=Paper=
{{Paper
|id=Vol-2870/paper66
|storemode=property
|title=Methods of Eliminating Features from Linguistic Equations
|pdfUrl=https://ceur-ws.org/Vol-2870/paper66.pdf
|volume=Vol-2870
|authors=Dmitry Sitnikov,Polina Sytnikova,Andrii Kovalenko
|dblpUrl=https://dblp.org/rec/conf/colins/SitnikovSK21
}}
==Methods of Eliminating Features from Linguistic Equations==
https://ceur-ws.org/Vol-2870/paper66.pdf
Methods of Eliminating Features from Linguistic Equations
Dmitry Sitnikova, Polina Sytnikovaa, Andrii Kovalenkoa
a
Kharkiv National University of Radio Electronics, Nauky Ave. 14, Kharkiv, City, 61166, Ukrain
Abstract
In this paper approaches to modelling relations between discrete linguistic features are
considered. Linguistic equations as a tool for describing complicated logic dependencies
between semantic and syntactic features have been investigated. Finite predicate equations
have been considered from the viewpoint of quick finding hidden dependencies in data. A way
to defining the tightness of links between discrete features has been suggested. For this purpose,
different types of substitution operators have been investigated. A class of finite predicates that
allows eliminating non-salient features without an increase in the size of the original formula
has been considered in relation with some linguistic examples. The results obtained can be used
not only in applied linguistic, but also in other fields where deductive inferences in knowledge
bases are important.
Keywords 1
Knowledge base, linguistic equations, finite predicates, logic equations, variable elimination
1. Introduction
To formalize information on objects and processes in databases, a variety of discrete mathematics
methods are used. In cases where such information represented by discrete information features has a
complicated logic structure, in particular, to represent it formally, logic equations with Boolean
variables are used. Logic methods of pattern recognition suppose composing and solving logic
equations with variables that take on values 1 and 0, depending on whether the given object has a certain
property. Solving such equations allow either identifying the object by the available set of values for
feature variables or determine unknown properties of the given object [1]. A natural generalization of
Boolean algebra equations are finite predicate algebra equations [2] that provide the possibility of
operating with arbitrary feature variables defined on finite sets (alphabets). Using such equations for
building logic inferences in knowledge bases allows extending the possibilities of logic methods for
pattern recognition [3].
For solving a variety of linguistic problems, some of which are describe in this paper, a solution is
suggested based on finite predicate algebra equations.
A universal way for solving finite algebra equations is the transformation of the predicate defined
by a system of logic equations and initial variable values to the perfect disjunctive normal form [1].
Nevertheless, such a procedure implies exhaustive search for a great number of intermediate solutions,
and its practical implementation requires significant time and memory resources.
We will show that for some quite general types of linguistic equations, when peculiarities of their
structure are taken into consideration, it is possible to develop much simpler methods their solution.
Such methods differ from, for example, heuristic algorithms [3], that suppose finding all sets of values
for semantic features, which slows down the solution finding process if the number of variables
increases and time increases exponentially.
COLINS-2021: 5th International Conference on Computational Linguistics and Intelligent Systems, April 22β23, 2021, Kharkiv, Ukraine
EMAIL: dmytro.sytnikov@nure.ua (D. Sitnikov); polina.sytnikova@nure.ua (P. Sytnikova); andrey.kovalenko@nure.ua (A. Kovalenko)
ORCID: 0000-0003-1240-7900 (D. Sitnikov); 0000-0002-6688-4641 (P. Sytnikova); 0000-0003-2882-5082 (A. Kovalenko)
Β©οΈ 2021 Copyright for this paper by its authors.
Use permitted under Creative Commons License Attribution 4.0 International (CC BY 4.0).
CEUR Workshop Proceedings (CEUR-WS.org)
2. Related Works
The scientific field NLP deals with natural language processing. Its origin goes back to the middle
of the last century. In the last decades it has become one of the most important artificial intelligence
technologies. The application of logic methods to the solution of applied linguistic problems is now
widely spread not only in linguistic research, but also in other scientific fields related to discrete
information processing. One of the main achievements of mathematical linguistic has lately become the
application of complicated logic methods to the investigation of natural language syntax [5].
The first stage of solving many practical problems is the construction of a mathematical model that
is often represented in the form of equations. Linguistic equations are used in many fields. For example,
in [6] the authors use linguistic equations for the description of fuzzy logic inference for solving the
problem of increasing effectiveness of electronic control detail production in the automobile industry.
In [7] linguistic equations are a basis for developing various types of fuzzy models with different types
of rules that describe relations in these models. An interesting application of NLP techniques is also
textual information processing in medical reports [8].
A broad field of using linguistic equations has led to the development of methods and algorithms of
their solution. For example, in [9] a system of linguistic equations of a special type where each equation
can contain operations of concatenation is considered. In [10] equations with formal languages, using
all Boolean operations and concatenation have been investigated, issues of solution existence and
uniqueness being considered.
Building a linguistic equation is always associated with the description of the set of linguistic
features related to a concrete task. In [11] a language for defining mathematical problems and its
association with natural language is analyzed by forming corresponding feature sets. In [12] the authors
have analyzed sets of linguistic features for developing a model of linguistic constructs for the analysis
of writing quality
In several research papers Boolean algebra tools are considered as an approach to solving linguistic
equations. For example, paper [13] is devoted to solving a system of Boolean equations with the
operations of union and negation. It is determined there whether such a system has solutions.
At present, since huge datasets that are available on the Internet, approaches to solving linguistic
problems change and require scalable methods for the analysis of data and texts. In paper [14] it is
proposed to use Big Data methods for improving ways for solving semantic problems related to natural
language texts.
3. Tightness of links between features
In many practical tasks associated with the semantic processing of natural language information it s
not necessary to obtain all semantic feature value sets, but it is required to obtain one or several value
sets for target features that are interesting for user. Often it is necessary to find variable value sets under
predefined initial conditions represented in the form of a fixed set of values for other features. When
such problems are solved the variables that are not included in the initial conditions and are not target
ones are eliminated from the equation by the application of existence quantifiers [1].
When knowledge bases with linguistic variables and corresponding inferences are considered
questions concerning determining the tightness of links between object features arise. Also, it is often
important to know if such links are salient. Probably a formal link between features is stronger if fewer
sets of variable values satisfy the equation. At that, if any sets of variable values satisfy the original
equation, one can conclude that there is no relation between these variables.
Besides, when practical problems are solved, the following questions arise:
1. How do concrete values of a given feature substituted in the logic equation affect links between
other features?
2. How strong is the logic dependence between two (or more) given features?
In order to obtain an answer to the first question, it is necessary to consider the predicates (and
equations correspondingly) that after the substitution of a certain feature value are transformed into
predicates with a stronger link between features, and the predicates for which the substitution of a given
feature value leads to weakening the logic link between features.
In order to obtain an answer to the second question, it is necessary to eliminate from the original
equation with the help of the existence quantifier all the variable except the considered ones and
investigate the resulting equation with a fewer number of variables, which describes all allowable sets
of investigated feature values.
The mentioned procedures will be considered in the next sections.
In order to answer the posed questions, it is necessary to consider different types of finite algebra
predicated and effective methods for eliminating variables from such equations.
4. Eliminating variables with the help of logic quantification operations and
simplifying finite predicate formulae.
Let predicate π depend on variables π₯, π¦, . . . , π§. Let us define the substitution operator π(π) (π is an
element from the domain for the variable π₯) that is applied to the predicate π in the following way:
π(π(π₯, π¦, . . . , π§) = π(π, π¦, . . . , π§).
Let us call this operator a limiting one, if the following condition holds
π(π, π¦, . . . , π§) β π(π₯, π¦, . . . , π§) (1)
for any π₯, π¦, . . . , π§.
Let us call this operator a spreading one, if the following condition holds:
π(π, π¦, . . . , π§) β π(π₯, π¦, . . . , π§) (2)
for any π₯, π¦, . . . , π§.
Let us call this operator a shifting one, if both conditions (1) and (2) do not hold.
Limiting operators strengthen the logic link between discrete features, spreading substitution
operators weaken such a link, shifting operators transform the link between the features in an arbitrary
way.
Let us represent the predicate π as follows:
π(π₯, π¦, β¦ , π§) = π₯ π1 π1 (π¦, β¦ , π§) β¨ π₯ π2 π2 (π¦, β¦ , π§) β¨ β¦ β¨ π₯ ππ ππ (π¦, β¦ , π§).
Then π1 (π) = π1 (π¦, β¦ , π§)
Obviously, the operator π1 (π) will be a limiting one, if π1 β ππ βπ = 1,2, . . . , π.
The operator π1 (π) will be a spreading one, if π1 β ππ βπ = 1,2, . . . , π. The operator π1 (π) will be
a shifting one, if both conditions do not hold.
Consider the application of the substitution operator π1 to a predicate π(π₯, π¦), where the variables
π₯, π¦ and π§ have the domains {π1 , π2 }, {π1 , π2 } and {π1 , π2 } correspondingly.
Suppose
π = π₯ π1 π¦ π1 π§ π1 β¨ π₯ π2 π¦ π1 π§ π2 β¨ π₯ π2 π¦ π1 π§ π1 .
Then
π1 (π) = π¦ π1 π§ π1 = (π₯ π1 β¨ π₯ π2 ) β§ π¦ π1 π§ π1 = π₯ π1 π¦ π1 π§ π1 β¨ π₯ π2 π¦ π1 π§ π1 .
The predicate π, except for those disjuncts contained in π1 (π), one more disjunct π₯ π2 π¦ π1 π§ π1 . It
means that the operator π1 is a limiting one for the predicate π. In terms of the introduced definitions,
for the given example π1 = π¦ π1 π§ π1 , π2 = π¦ π1 π§ π2 β¨ π¦ π1 π§ π1 and, obviously, π1 β π2 .
Consider then the predicate
π = π₯ π1 π¦ π1 π§ π1 β¨ π₯ π1 π¦ π1 π§ π2 β¨ π₯ π2 π¦ π1 π§ π1 .
Then
π1 (π) = π¦ π1 π§ π1 β¨ π¦ π1 π§ π2 = (π₯ π1 β¨ π₯ π2 ) β§ (π¦ π1 π§ π1 β¨ π¦ π1 π§ π2 ) =
= π₯ π1 π¦ π1 π§ π1 β¨ π₯ π2 π¦ π1 π§ π2 β¨ π₯ π2 π¦ π1 π§ π1 β¨ π₯ π2 π¦ π1 π§ π2 .
The operator π1 for this predicate is a spreading one. For the given example π1 = π¦ π1 π§ π1 β¨ π¦ π1 π§ π2 ,
Π° π2 = π¦ π1 π§ π1 . It means π1 β π2 .
In the case when the predicate P, for example, is represented in the form
π = π₯ π1 π¦ π1 π§ π1 β¨ π₯ π1 π¦ π2 π§ π2 β¨ π₯ π2 π¦ π1 π§ π2 .
π1 (π) = π¦ π1 π§ π1 β¨ π¦ π2 π§ π2 = (π₯ π1 β¨ π₯ π2 ) β§ (π¦ π1 π§ π1 β¨ π¦ π2 π§ π2 ) =
= π₯ π1 π¦ π1 π§ π1 β¨ π₯ π1 π¦ π2 π§ π2 β¨ π₯ π2 π¦ π1 π§ π1 β¨ π₯ π2 π¦ π2 π§ π2
i.e., the operator π1 is a shifting one.
The general method for variable elimination looks as follows [1]. Consider the finite predicate
algebra equation
π(π₯1 , π₯2 , . . . π₯π , π₯π+1 , . . . , π₯π , π₯π+1 , . . . , π₯π ) = 1, (3)
where each variable π₯1 , π₯2 , . . . , π₯π has the domain π΄π = {ππ1 , ππ2 , . . . , ππππ }, π = 1, π, at that, for the
system of features the laws of truthfulness should hold:
π π π
π₯π π1 β¨ π₯π π2 β¨ β¦ β¨ π₯π ππ = 1, π = Μ
Μ
Μ
Μ
Μ
1, π.
Also the laws of falseness should hold:
π π
π₯π ππ β§ π₯π ππ = 0, π β π, π, π = Μ
Μ
Μ
Μ
Μ
Μ
1, ππ .
Suppose also that for the given values {π1π1 , π2π2 , . . . , ππππ } of the variables π₯1 , π₯2 , . . . , π₯π it is
necessary to compute the values of variables π₯π+1 , . . . , π₯π such that the equation (3) will be true for
some values of the variables π₯π+1 , . . . , π₯π .
Let us define this problem mathematically:
βπ₯π+1 . . . βπ₯π π(π1π1 , . . . ππππ , π₯π+1 , . . . , π₯π , π₯π+1 , . . . , π₯π ) = 1, (4)
which can be represented in the finite predicate language as follows:
ππ+1 ππ
β¨π=1 ππ+1,π . . . β¨π=1 πππ π(π1π1 , . . . ππππ , ππ+1,π , . . . , ππ,π , π₯π+1 , . . . , π₯π ) = 1,
where only for the variables π₯π+1 , . . . , π₯π possible sets of values should be found.
The system of linguistic equations of the form
π¦ π΅π = ππ (π₯1 , π₯2 , . . . , π₯π ), π = 1, π (5)
that satisfies the condition
ππ (π₯1 , π₯2 , . . . , π₯π ) β§ ππ (π₯1 , π₯2 , . . . , π₯π ) = 0, (6)
π β π, π, π, = 1, π
can be transformed to the following form:
β¨ππ=1 π¦ π΅π β§ ππ (π₯1 , π₯2 , . . . , π₯π ) = 1. (7)
Nevertheless, the algorithm of variable elimination with the help of the existence quantifier has a
high complexity if arbitrary predicates are considered.
Suppose the predicate π(π₯1 , π₯2 , . . . , π₯π ) has the following form:
π = π1 (π₯π11 , π₯π12 , . . . , π₯π1π )π2 (π₯π21 , π₯π22 , . . . , π₯π2π ) β§. . .β§ ππ (π₯ππ1 , π₯ππ2 , . . . , π₯πππ ),
where
{π₯π11 , π₯π12 , . . . , π₯π1π } β© {π₯π21 , π₯π22 , . . . , π₯π2π } β©. . .
β© {π₯ππ1 , π₯ππ2 , . . . , π₯πππ } = {π₯π1 , π₯π2 , . . . , π₯ππ } = π΄.
Suppose also we should eliminate variables the variables from the set A.
Consider the application of the existence quantifier to the variable π₯π1 from the set π΄
π π π
βπ₯π1 (π) =β¨ π1 1 π2 1 . . . ππ1 ,
π1
π
where ππ 1 = ππ1 π1 (ππ ), π = 1, π.
For example, π = π1 (π₯1 , π₯2 , π₯3 , π₯4 ) β§ π2 (π₯2 , π₯3 , π₯5 ) β§ π3 (π₯2 , π₯3 , π₯6 , π₯7 ). Let the domain for π₯2
consist of 3 values, and the domain for π₯3 from 2 values. Then
βπ₯2 (π) = π1 (π₯1 , π21 , π₯3 , π₯4 )π2 (π21 , π₯3 , π₯5 )π3 (π21 , π₯3 , π₯6 , π₯7 ) β¨
β¨ π1 (π₯1 , π22 , π₯3 , π₯4 )π2 (π22 , π₯3 , π₯5 )π3 (π22 , π₯3 , π₯6 , π₯7 ) β¨
β¨ π1 (π₯1 , π23 , π₯3 , π₯4 )π2 (π23 , π₯3 , π₯5 )π3 (π23 , π₯3 , π₯6 , π₯7 ) =
π π π
=β¨ π1 1 (π₯1 , π₯3 , π₯4 )π2 1 (π₯3 , π₯5 )π3 1 (π₯3 , π₯6 , π₯7 ).
π1
The quantification looks as follows:
π π π π π π π π
βπ₯π2 (π) (βπ₯π1 (π)) =β¨ β¨ π1 1 2 π2 1 2 β¦ ππ1 2 , Π³Π΄Π΅ ππ 1 2 = ππ2 π2 (ππ1 π1 (ππ )), π = 1, π.
π2 π1
Then for the considered example
βπ₯3 (π)βπ₯2 (π) = π1 (π₯1 , π21 , π31 , π₯4 )π2 (π21 , π31 , π₯5 )π3 (π21 , π31 , π₯6 , π₯7 ) β¨
π1 (π₯1 , π21 , π31 , π₯4 )π2 (π21 , π31 , π₯5 )π3 (π21 , π31 , π₯6 , π₯7 ) β¨
β¨ π1 (π₯1 , π22 , π31 , π₯4 )π2 (π22 , π31 , π₯5 )π3 (π22 , π31 , π₯6 , π₯7 ) β¨
β¨ π1 (π₯1 , π23 , π31 , π₯4 )π2 (π23 , π31 , π₯5 )π3 (π23 , π31 , π₯6 , π₯7 ) β¨
β¨ π1 (π₯1 , π21 , π32 , π₯4 )π2 (π21 , π32 , π₯5 )π3 (π21 , π32 , π₯6 , π₯7 ) β¨
β¨ π1 (π₯1 , π22 , π32 , π₯4 )π2 (π22 , π32 , π₯5 )π3 (π22 , π32 , π₯6 , π₯7 ) β¨
β¨ π1 (π₯1 , π23 , π32 , π₯4 )π2 (π23 , π32 , π₯5 )π3 (π23 , π32 , π₯6 , π₯7 ) =
=β¨ β¨ π1 (π₯1 , π2π1 , π3π2 , π₯4 )π2 (π2π1 , π3π2 , π₯5 )π3 (π2π1 , π3π2 , π₯6 , π₯7 ) =
π2 π1
π π π π π π
=β¨ β¨ π1 1 2 (π₯1 , π₯4 )π2 1 2 (π₯5 )π3 1 2 (π₯6 , π₯7 ).
π2 π1
It can be seen from the above example that, when the existence quantifier is applied sequentially to
the variables from the set A (i.e., to the variables common for all the predicates), the general formula
for the original predicate ππ operations more. Nonetheless, there exist many problems for which the
conditions are defined by the predicate that has such a structure that the complexity of problem solving
with the help of eliminating non-salient variables and finding values of target variables is much lower
than in the general case. There exist also cases where it is possible to simplify a predicate obtained at
an intermediary stage of solving the equation. Let us consider such cases.
Consider a particular case of the defined task, where the set A consists of one element:
π = π1 (π₯1 , . . . , π₯π1 , π₯π )π2 (π₯π1 +1 , . . . , π₯π2 , π₯π ) β§. . .β§ ππ (π₯ππβ1 +1 , . . . , π₯ππ , π₯π ), (8)
where
{π₯1 , . . . , π₯π1 } β© {π₯π1 +1 , . . . , π₯π2 } β©. . .β© {π₯ππβ1 +1 , . . . , π₯ππ } = β
.
Then, as a result of the application of the existence quantifier to the variable π₯π , we obtain the
disjunctions of predicate conjunctions the variables of which do not intersect:
βπ₯π (π) =β¨ π1 (π₯1 , . . . , π₯π1 , πππ ) β§. . .β§ ππ (π₯ππβ1 +1 , . . . , π₯ππ , πππ ) = (9)
π
=β¨ π1π (π₯1 , . . . , π₯π1 ). . . πππ (π₯ππβ1 +1 , . . . , π₯ππ ).
π
Let us investigate the possibility of minimizing the obtained predicate.
Statement 1. Let π = π1 (π₯1 , . . . , π₯π1 ) β§. .. β§ ππ (π₯ππβ1 +1 , . . . , π₯ππ ), πΊ = πΊ1 (π₯1 , . . . , π₯π1 ) β§. .. β§
πΊπ (π₯ππβ1 +1 , . . . , π₯ππ ). The implication
πβπΊ (10)
holds if and only if βπ = 1, π ππ β πΊπ .
Proof. Necessity. Let ππ β πΊπ βπ = 1, π, i.e., πΊπ = ππ β¨ ππ . Then πΊ =β§ πΊπ =β§ (ππ β¨ ππ ) =β§ ππ β¨
π π π
β§ ππ πΊπ β¨β§ ππ = π β¨ π¬.
π,π π
Hence the implication (10) is true.
Sufficiency. Let the implication (10) is true. Suppose, βπ β {1. . . π} such that ππ β πΊπ . Then, if the
predicate ππ contains such an elementary conjunction πΆπ that it is not present in the predicate πΊπ , since
the domains for the predicates do not intersect, the predicate πΊ will not contain πΆπ as well, whereas it
is present in the predicate π. In this case the implication (10) is false, which contradicts the premise.
Consequence. Let the predicate π satisfies the condition (8), i.e., the application of the operator
π π
βπ₯π (π) is defined by the formula (9). Then, if βπ = 1, π ππ1 β ππ2 , the addend π2 is simplified and
formula (9) is minimized.
Statement 2. Let ππ = πππ (π), ππ = πππ (π). The implication
ππ β ππ (11)
is true for the predicate π one of the following conditions holds:
πππ is a shifting operator, and πππ is a spreading operator;
πππ is a limiting operator, and πππ is a spreading operator;
πππ is a limiting operator, and πππ is a shifting operator.
Proof. Necessity. Suppose the implication (11) is true. Suppose πππ is a spreading operator, and πππ
is a limiting operator. Then πππ (π) β π, π β πππ (P), which contradicts condition (11). Suppose that
πππ is a spreading operator, and πππ is a shifting one. Since the substitution operators πππ and πππ are
π
applied to the same variable π₯π , the predicate π contains addends of the form π₯π ππ π΄π and
πππ
π
π₯π , where β¨ π΅π ββ¨ π΄π . Then πππ (π) =β¨ π΄π β¨ πΆ; πππ (π) =β¨ π΅π β¨ πΆ, where πΆ does not depend on the
π π π π
variable π₯π . Hence, ππ β ππ , which contradicts the premise.
Suppose πππ and πππ are shifting operators. Then the analogous considerations demonstrate the fact
that π΄π and π΅π are different and the condition (11) does not hold.
Let πππ is a shifting operator, and πππ is a limiting operator. Then in elementary conjunctions
π
containing the predicate π₯π ππ , are not present in the disjunctive normal form for the predicate P, and the
predicate π can be presented as follows:
π
π =β¨ π₯π ππ π΄π β¨ π΅π β¨ πΆ,
π π
π π
where π΅π contains the "recognition" of the variable π₯π , besides π₯π ππ and π₯π ππ , and πΆ does not contain
"recognitions" of this variable. Then πππ (π) =β¨ π΄π β¨ πΆ, πππ (π) = πΆ and πππ (π) β πππ (π), which
π
contradicts the premise.
If both operators are limiting or spreading ones, their application to the predicate π is the same.
Sufficiency. Follows from the first part of the proof.
π π π π π π π π π
Example. Let π = π₯1 11 π₯2 21 π₯3 31 β¨ π₯1 12 π₯2 21 π₯3 33 β¨ π₯1 12 π₯2 22 π₯3 33 . Then π21 (π) =
π11 π31 π12 π33 π12 π33
= π₯1 π₯3 β¨ π₯1 π₯3 is a spreading substitution operator, and π22 (π) = π₯1 π₯3 is a shifting
operator and
π21 (π) β π22 (π).
π π π π π π π π
Let π = π₯1 12 π₯2 21 π₯3 33 β¨ π₯1 12 π₯2 22 π₯3 33 β¨ π₯3 31 . Then π11 (π) = π₯3 31 is a limiting operator, and
π π π π π
π12 (π) = π₯1 12 π₯3 33 β¨ π₯2 22 π₯3 33 β¨ π₯3 31 is a shifting one, at that, π11 (π) β π12 (π).
π11 π12 π22 π31 π π π π
Let π = π₯1 β¨ π₯1 π₯2 π₯3 β¨ π₯1 12 π₯2 22 π₯3 33 . Then π32 (π) = π₯1 11 is a limiting operator, and
π π π
π31 (π) = π₯1 11 β¨ π₯1 12 π₯2 22 is a spreading operator and π32 (π) β π31 (π).
5. A method of feature elimination from predicates represented in the
disjunctive and conjunctive normal forms.
Let us consider some types of predicate equations whose structure allows us to substantially simplify
the process of variable elimination.
Suppose we have a model represented in the form of systems of logic equations that are written in
the disjunctive normal form where all conjunctions of different elements on the left side are equal to
zero. Then, by transforming the system to a single equation ((5)-(7)), we obtain an equation in the
disjunctive normal form. By denoting πΊππ π th disjunction in the equation corresponding to the description
of πth object in the subject field, the following equation is obtained:
π = β¨ πΊππ . (12)
π,π
Then βπ₯π (π) = β¨ (πΊππ ). The application of the quantifier βπ₯π (πΊππ ) does not change πΊππ , if πΊππ des
π,π
not contain the variable π₯π . Thus, the application of the quantifier in this case is equivalent to eliminating
the recognition of the given variable from the elementary conjunction. It follows from the described
properties that in the case of disjunctive normal form eliminating variables from equation (7) by the
application of the existence quantifier simplifies the given equation, since the number of recognitions
(elementary predicates with one variable that are equal to 1 if and only if the value of the variable is the
same as the given element) does not increase (very often decreases).
Let a model be represented in the conjunctive normal forms in which every elementary disjunction
is a unary predicate. Suppose also that property (6) holds. In this case this condition means that for any
two conjunctive normal forms on the right side of the equations the following statement is true: we can
find two elementary disjunctions from different conjunctive normal forms the multiplication of which
is zero. Then for solving linguistic equation of this type it is possible to apply the existence quantifier
to the intermediary variables to eliminate them. After transforming the given system of equations to a
single logic equation we obtain an expression that is written as the disjunction of conjunctive normal
forms containing elementary conjunctions represented by unary predicates, which substantially
simplifies the process of eliminating the non-salient variables.
Let us denote elementary disjunctions in the form of unary predicates as π·ππ (π = 1, π). The
application of the existence quantifier to the disjunction of the conjunctive normal forms means the
application of the quantifier to this variable in every conjunctive normal form ο i , for which the
following formula is true:
π·π = βππ=1 π·ππ .
Every disjunctive normal form π·ππ contains the disjunction of a certain number of recognition
predicates for π₯π . Taking into consideration the above notation, it is possible to write down the following
equation:
βπ₯π π·π = π·π1 π·π2 . . . π·π(πβ1) (βπ₯π π·ππ )π·π(π+1) . . . π·ππ β‘ π·π1 π·π2 . . . π·π(πβ1) π·π(π+1) . . . π·ππ ,
where π = 1, π, π is the cardinality of the subject field. If the conjunctive normal form π·π does not
contain some variable π₯π , the application of the existence quantifier does not change π·π .
Thus, the application of the existence quantifier to the intermediary variables does not lead to any
increase in the number of recognitions. In some cases, eliminating variable with the help of the
quantifier does not change the original formula. Nevertheless, as a rule, the application of the described
method leads to a substantial decrease in the number of formula terms (recognitions). Hence, the using
of the quantifier in the considered cases does not complicate the original model.
Let us represent a generalized method for solving systems of linguistic equations with target
variables and initial conditions.
1. Check if the conjunctions of any 2 predicates on the right sides of equations are zeros.
2. Represent the original system in the form of a single equation.
3. Substitute the initial values of selected variables in the obtained equation.
4. Eliminate all variables except for the target ones by the application of the existence quantifier.
5. The ordered sets of values for the target variables that satisfy the equation obtained at the previous
stage is the solution for this problem.
Let us consider an example of solving a system of linguistic equations the right sides of which are
disjunctive normal forms. Consider the following system of logic equations:
π π π π π π π π
π¦ π΄ = π₯1 11 π₯2 21 π₯3 32 β¨ π₯1 12 π₯2 21 π₯3 33 β¨ π₯1 13 π₯2 23 ,
π π π π
{π¦ π΅ = π₯1 11 π₯2 22 β¨ π₯2 21 π₯3 31 , (13)
π π π π π
π¦ πΆ = π₯1 11 π₯221 π₯3 32 β¨ π₯1 12 π₯2 21 π₯3 32 .
where π¦ = {π΄, π΅, πΆ}, π₯1 β {π11 , π12 , π13 }, π₯2 β {π21 , π22 , π23 }, π₯3 β {π31 , π32 , π33 }.
It is required for the initial condition π₯2 = π21 to find values of the target feature π₯1 .
Let us solve the problem step by step in accordance with the algorithm.
1. Check the fact that the paired conjunctions of the right sides of equations are zeros.
π π π π π π π π π π π π π π
π₯1 11 π₯2 21 π₯3 32 β§ π₯1 11 π₯2 22 = 0, π₯1 12 π₯2 21 π₯3 33 β§ π₯1 11 π₯2 22 = 0, π₯1 13 π₯2 23 β§ π₯1 11 π₯2 22 = 0,
π π π π π π π π π π π π π π
π₯1 11 π₯2 21 π₯3 32 β§ π₯2 21 π₯3 31 = 0, π₯1 12 π₯2 21 π₯3 33 β§ π₯2 21 π₯3 31 = 0, π₯1 13 π₯2 23 β§ π₯2 21 π₯3 31 = 0.
Hence, the conjunction of the right sides of the second equation is zero.
Further,
π π π π π π π π
π₯1 11 π₯2 22 β§ π₯1 11 π₯221 π₯3 32 = 0, π₯2 21 π₯3 31 β§ π₯1 11 π₯221 π₯3 32 = 0,
π11 π22 π12 π21 π32 π21 π31 π12 π21 π32
π₯1 π₯2 β§ π₯1 π₯2 π₯3 = 0, π₯2 π₯3 β§ π₯1 π₯2 π₯3 = 0.
i.e., the conjunction of the right sides of the second and third equations is zero. By analogy, the
conjunction of the right sides of the first and third equations is zero.
2. Let us represent the system (13) in the form of a single equation:
π π π π π π π π π π π π
π¦ π΄ β§ (π₯1 11 π₯2 21 π₯3 32 β¨ π₯1 12 π₯2 21 π₯3 33 β¨ π₯1 13 π₯2 23 ) β¨ π¦ π΅ β§ (π₯1 11 π₯2 22 β¨ π₯2 21 π₯3 31 ) β¨
π π π π π
β¨ π¦ πΆ β§ (π₯1 11 π₯221 π₯3 32 β¨ π₯1 12 π₯2 21 π₯3 32 ) = 1.
3. Substitute the initial value of the variable π₯2 = π21 in the obtained equation to get the following:
π π π π π π π π π
π¦ π΄ β§ (π₯1 11 π₯3 32 β¨ π₯1 12 π₯3 33 ) β¨ π¦ π΅ β§ (π₯3 31 ) β¨ π¦ πΆ β§ (π₯1 11 π₯3 32 β¨ π₯1 12 π₯3 32 ) = 1
4. Use the quantifiers for a sequential elimination of the variables π¦ and π₯3 from the last equation:
π π π π π π π π π
βπ¦ (π¦ π΄ β§ (π₯1 11 π₯3 32 β¨ π₯1 12 π₯3 33 ) β¨ π¦ π΅ β§ (π₯3 31 ) β¨ π¦ πΆ β§ (π₯1 11 π₯3 32 β¨ π₯1 12 π₯3 32 )) =
π π π π π π π π π
= (π₯1 11 π₯3 32 β¨ π₯1 12 π₯3 33 ) β¨ (π₯3 31 ) β¨ (π₯1 11 π₯3 32 β¨ π₯1 12 π₯3 32 ) = 1.
Further,
π π π π π π π π π
βπ₯3 (π₯1 11 π₯3 32 β¨ π₯1 12 π₯3 33 ) β¨ (π₯3 31 ) β¨ (π₯1 11 π₯3 32 β¨ π₯1 12 π₯3 32 ) =
π π π π
= (π₯1 11 β¨ π₯1 12 ) β¨ 1 β¨ (π₯1 11 β¨ π₯1 12 ) = 1.
We will get the identity 1 = 1. This means that for the given initial condition the feature π₯1 can take
on any value from its domain.
Consider an example of solving a system of linguistic equations the right sides of which are
conjunctive normal forms. Suppose we have the following system of equations:
π π π π π
π¦ π΄ = (π₯1 11 β¨ π₯1 12 )(π₯2 21 β¨ π₯2 23 )π₯3 31 , (14)
π π π π π
{ π¦ π΅ = (π₯1 11 β¨ π₯1 13 )(π₯2 21 β¨ π₯2 22 )π₯3 32 ,
π π
π¦ πΆ = π₯1 13 π₯2 23 .
where
π¦ = {π΄, π΅, πΆ}, π₯1 β {π11 , π12 , π13 }, π₯2 β {π21 , π22 , π23 }, π₯3 β {π31 , π32 , π33 }.
It is required for the initial condition π₯3 = π31 to find values of the target feature π₯2 .
Solving the defined problem will be carried out in accordance with the suggested algorithm:
1. Check the fact that the paired conjunctions of the right sides of the equations (14) are zeros. The
π
conjunction of the right-hand sides of the first and second equations is equal to zero, since π₯3 31 β§
π
β§ π₯3 32 = 0. The conjunction of the first and third sides of the equations is zero, since
π π π
(π₯1 11 β¨ π₯1 12 )π₯1 13 = 0. Finally, the conjunction of the right sides of the second and third equations
π π π
is zero, since (π₯2 21 β¨ π₯2 22 )π₯2 23 = 0.
2. Let us represent the system (14) in the form of a single equation:
π π π π π π π π π π π π
π¦ π΄ β§ (π₯1 11 β¨ π₯1 12 )(π₯2 21 β¨ π₯2 23 )π₯3 31 β¨ π¦ π΅ β§ (π₯1 11 β¨ π₯1 13 )(π₯2 21 β¨ π₯2 22 )π₯3 32 β¨ π¦ πΆ β§ π₯1 13 π₯2 23 = 1.
3. Substitute in the resulting equation the initial value of the variable π₯3 = π31 . We get the following
result:
π π π π π π
π¦ π΄ (π₯1 11 β¨ π₯1 12 )(π₯2 21 β¨ π₯2 23 ) β¨ π¦ πΆ π₯1 13 π₯2 23 = 1.
4. Use the existence quantifiers for a sequential elimination of the variables π¦ and π₯1 from the last
equation:
π π π π π π π π π π π π
βπ¦(π¦ π΄ (π₯1 11 β¨ π₯1 12 )(π₯2 21 β¨ π₯2 23 ) β¨ π¦ πΆ π₯1 13 π₯2 23 ) = (π₯1 11 β¨ π₯1 12 )(π₯2 21 β¨ π₯2 23 ) β¨ π₯1 13 π₯2 23 = 1,
π π π π π π π π π
βπ₯1 (π₯1 11 β¨ π₯1 12 )(π₯2 21 β¨ π₯2 23 ) β¨ π₯1 13 π₯2 23 = (π₯2 21 β¨ π₯2 23 ) β¨ π₯2 23 = 1.
The application of the Boolean identity π β¨ π = π gives us the possibility to obtain the following
equation:
π π
π₯2 21 β¨ π₯2 23 = 1. (15)
From (15) π₯2 can be found directly: π₯2 β {π21 , π23 }.
6. A method for feature elimination from splitable predicates
Consider a class of problems that can be described with the help of logic equations having a more
complicated structure.
The book [3] has considered a task related to mathematical description of the Russian language
morphology, and a general approach to solving this problem has been described. This approach is
illustrated on the example of a mathematical description of noun declensions. In Ukrainian, the models
will be similar as far as mathematical formulae are concerned, although they will differ substantially
from the described ones in the sense of dependences between semantic features, and this matter should
be carefully investigated as a very prospective research field. In the English language, we do not observe
such a variety of syntactic features, but they are often replaced with special word collocations and a
variety of particles that completely change the sense of a verb. This is a great field for further research.
For an unambiguous definition of the first letter in an ending for the main forms of words for the
substantive declension (the first letter can take on one of the letters {Π°, Π΅, Ρ, ΠΈ, ΠΎ, Ρ, Ρ, Ρ, Ρ, _} a complete
and nonreducible set of features has been determined:π₯1 is the case with the values ΠΈ, Ρ, Π΄, Π², Ρ, ΠΏ
(nominative,β¦,prepositional); π₯2 is the gender with the values ΠΌ, ΠΆ, Ρ; π₯3 is the number (plural or
singular) Π΅ and ΠΌ; π₯4 is the feature of animacy with the values ΠΎ and Π½, π₯5 is the feature of stress with
the values Ρ and Π±, π₯6 is the sign with the values Ρ and Π½, π₯7 is the last letter of the basis for the word
form with the values Π±, Π², Π³, Π΄, Π΅, ΠΆ, Π·, ΠΈ, ΠΉ, ΠΊ, Π», ΠΌ, Π½, ΠΎ, ΠΏ, Ρ, Ρ, Ρ, Ρ, Ρ, Ρ
, Ρ, Ρ, Ρ, Ρ, Ρ, Ρ, Ρ; π₯8 is
the type of the basis of the word form with the values Ρ is hard, ΠΌ is soft.
Thus, the problem definition should describe links between the linguistic variables as follows:
πΏ(π₯1 , π₯2 , . . . , π₯8 , π¦1 ) = 1.
The first ending letter π¦1 and 8 semantic features π₯1 , π₯2 , . . . , π₯8 are interconnected with the help of
the finite predicate. Since this set of features is complete, the given equation defines the following
function:
π¦1π = πΉπ (π₯1 , π₯2 , . . . , π₯8 ). (16)
The predicate πΉπ is written in the form of finite algebra formulae and defines recognitions of the
variable π¦1 from the set {Π°, Π΅, Ρ, ΠΈ, ΠΎ, Ρ, Ρ, Ρ, Ρ, _}.
Let us illustrate the general task using the described example. An ending starts from the letter Ρ in
word forms with a soft basis. The form should end in Π±, Π², Π΄, ΠΆ, Π», ΠΌ, Π½, ΠΏ, Ρ, Ρ , Ρ, Ρ, having the basis
ending in Π°, Π΅, ΠΈ, ΠΉ, ΠΎ, Ρ, Ρ, Ρ, Ρ, 1) for the singular a) for the feminine nominative case, b) for the
genitive case with the masculine and neuter, and animality, 2) for the plural a) in the nominative and
accusative case with inanimation and neuter gender, b) in the dative, instrumental and prepositional
cases. These rules can be written with the help of finite algebra equations as follows:
π¦1Ρ = (π₯8ΠΌ (π₯7Π± β¨ π₯7Π² β¨ π₯7Π΄ β¨ π₯7Π· β¨ π₯7Π» β¨ π₯7ΠΌ β¨ π₯7Π½ β¨
Ρ Ρ
β¨ π₯7ΠΏ β¨ π₯7 β¨ π₯7Ρ β¨ π₯7Ρ β¨ π₯7 ) β¨ π₯7Π° β¨ π₯7Π΅ β¨ π₯7ΠΈ β¨ π₯7ΠΉ β¨ π₯7ΠΎ β¨
Ρ Ρ
β¨ π₯7 β¨ π₯7Ρ β¨ π₯7Ρ β¨ π₯7Ρ ) β§ π₯3Π΅ (π₯1ΠΈ π₯2ΠΆ β¨ π₯1 (π₯2ΠΌ β¨ π₯2Ρ ) β¨
Π΄
β¨ π₯1 π₯2 π₯4 ) β¨ π₯3 ((π₯1 β¨ π₯1 π₯4 )π₯2 β¨ π₯1 β¨ π₯1Ρ β¨ π₯1ΠΏ )).
Π² ΠΌ ΠΎ ΠΌ ΠΈ Π² Π½ Ρ
From the given example and the way the equations are built we can track the following structure of
the model: each addend of the given predicate consists of multipliers whose domains do not intersect.
At that each multiplier is represented either in the disjunctive normal form or has the same structure as
the entire predicate. Thus, it can be split into addends that, in their turn, consist of similar multipliers
with the domains that do not intersect. Mathematically, we can represent the predicate πΉπ in the
following form:
πΉπ = π΄1 β¨ π΄2 β¨. . .β¨ π΄π , (17)
where each π΄π , π = 1, π can be represented as follows:
π΄π = π΄1π (π₯1 , . . . , π₯π1 ) β§ π΄2π (π₯π1 +1 , . . . , π₯π2 ) β§. . .β§ π΄ππ (π₯ππβ1 +1 , . . . , π₯ππ ). (18)
π
The predicates π΄π , π = 1, π, π = 1, π are written in the disjunctive normal form or can be represented
with the help of formulae (17), (18). We call such predicates splitable.
In the paper [4] we consider quite a large class of predicates for which it is possible to find an
effective algorithm of eliminating variables without any increase in the size of the original formula. In
paper [4] the following properties of the existence quantifier have been considered:
1. βπ₯π₯ π = 1.
2. βπ₯(π(π₯) β¨ π(π₯)) = βπ₯π(π₯) β¨ βπ₯π(π₯).
3. βπ₯(π(π₯) β§ π(π¦) = βπ₯π(π₯) β§ π(π¦).
4. βπ¦(π(π₯) β π(π¦)) = π(π₯) β βπ¦π(π¦).
5. βπ¦(π(π₯) β π(π¦)) = π(π₯) β βπ¦π(π¦).
Suppose ππ (π₯) β§ ππ (π₯) = 0, π β π, π, π = 1,2, β¦ , π,
then
βπ¦ ((π1 (π₯) β π1 (π¦)) β§ (π2 (π₯) β π2 (π¦)) β§ β¦ β§ (ππ (π₯) β ππ (π¦))) =
= (π1 (π₯) β βπ¦π1 (π¦)) β§ (π2 (π₯) β βπ¦π2 (π¦)) β§. . .β§ (ππ (π₯) β βπ¦ππ (π¦)).
6. If the identity ππ (π₯) β‘ 0 does not hold for any π = 1,2, β¦ , π and ππ (π₯) β§ ππ (π₯) = 0 for
π β π, π, π = 1,2, . . . , π, then:
βπ₯((π1 (π₯) β π1 (π¦)) β§ (π2 (π₯) β π2 (π¦)) β§. . .β§ (ππ (π₯) β ππ (π¦))) = π1 (π¦) β¨ π2 (π¦) β¨. . .β¨ ππ (π¦).
The above properties allow formulating rules for building a class π₯π₯ of finite predicates defined on
the set of variables {π₯, π¦, . . . , π§}. The subset π₯π₯ of the set π΄ is defined as follows:
1. All recognitions π₯ π , π₯ π , . . . , π₯ π , (π, π, . . . , π are symbols from the domain for the variable π₯)
belong to π₯π₯ .
2. All predicates that do not depend on the variable π₯ belong to π₯π₯ .
3. If the predicates π1 and π2 belong to π₯π₯ , the predicate π = π1 β¨ π2 belongs to π₯π₯ .
4. If the predicate π1 belongs to π₯π₯ , and the predicate π2 does not depend on π₯, then the predicate
π = π1 β§ π2 belongs to π₯π₯ .
5. If the predicate π1 does not depend on π₯, and the predicate π2 belongs to π₯π₯ , then the predicate
π = π1 β π2 belongs to π₯π₯ .
6. Suppose the predicates π1 , π2 , . . . , ππ do not depend on π₯; ππ β§ ππ = 0 for π β π, π, π = 1,2, . . . , π,
the predicates π1 , π2 , . . . , ππ belong to π₯π₯ ; then π = (π1 β π1 ) β§ (π2 β π2 ) β§. . .β§ (ππ β ππ )
belongs to π₯π₯ .
7. If the predicates π1 , π2 , . . . , ππ depend only on π₯; ππ β§ ππ = 0 for π β π, π, π = 1,2, . . . , π; for any
π = 1,2, . . . , π the identity ππ β‘ 0 is not true; the predicates π1 , π2 , . . . , ππ do not depend on π₯; then
the predicate π = (π1 β π1 ) β§ (π2 β π2 ) β§. . .β§ (ππ β ππ ) belongs to π₯π₯ .
Eliminating features with the help of the existence quantifier gives us all ordered sets of possible
feature values for which there exists at least one allowable set of values of the other features. If we wish
to obtain sets of values of the target features that satisfy the equation irrespectively of which values we
have for the other features, the variables should be eliminated with the help of the universal quantifier.
The property of predicates to be splitable simplifies significantly the procedure of eliminating
variables. When applying this procedure, we will use the following properties of the existence
quantifier:
1. Addictiveness property:
βπ₯π (π΄1 β¨ π΄2 β¨. . .β¨ π΄π ) = βπ₯π π΄1 β¨ βπ₯π π΄2 β¨. . .β¨ βπ₯π π΄π ;
2. The application of the existence quantifier to the conjunction of predicates when only one
predicate depends on the variable to be eliminated is identical to the application of this quantifier to
the given predicate, whereas the other predicates do not change:
βπ₯π (π΄1 (π₯1 , . . . , π₯π1 ) β§. . .β§ π΄π (π₯ππβ1 +1 , . . . π₯π , . . . , π₯ππ ) β§. . .β§ π΄π (π₯ππβ1 +1 , . . . , π₯ππ )) =
= (π΄1 (π₯1 , . . . , π₯π1 ) β§. . .β§ π΄π (π₯ππβ1 +1 , . . . , π₯ππ )) β§ βπ₯π π΄π (π₯ππβ1 +1 , . . . π₯π , . . . , π₯ππ ).
Thus, owing to the described structure of splitable predicates and the above properties of the
existence quantifier, the method of eliminating variables is significantly simplified and can be split into
the following steps:
Step 1. Split the original predicate into addends.
Step 2. Split the obtained addends into multipliers.
Step 3. Select the multiplier that depends on the variable to be eliminated.
Step 4. If this multiplier is a disjunctive normal form, eliminate the variable from it. This process
can be split into the following steps:
a) find the recognition of this variable in every elementary conjunction;
b) if the elementary conjunction consists of a single recognition of the given variable, then the entire
disjunctive normal form is replaced with 1;
c) if not, replace this recognition predicate with 1, which is identical to eliminating it from the
formula.
Step 5. If the addend is represented not in the disjunctive normal form but has a complex structure,
then perform all actions starting from step 1.
Let us illustrate how the method works on the above example.
Let
Ρ Ρ
π = (π₯8ΠΌ (π₯7Π± β¨ π₯7Π² β¨ π₯7Π΄ β¨ π₯7Π· β¨ π₯7Π» β¨ π₯7ΠΌ β¨ π₯7Π½ β¨ π₯7ΠΏ β¨ π₯7 β¨ π₯7Ρ β¨ π₯7Ρ β¨ π₯7 ) β¨ π₯7Π° β¨ π₯7Π΅ β¨ π₯7ΠΈ β¨ π₯7ΠΉ β¨ π₯7ΠΎ β¨
Ρ Ρ
β¨ π₯7 β¨ π₯7Ρ β¨ π₯7Ρ β¨ π₯7Ρ ) β§ π₯3Π΅ (π₯1ΠΈ π₯2ΠΆ β¨ π₯1 (π₯2ΠΌ β¨ π₯2Ρ ) β¨ π₯1Π² π₯2ΠΌ π₯4ΠΎ ) β¨ π₯3ΠΌ ((π₯1ΠΈ β¨ π₯1Π² π₯4Π½ )π₯2Ρ β¨
β¨ π₯1Π΄ β¨ π₯1Ρ β¨ π₯1ΠΏ )).
Find βπ₯2 (π).
a) select in the predicate π all the addends (in this example there is a single addend)
Ρ Ρ
(π₯8ΠΌ (π₯7Π± β¨ π₯7Π² β¨ π₯7Π΄ β¨ π₯7Π· β¨ π₯7Π» β¨ π₯7ΠΌ β¨ π₯7Π½ β¨ π₯7ΠΏ β¨ π₯7 β¨ π₯7Ρ β¨ π₯7Ρ β¨ π₯7 ) β¨ π₯7Π° β¨ π₯7Π΅ β¨ π₯7ΠΈ β¨ π₯7ΠΉ β¨ π₯7ΠΎ β¨
Ρ Ρ
β¨ π₯7 β¨ π₯7Ρ β¨ π₯7Ρ β¨ π₯7Ρ ) β§ π₯3Π΅ (π₯1ΠΈ π₯2ΠΆ β¨ π₯1 (π₯2ΠΌ β¨ π₯2Ρ ) β¨ π₯1Π² π₯2ΠΌ π₯4ΠΎ ) β¨ π₯3ΠΌ ((π₯1ΠΈ β¨ π₯1Π² π₯4Π½ )π₯2Ρ β¨ π₯1Π΄ β¨ π₯1Ρ β¨ π₯1ΠΏ )).
b) select multipliers in the obtained addend:
Ρ Ρ
(π₯8ΠΌ (π₯7Π± β¨ π₯7Π² β¨ π₯7Π΄ β¨ π₯7Π· β¨ π₯7Π» β¨ π₯7ΠΌ β¨ π₯7Π½ β¨ π₯7ΠΏ β¨ π₯7 β¨ π₯7Ρ β¨ π₯7Ρ β¨ π₯7 ) β¨
Ρ
β¨ π₯7Π° β¨ π₯7Π΅ β¨ π₯7ΠΈ β¨ π₯7ΠΉ β¨ π₯7ΠΎ β¨ π₯7 β¨ π₯7Ρ β¨ π₯7Ρ β¨ π₯7Ρ )
Ρ
(π₯3Π΅ (π₯1ΠΈ π₯2ΠΆ β¨ π₯1 (π₯2ΠΌ β¨ π₯2Ρ ) β¨ π₯1Π² π₯2ΠΌ π₯4ΠΎ ) β¨ π₯3ΠΌ ((π₯1ΠΈ β¨ π₯1Π² π₯4Π½ )π₯2Ρ β¨ π₯1Π΄ β¨ π₯1Ρ β¨ π₯1ΠΏ )).
c) it is obvious that only the second multiplier depends on π₯2 . It has a complex structure, therefore
we should split it into addends:
Ρ
π₯3Π΅ (π₯1ΠΈ π₯2ΠΆ β¨ π₯1 (π₯2ΠΌ β¨ π₯2Ρ ) β¨ π₯1Π² π₯2ΠΌ π₯4ΠΎ )
π₯3ΠΌ ((π₯1ΠΈ β¨ π₯1Π² π₯4Π½ )π₯2Ρ β¨ π₯1Π΄ β¨ π₯1Ρ β¨ π₯1ΠΏ )
d) split every addend into multipliers:
β’ π₯3Π΅ ;
Ρ
β’ (π₯1ΠΈ π₯2ΠΆ β¨ π₯1 (π₯2ΠΌ β¨ π₯2Ρ ) β¨ π₯1Π² π₯2ΠΌ π₯4ΠΎ );
β’ π₯3ΠΌ ;
β’ ((π₯1ΠΈ β¨ π₯1Π² π₯4Π½ )π₯2Ρ β¨ π₯1Π΄ β¨ π₯1Ρ β¨ π₯1ΠΏ ).
e) the first and third multipliers do not depend on π₯2 . Split the rest into addends to get:
β’ π₯1ΠΈ π₯2ΠΆ ;
Ρ
β’ π₯1 (π₯2ΠΌ β¨ π₯2Ρ );
β’ π₯1Π² π₯2ΠΌ π₯4ΠΎ ;
β’ (π₯1ΠΈ β¨ π₯1Π² π₯4Π½ )π₯2Ρ;
β’ π₯1Π΄ β¨ π₯1Ρ β¨ π₯1ΠΏ .
f) the first and third addends are disjunctive normal forms. Apply the operation βπ₯2 (π) to get:
βπ₯2 (π₯1ΠΈ π₯2ΠΆ ) = π₯1ΠΈ ; βπ₯2 (π₯1Π² π₯2ΠΌ π₯4ΠΎ ) = π₯1Π² π₯4ΠΎ .
g) the second and fourth addends should be split into multipliers to get:
Ρ
β’ π₯1 ;
β’ (π₯2ΠΌ β¨ π₯2Ρ );
β’ (π₯1ΠΈ β¨ π₯1Π² π₯4Π½ );
β’ π₯2Ρ .
h) apply the operation βπ₯2 :
βπ₯2 (π₯2ΠΌ β¨ π₯2Ρ ) = 1; βπ₯2 (π₯2Ρ ) = 1.
i) taking into consideration the fact that all the other multipliers have not changed, we get the result
in the following form:
βπ₯2 (π) = (π₯8ΠΌ (π₯7Π± β¨ π₯7Π² β¨ π₯7Π΄ β¨ π₯7Π· β¨ π₯7Π» β¨ π₯7ΠΌ β¨ π₯7Π½ β¨ π₯7ΠΏ β¨
Ρ Ρ
β¨ π₯7 β¨ π₯7Ρ β¨ π₯7Ρ β¨ π₯7 ) β¨ π₯7Π° β¨ π₯7Π΅ β¨ π₯7ΠΈ β¨ π₯7ΠΉ β¨ π₯7ΠΎ β¨
Ρ Ρ
β¨ π₯7 β¨ π₯7Ρ β¨ π₯7Ρ β¨ π₯7Ρ ) β§ π₯3Π΅ (π₯1ΠΈ β¨ π₯1 β¨ π₯1Π² π₯4ΠΎ ) β¨
β¨ π₯3ΠΌ ((π₯1ΠΈ β¨ π₯1Π² π₯4Π½ ) β¨ π₯1Π΄ β¨ π₯1Ρ β¨ π₯1ΠΏ )).
We can see that eliminating variables from splitable predicates simplifies their structure, whereas
universal methods lead to an increase in the size of the original formula.
Thus, a generalized algorithm of finding values of target variables under predefined initial conditions
from the system (5), where predicates on the right side are splitable can be described as follows:
Step 1. Check the correctness of the model, i.e., whether conditions (6) for right sides hold.
Step 2. Transform the system to a single equation in accordance with formula (7).
Step 3. Substitute the initial values.
Step 4. Eliminate the non-salient variables with the help of the existence quantifier.
Step 5. Find the values of the target variables that satisfy the resulting equation.
In the second example it is demonstrated how it is possible to find the dependence between some
linguistic variables if a problem is described in the form of a linguistic variable system. The second
example shows how you can find the relationship between some variables if the problem is described
by a system of equations. For example, it is necessary to express this dependence between noun gender
and particular case and number values. We have considered a relation between a noun gender and
particular values for case and number. In the proposed example we have considered the following initial
values: accusative case and singular form: π₯3 = Π΅ and π₯1 = Π².
For simplification of complex deductions only two equations have been considered, although the
method itself is universal. The main advantage of this method lies in the fact that the original formulae
are simplified at every step.
Consider this example. Suppose we have the following model:
Ρ Ρ Ρ
π¦1Ρ = (π₯8Ρ (π₯7Π± β¨ π₯7Π² β¨ π₯7Π΄ β¨ π₯7Π· β¨ π₯7Π» β¨ π₯7ΠΌ β¨ π₯7Π½ β¨ π₯7ΠΏ β¨ π₯7 β¨ π₯7Ρ β¨ π₯7Ρ β¨ π₯7 ) β¨ π₯7 ) β§
Ρ
β§ (π₯3Π΅ π₯1 π₯2ΠΆ β¨ π₯3ΠΌ (π₯1ΠΈ β¨ π₯1Π² π₯4Π½ )(π₯2ΠΌ β¨ π₯2ΠΆ )) ,
Ρ Ρ
π¦1Ρ = (π₯8ΠΌ (π₯7Π± β¨ π₯7Π² β¨ π₯7Π΄ β¨ π₯7Π· β¨ π₯7Π» β¨ π₯7ΠΌ β¨ π₯7Π½ β¨ π₯7ΠΏ β¨ π₯7 β¨ π₯7Ρ β¨ π₯7Ρ β¨ π₯7 ) β¨ π₯7Π° β¨ π₯7Π΅ β¨
Ρ
{ β¨ π₯7ΠΈ β¨ π₯7ΠΉ β¨ π₯7ΠΎ β¨ π₯7 β¨ π₯7Ρ β¨ π₯7Ρ β¨ π₯7Ρ ) β§ π₯3Π΅ (π₯1Π΄ (π₯2ΠΌ β¨ π₯2Ρ ) β¨ π₯1Π² π₯2ΠΆ ).
Find values of the target variable π₯2 under the initial conditions: 1) π₯3 = Π΅; 2) π₯1 = Π².
Check the correctness of the model and transform the system of equations to a single equation:
Ρ Ρ
(π¦1Ρ ) β§ ((π₯8Ρ (π₯7Π± β¨ π₯7Π² β¨ π₯7Π΄ β¨ π₯7Π· β¨ π₯7Π» β¨ π₯7ΠΌ β¨ π₯7Π½ β¨ π₯7ΠΏ β¨ π₯7 β¨ π₯7Ρ β¨ π₯7Ρ β¨ π₯7 ) β¨ π₯7Ρ ) β§
Ρ
β§ (π₯3Π΅ π₯1 π₯2ΠΆ β¨ π₯3ΠΌ (π₯1ΠΈ β¨ π₯1Π² π₯4Π½ )(π₯2ΠΌ β¨ π₯2ΠΆ ))) β¨ (π¦1Ρ ) β§ ((π₯8ΠΌ (π₯7Π± β¨ π₯7Π² β¨ π₯7Π΄ β¨ π₯7Π· β¨ π₯7Π» β¨ π₯7ΠΌ β¨ π₯7Π½ β¨ π₯7ΠΏ β¨
Ρ Ρ Ρ
β¨ π₯7 β¨ π₯7Ρ β¨ π₯7Ρ β¨ π₯7 ) β¨ π₯7Π° β¨ π₯7Π΅ β¨ π₯7ΠΈ β¨ π₯7ΠΉ β¨ π₯7ΠΎ β¨ π₯7 β¨ π₯7Ρ β¨ π₯7Ρ β¨ π₯7Ρ ) β§
Π΅ Π΄ ΠΌ Ρ Π² ΠΆ
β§ π₯3 (π₯1 (π₯2 β¨ π₯2 ) β¨ π₯1 π₯2 )) = 1.
Substitute the initial values to get:
Ρ Ρ
(π¦1Ρ ) β§ ((π₯8ΠΌ (π₯7Π± β¨ π₯7Π² β¨ π₯7Π΄ β¨ π₯7Π· β¨ π₯7Π» β¨ π₯7ΠΌ β¨ π₯7Π½ β¨ π₯7ΠΏ β¨ π₯7 β¨ π₯7Ρ β¨ π₯7Ρ β¨ π₯7 ) β¨ π₯7Π° β¨ π₯7Π΅ β¨ π₯7ΠΈ β¨ π₯7ΠΉ β¨
Ρ
β¨ π₯7ΠΎ β¨ π₯7 β¨ π₯7Ρ β¨ π₯7Ρ β¨ π₯7Ρ ) β§ (π₯2ΠΆ )) = 1.
Eliminate all the non-salient values, i.e., π¦1 , π₯7 , π₯8 :
Ρ Ρ
βπ¦((π¦1Ρ ) β§ ((π₯8ΠΌ (π₯7Π± β¨ π₯7Π² β¨ π₯7Π΄ β¨ π₯7Π· β¨ π₯7Π» β¨ π₯7ΠΌ β¨ π₯7Π½ β¨ π₯7ΠΏ β¨ π₯7 β¨ π₯7Ρ β¨ π₯7Ρ β¨ π₯7 ) β¨ π₯7Π° β¨ π₯7Π΅ β¨
Ρ
β¨ π₯7ΠΈ β¨ π₯7ΠΉ β¨ π₯7ΠΎ β¨ π₯7 β¨ π₯7Ρ β¨ π₯7Ρ β¨ π₯7Ρ ) β§ (π₯2ΠΆ ))) = ((π₯8ΠΌ (π₯7Π± β¨ π₯7Π² β¨ π₯7Π΄ β¨ π₯7Π· β¨ π₯7Π» β¨ π₯7ΠΌ β¨ π₯7Π½ β¨ π₯7ΠΏ β¨
Ρ Ρ Ρ
β¨ π₯7 β¨ π₯7Ρ β¨ π₯7Ρ β¨ π₯7 ) β¨ π₯7Π° β¨ π₯7Π΅ β¨ π₯7ΠΈ β¨ π₯7ΠΉ β¨ π₯7ΠΎ β¨ π₯7 β¨ π₯7Ρ β¨ π₯7Ρ β¨ π₯7Ρ ) β§ (π₯2ΠΆ )) = 1.
Ρ Ρ
βπ₯7 (((π₯8ΠΌ (π₯7Π± β¨ π₯7Π² β¨ π₯7Π΄ β¨ π₯7Π· β¨ π₯7Π» β¨ π₯7ΠΌ β¨ π₯7Π½ β¨ π₯7ΠΏ β¨ π₯7 β¨ π₯7Ρ β¨ π₯7Ρ β¨ π₯7 ) β¨ π₯7Π° β¨ π₯7Π΅ β¨ π₯7ΠΈ β¨
Ρ
β¨ π₯7ΠΉ β¨ π₯7ΠΎ β¨ π₯7 β¨ π₯7Ρ β¨ π₯7Ρ β¨ π₯7Ρ ) β§ (π₯2ΠΆ ))) = ((π₯8ΠΌ ) β§ (π₯2ΠΆ )) = 1.
βπ₯8 ((π₯8ΠΌ ) β§ (π₯2ΠΆ )) = (π₯2ΠΆ ) = 1.
Thus, the result is π₯2ΠΆ = 1. Hence, for the given initial values the variable π₯2 takes on the value {ΠΆ}.
7. Conclusions
Logic inferences in a variety of knowledge bases can be done with the help of logic equations. The
main advantage of such models is the absence of a predefined input or output. The input and output
depend on the problem under consideration. Also, logic equations allow describing much more complex
data structures than relational databases or decision trees. The main problem is algorithmic difficulties
in solving such equations. Eliminating feature variables sometimes becomes quite a time-consuming
procedure. In this paper we have tried to show that there are quite large classes of equations that allow
us to eliminate variables without an increase in the size of the original formula. Real-world linguistic
problems very often can be solved using methods described in this paper.
This research demonstrates the fact that for a large class of finite predicates eliminating non-salient
feature variables with the help of quantifiers. Also, the tightness of links between discrete features has
been investigated. It should be noted that the results obtained can be used not only for linguistic
problems but also for any knowledge bases with a complex structure.
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