example 14.
Let us take a nested named set X = (X, r, N), in which the set X contains exactly two different non-empty named sets, X1 and X2 with anon-empty intersection, whereas the set of names N contains only one named set. Building the first-level disjunctive intersection X ∩d X = (Z, h, Q), we see that the support, Z, contains three named sets, X1, X2, and X1 ∩ X2, which is equal to neither to X1 nor X2. Therefore, Z is not equal to X, and X ∩d X is not equal to X.
One more binary operation with nested named sets is first-level conjunctive union.
Definition 15. The first-level conjunctive intersection of two nested named sets X = (X, f, N) and Y = (Y, g, M) is defined as the named sets Z = (Z, h, Q), in which:
The relation h, is constructed in the following way.
If X ∩ Yj if it is not empty, Xi is connected to Nk by f and Yj is connected to Mr by g, then Xi ∩ Yj is connected to Nk ∩ Mr by h.
Let us suppose that in non-empty named sets X = (X, r, N) and Y = (Y, q, M), each of the elements from X has a non-empty intersection with some elements from Y, and each of the elements from Y has a non-empty intersection with some elements from X.
Theorem 24 (Functionality Law). The first-level conjunctive intersection Y ∩c X is a functional named set if and only if both X and Y are functional named sets.
Proof. Sufficiency. Let us take two named sets X = (X, r, N) and Y = (Y, q, M), and suppose that both of them are functional. First, we also suppose that these named sets are not empty. Building their first-level conjunctive intersection X ∩c Y= (Z, h, Q), we have:
The relation h is constructed in the following way.
If Xi is connected to Nk by r and Yj is connected to Mr by q, then when Xi∩Yj is not empty, it is connected to Nk ∩ Mr by h.
As the named set X is functional, any element a = Xi, from X is connected to not more than one element b = Nk from N by the relation r. As the named set Y is functional, any element d = Yj from Y is connected to not more than one element c = Mr from M by the relation q.
Consequently, the element Xi ∩ Yj is connected to not more than one element Nk ∩ Mr, from Q by the relation h. This means that the first-level conjunctive intersection X ∩cY is a functional named set.
Sufficiency is proved.
Necessity. Let us take two named sets X = (X, r, N) and Y = (Y, q, M) and suppose that X is not functional, whereas both named sets are not empty. By definition, this means that there is an element a = Xi from X that has two names b and c, i.e., a is connected to b and c by the relation r. By the initial assumption, there is an element d = Yj from Y such that a ∩ d ≠ Ø. Then, by the construction of their first-level conjunctive intersection X ∩cY= (Z, h, Q), the element a ∩ d = Xi ∩ Yj from Z is connected to b ∩ c from Q. As a is an arbitrary element from X, this means that Y ∩c X is not a functional named set.
The case when Y is not a functional named set is treated in the same way.
Thus, if the first-level conjunctive intersection Y ∩c X is a functional named set, then both of the named sets X and Y must be functional.
Theorem is proved. □
Let us suppose that in non-empty named sets X = (X, r, N) and Y = (Y, q, M), each of the elements from N has a non-empty intersection with some elements from M, and each of the elements from M has a non-empty intersection with some elements from N.
Theorem 25 (Cofunctionality Law). For any nested named sets X and Y, their first-level conjunctive intersection Y ∩c X is a cofunctional named set if and only if both X and Y are cofunctional named sets.
Proof is similar to the proof of Theorem 6.
Remark 15. It is also possible to prove Theorem 25 using the duality relation between a named set and its inverse [
1].
Let us assume that in non-empty named sets, X = (X, r, N) and Y = (Y, q, M), each of the elements from X has a non-empty intersection with some elements from Y, each of the elements from Y has a non-empty intersection with some elements from X, each of the elements from N has a non-empty intersection with some elements from M, and each of the elements from M has a non-empty intersection with some elements from N. Then, Theorems 24 and 25 imply the following result.
Corollary 7 (Individualization Law). For any nested named sets X and Y, their first-level conjunctive intersection Y ∪c X is an individually named set if and only if both X and Y are individually named sets.