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. Knowledge base, linguistic equations, finite predicates, logic equations, variable elimination 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].
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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.
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.
1. How do concrete values of a given feature substituted in the logic equation affect links between other 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.
4. Eliminating variables with the help of logic quantification operations and simplifying finite predicate formulae.
Let predicate depend on variables , , . . . , . Let us define the substitution operator ( ) ( is an element from the domain for the variable ) that is applied to the predicate in the following way: ( ( , , . . . , ) = ( , , . . . , ).
( , , . . . , ) → ( , , . . . , ) ( , , . . . , ) ← ( , , . . . , ) (1) (2)
Then 1( ) = 1( , … , ) ( , , … , ) = 1 1( , … , ) ∨ 2 2( , … , ) ∨ … ∨ ( , … , ).
Obviously, the operator 1( ) will be a limiting one, if 1 → ∀ = 1,2, . . . , .
The operator 1( ) will be a spreading one, if 1 ← ∀ = 1,2, . . . , . The operator 1( ) will be a shifting one, if both conditions do not hold. , and have the domains { 1, 2}, { 1, 2} and { 1, 2} correspondingly.
Consider the application of the substitution operator 1 to a predicate ( , ), where the variables = 1 1 1 ∨ 2 1 2 ∨ 2 1 1. for any , , . . . , . for any , , . . . , . way.
1( ) = 1 1 = ( 1 ∨ 2)∧ 1 1 = 1 1 1 ∨ 2 1 1. for the given example 1 = 1 1, 2 = 1 2 ∨ 1 1 and, obviously, 1 → 2.
The predicate , except for those disjuncts contained in 1( ), one more disjunct 2 1 1. It means that the operator 1 is a limiting one for the predicate . In terms of the introduced definitions,
= 1 1 1 ∨ 1 1 2 ∨ 2 1 1. 1( ) = 1 1 ∨ 1 2 = ( 1 ∨ 2) ∧ ( 1 1 ∨ 1 2) = = 1 1 1 ∨ 2 1 2 ∨ 2 1 1 ∨ 2 1 2. а 2 = 1 1. It means 1 ← 2.
The operator 1 for this predicate is a spreading one. For the given example 1 = 1 1 ∨ 1 2,
= 1 1 1 ∨ 1 2 2 ∨ 2 1 2. 1( ) = 1 1 ∨ 2 2 = ( 1 ∨ 2) ∧ ( 1 1 ∨ 2 2) =
= 1 1 1 ∨ 1 2 2 ∨ 2 1 1 ∨ 2 2 2 i.e., the operator 1is a shifting one.
Also the laws of falseness should hold: 1 ∨ 2 ∨ … ∨
= 1, = ̅1̅̅,̅̅. ∧ = 0, ≠ , , = ̅1̅̅,̅̅̅.
Suppose also that for the given values { 1 1, 2 2, . . . , } of the variables 1, 2,. . . , it is necessary to compute the values of variables +1,. . . , such that the equation (3) will be true for some values of the variables +1,. . . , .
Let us define this problem mathematically: ∃ +1.. . ∃ ( 1 1, . . .
, +1,. . . , , +1,. . . , ) = 1, which can be represented in the finite predicate language as follows: ∨ +1 +1, . . . ∨ =1 ( 1 1, . . . =1 , +1, ,. . . , , , +1,. . . , ) = 1, where only for the variables +1,. . . , possible sets of values should be found.
( 1, 2,. . . , +1,. . . , , +1,. . . , ) = 1, system of features the laws of truthfulness should hold: where each variable 1, 2,. . . , has the domain = { 1, 2, . . . , }, = 1, , at that, for the (3) (4) (5) (6) (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 ( 1, 2,. . . , ) has the following form: where = 1( 11 , 12 , . . . , 1 ) 2( 21 , 22 , . . . , 2 )∧. . .∧ ( 1 , 2, . . . , ), that satisfies the condition can be transformed to the following form:
= ( 1, 2, . . . , ), = 1, ( 1, 2,. . . , )∧ ( 1, 2, . . . , ) = 0, ≠ , , , = 1,
∨ =1 ∧ ( 1, 2,. . . , ) = 1. =∨
1
∃ 2( )(∃ 1( )) =∨2∨1
Consider the application of the existence quantifier to the variable 1 from the set { 11 ∩ { , 12 , . . . , 1 } ∩ { 21 , 22
,. . . , 2 } ∩.. . 1 , 2, . . . , } = { 1
, 2, . . . , } = . ∃ 1( ) =∨1 1 2 1. . . 1, 1 where
1 = 1 1 For example,
( ), = 1, . consist of 3 values, and the domain for 3 from 2 values. Then = 1( 1, 2, 3, 4)∧ 2( 2, 3, 5)∧ 3( 2, 3, 6, 7). Let the domain for 2
1 1( 1, 3, 4) 2 1( 3, 5) 3 1( 3, 6, 7). ∃ 2( ) = 1( 1, 21, 3, 4) 2( 21, 3, 5) 3( 21, 3, 6, 7)∨ ∨ 1( 1, 22, 3, 4) 2( 22, 3, 5) 3( 22, 3, 6, 7)∨ ∨ 1( 1, 23, 3, 4) 2( 23, 3, 5) 3( 23, 3, 6, 7) = 1 2 1 2 1 2 … 1 2, где 1 2 = 2 2 ( 1 1 ( )), = 1, . ∃ 3( )∃ 2( ) = 1( 1, 21, 31, 4) 2( 21, 31, 5) 3( 21, 31, 6, 7)∨ 1( 1, 21, 31, 4) 2( 21, 31, 5) 3( 21, 31, 6, 7)∨ ∨ 1( 1, 22, 31, 4) 2( 22, 31, 5) 3( 22, 31, 6, 7)∨ ∨ 1( 1, 23, 31, 4) 2( 23, 31, 5) 3( 23, 31, 6, 7)∨ ∨ 1( 1, 21, 32, 4) 2( 21, 32, 5) 3( 21, 32, 6, 7)∨ ∨ 1( 1, 22, 32, 4) 2( 22, 32, 5) 3( 22, 32, 6, 7)∨ for the original predicate 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.
= 1( 1, . . . , 1 , ) 2( 1+1, . . . , 2
, ) ∧. . .∧ ( −1+1, . . . , , ), disjunctions of predicate conjunctions the variables of which do not intersect:
{ 1, . . . , 1} ∩ { 1+1, . . . , 2} ∩. . .∩ { −1+1, . . . , } = ∅.
Then, as a result of the application of the existence quantifier to the variable , we obtain the ← → (8) (9) (10) (11)
∃ ( ) =∨ 1( 1, . . . , 1
, ) ∧. . .∧ ( −1+1, . . . , , ) = =∨ 1( 1, . . . , 1). . . ( −1+1, . . . , ).
( −1+1, . . . , ). The implication holds if and only if ∀ = 1, ← .
Sufficiency. Let the implication (10) is true. Suppose, ∃ ∈ {1. . . } such that → . Then, if the predicate contains such an elementary conjunction that it is not present in the predicate , since the domains for the predicates do not intersect, the predicate will not contain as well, whereas it is present in the predicate . In this case the implication (10) is false, which contradicts the premise.
Consequence. Let the predicate satisfies the condition (8), i.e., the application of the operator 1 → 2 , the addend 2 is simplified and ∃ ( ) is defined by the formula (9). Then, if ∀ = 1, formula (9) is minimized.
Statement 2. Let = ( ), = ( ). The implication is true for the predicate one of the following conditions holds: is a shifting operator, and is a spreading operator; is a limiting operator, and is a spreading operator; is a limiting operator, and is a shifting operator.
variable . Hence, → , which contradicts the premise.
Proof. Necessity. Suppose the implication (11) is true. Suppose is a spreading operator, and is a limiting operator. Then ( ) → ,
→ (P), which contradicts condition (11). Suppose that is a spreading operator, and is a shifting one. Since the substitution operators and are applied to the same variable , the predicate contains addends of the form and , where ∨ →∨ . Then ( ) =∨ ∨ ; ( ) =∨ ∨ , where does not depend on the Suppose and
are shifting operators. Then the analogous considerations demonstrate the fact that and are different and the condition (11) does not hold.
is a shifting operator, and containing the predicate predicate can be presented as follows:
is a limiting operator. Then in elementary conjunctions , are not present in the disjunctive normal form for the predicate P, and the =∨ ∨ ∨ , where contains the "recognition" of the variable , besides "recognitions" of this variable. Then ( ) =∨ ∨ , ( ) = and ( ) → ( ), which and , and does not contain contradicts the premise. operator and
If both operators are limiting or spreading ones, their application to the predicate is the same.
= 1 12 3 33 is a spreading substitution operator, and 22( ) = 1 12 3 33 is a shifting 21( ) → 22( ).
Let = 1
12 2 22 3 33 ∨ 3 31. Then 11( ) = 3 31 is a limiting operator, and 12( ) = 1 12 2 22 is a spreading operator and 32( ) → 31( ).
11 is a limiting operator, and 5. A method of feature elimination from predicates represented in the disjunctive and conjunctive normal forms.
= ∨ . zero. Then, by transforming the system to a single equation ((5)-(7)), we obtain an equation in the disjunctive normal form. By denoting th disjunction in the equation corresponding to the description of th object in the subject field, the following equation is obtained:
( ). The application of the quantifier ∃ ( ) does not change , if des not contain the variable . 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 ( = 1, ). 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 i , for which the following formula is true:
contains the disjunction of a certain number of recognition predicates for . Taking into consideration the above notation, it is possible to write down the following equation: ∃ = 1 2. . . ( −1)(∃ ) ( +1). . . ≡ 1 2. . . ( −1) ( +1). . . , contain some variable , the application of the existence quantifier does not change . where = 1, , is the cardinality of the subject field. If the conjunctive normal form does not
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.
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: { = 1 11 2 22 ∨ 2
21 3 31, = 1 11 221 3 32 ∨ 1
12 2 21 3 32. = 1 11 2 21 3 32 ∨ 1 12 2 21 3 33 ∨ 1 13 2 23, (13)
where
= { , , }, 1 ∈ { 11, 12, 13}, 2 ∈ { 21, 22, 23}, 3 ∈ { 31, 32, 33}.
It is required for the initial condition 2 = 21 to find values of the target feature 1.
1
11 2 21 3 32 ∧ 1 11 2 22 = 0, 1 12 2 21 3 33 ∧ 1 11 2 22 = 0, 1 13 2 23 ∧ 1 11 2 22 = 0, 1 11 2 21 3 32 ∧ 2 21 3 31 = 0, 1 12 2 21 3 33 ∧ 2 21 3 31 = 0, 1 13 2 23 ∧ 2 21 3 31 = 0.
1 11 2 22 ∧ 1
11 221 3 32 = 0, 2 21 3 31 ∧ 1 11 221 3 32 = 0, 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: ∧ ( 1 11 2 21 3 32 ∨ 1 3. Substitute the initial value of the variable 2 = 21 in the obtained equation to get the following: ∧ ( 1 11 3 32 ∨ 1 12 3 33) ∨ ∧ ( 3 31) ∨ ∧ ( 1 11 3 32 ∨ 1 12 3 32) = 1 4. Use the quantifiers for a sequential elimination of the variables and 3 from the last equation: ∃ ( ∧ ( 1 11 3 32 ∨ 1 12 3 33) ∨ ∧ ( 3 31) ∨ ∧ ( 1 11 3 32 ∨ 1 12 3 32)) = = ( 1 11 3 32 ∨ 1 12 3 33) ∨ ( 3 31) ∨ ( 1
11 3 32 ∨ 1 12 3 32) = 1. ∃ 3( 1 11 3 32 ∨ 1 12 3 33) ∨ ( 3 31) ∨ ( 1
11 3 32 ∨ 1 12 3 32) = = ( 1 11 ∨ 1 12) ∨ 1 ∨ ( 1 11 ∨ 1 12) = 1. on any value from its domain.
We will get the identity 1 = 1. This means that for the given initial condition the feature 1 can take 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: = ( 1 11 ∨ 1 12)( 2 21 ∨ 2 23) 3 31, (14) { = ( 1 11 ∨ 1 13)( 2 21 ∨ 2 22) 3 32 ,
= { , , }, 1 ∈ { 11, 12, 13}, 2 ∈ { 21, 22, 23}, 3 ∈ { 31, 32, 33}.
It is required for the initial condition 3 = 31 to find values of the target feature 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 conjunction of the right-hand sides of the first and second equations is equal to zero, since 3 31 ∧ ∧ 3 32 = 0. The conjunction of the first and third sides of the equations is zero, since ( 1 11 ∨ 1 12) 1 13 = 0. Finally, the conjunction of the right sides of the second and third equations is zero, since ( 2 21 ∨ 2 22) 2 23 = 0. 2. Let us represent the system (14) in the form of a single equation:
∧ ( 1 11 ∨ 1 12)( 2 21 ∨ 2 23) 3 31 ∨ ∧ ( 1 11 ∨ 1 13)( 2 21 ∨ 2 22) 3 32 ∨ ∧ 1 13 2 23 = 1. 3. Substitute in the resulting equation the initial value of the variable 3 = 31. We get the following result:
( 1 11 ∨ 1 12)( 2 21 ∨ 2 23) ∨ 1 13 2 23 = 1. 4. Use the existence quantifiers for a sequential elimination of the variables and 1from the last equation: ∃ ( ( 1 11 ∨ 1 12)( 2 21 ∨ 2 23) ∨ 1 13 2 23) = ( 1 11 ∨ 1 12)( 2 21 ∨ 2 23 ) ∨ 1 13 2 23 = 1, ∃ 1( 1 11 ∨ 1 12)( 2 21 ∨ 2 23) ∨ 1 13 2 23 = ( 2 21 ∨ 2 23) ∨ 2 23 = 1.
The application of the Boolean identity ∨ = gives us the possibility to obtain the following equation: 2 21 ∨ 2 23 = 1. (15)
From (15) 2 can be found directly: 2 ∈ { 21, 23}.
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 {а, е, ё, и, о, у, ы, ю, я, _} a complete and nonreducible set of features has been determined: 1 is the case with the values и, р, д, в, т, п (nominative,…,prepositional); 2 is the gender with the values м, ж, с; 3 is the number (plural or singular) е and м; 4 is the feature of animacy with the values о and н, 5 is the feature of stress with the values у and б, 6 is the sign with the values с and н, 7 is the last letter of the basis for the word form with the values б, в, г, д, е, ж, з, и, й, к, л, м, н, о, п, р, с, т, у, ф, х, ц, ч, ш, щ, ы, ю, я; 8 is the type of the basis of the word form with the values т is hard, м is soft.
Thus, the problem definition should describe links between the linguistic variables as follows: ( 1, 2, . . . , 8, 1) = 1.
1 = ( 1, 2, . . . , 8).
The first ending letter 1 and 8 semantic features 1, 2, . . . , 8 are interconnected with the help of the finite predicate. Since this set of features is complete, the given equation defines the following function:
The predicate is written in the form of finite algebra formulae and defines recognitions of the variable 1 from the set {а, е, ё, и, о, у, ы, ю, я, _}.
Let us illustrate the general task using the described example. An ending starts from the letter я in word forms with a soft basis. The form should end in б, в, д, ж, л, м, н, п, р, с , т, ф, having the basis ending in а, е, и, й, о, у, ы, ю, я, 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: (16) (17) (18) = 1 ∨ 2 ∨. . .∨ , following form:
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 in the where each , = 1, can be represented as follows: = 1( 1, . . . , 1) ∧ 2( 1+1, . . . , 2) ∧. . .∧ ( −1+1, . . . , ).
The predicates , = 1, , = 1, 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: 1я = ( 8м( 7б ∨ 7в ∨ 7д ∨ 7з ∨ 7л ∨ 7м ∨ 7н ∨ ∨ 7п ∨ 7р ∨ 7с ∨ 7т ∨ 7ф
) ∨ 7а ∨ 7е ∨ 7и ∨ 7й ∨ 7о ∨ ∨ 1в 2м
у ∨ 7 ∨ 7ы ∨ 7ю ∨ 7я) ∧ 3е( 1и
2ж ∨ 1р( 2м ∨ 2с ) ∨ 4о) ∨ 3м(( 1и ∨ 1в 4н) 2с ∨ 1д ∨ 1т ∨ 1п)). 1. ∃ = 1. 2. ∃ ( ( ) ∨ ( )) = ∃ 3. ∃ ( ( ) ∧ ( ) = ∃ ( ) ∨ ∃
( ).
( ) ∧ ( ). 4. ∃ ( ( ) → ( )) = ( ) → ∃ 5. ∃ ( ( ) → ( )) = ( ) → ∃ ( ).
( ).
Suppose ( ) ∧ ( ) = 0, ≠ , , = 1,2, … , , then
∃ (( 1( ) → 1( )) ∧ ( 2( ) → 2( )) ∧ … ∧ ( ( ) → ( ))) = = ( 1( ) → ∃ 1( )) ∧ ( 2( ) → ∃ 2( )) ∧. . .∧ ( ( ) → ∃ ( )). 6. If the identity ( ) ≡ 0 does not hold for any = 1,2, … , and ( ) ∧ ( ) = 0 for ≠ , , = 1,2, . . . , , then:
The above properties allow formulating rules for building a class of finite predicates defined on the set of variables { , , . . . , }. The subset of the set is defined as follows: 1. All recognitions , , . . . , , ( , , . . . , are symbols from the domain for the variable ) belong to . = 1 ∧ 2 belongs to . 2. All predicates that do not depend on the variable belong to . 3. If the predicates 1 and 2 belong to , the predicate = 1 ∨ 2 belongs to . 4. If the predicate 1 belongs to , and the predicate 2 does not depend on , then the predicate 5. If the predicate 1 does not depend on , and the predicate 2 belongs to , then the predicate = 1 → 2 belongs to . 6. Suppose the predicates 1, 2,..., do not depend on ; ∧ = 0 for ≠ , , = 1,2,..., , the predicates 1, 2, ..., belong to ; then = ( 1 → 1)∧ ( 2 → 2)∧...∧ ( → ) belongs to . 7. If the predicates 1, 2,.. ., depend only on ; ∧ = 0 for ≠ , , = 1,2,.. ., ; for any = 1,2,..., the identity ≡ 0 is not true; the predicates 1, 2,... , do not depend on ; then the predicate = ( 1 → 1)∧ ( 2 → 2)∧...∧ ( → )belongs to .
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: ∃ ( 1 ∨ 2 ∨...∨ ) = ∃ 1 ∨ ∃ 2 ∨...∨ ∃ ; 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: ∃ ( 1( 1,.. ., 1)∧...∧ ( −1+1,. .. ,..., )∧...∧ ( −1+1,... , )) = = ( 1( 1,..., 1)∧.. .∧ ( −1+1,..., ))∧ ∃ ( −1+1,... ,. .., ).
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 4. If this multiplier is a disjunctive normal form, eliminate the variable from it. This process can be split into the following steps:
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.
= ( 8м( 7б ∨ 7в ∨ 7д ∨ 7з ∨ 7л ∨ 7м ∨ 7н ∨ 7п ∨ 7р ∨ 7с ∨ 7т ∨ 7ф)∨ 7а ∨ 7е ∨ 7и ∨ 7й ∨ 7о ∨ ∨ 7у ∨ 7ы ∨ 7ю ∨ 7я)∧ 3е( 1и 2ж ∨ 1р( 2м ∨ 2с)∨ 1в 2м 4о)∨ 3м(( 1и ∨ 1в 4н) 2с ∨ ∨ 1д ∨ 1т ∨ 1п)).
Find ∃ 2( ). a) select in the predicate all the addends (in this example there is a single addend) ( 8м( 7б ∨ 7в ∨ 7д ∨ 7з ∨ 7л ∨ 7м ∨ 7н ∨ 7п ∨ 7р ∨ 7с ∨ 7т ∨ 7ф)∨ 7а ∨ 7е ∨ 7и ∨ 7й ∨ 7о ∨ ∨ 7у ∨ 7ы ∨ 7ю ∨ 7я)∧ 3е( 1и 2ж ∨ 1р( 2м ∨ 2с)∨ 1в 2м 4о)∨ 3м(( 1и ∨ 1в 4н) 2с ∨ 1д ∨ 1т ∨ 1п)). b) select multipliers in the obtained addend: ( 8м( 7б ∨ 7в ∨ 7д ∨ 7з ∨ 7л ∨ 7м ∨ 7н ∨ 7п ∨ 7р ∨ 7с ∨ 7т ∨ 7ф)∨
у ∨ 7а ∨ 7е ∨ 7и ∨ 7й ∨ 7о ∨ 7 ∨ 7ы ∨ 7ю ∨ 7я) ( 3е( 1и 2ж ∨ 1р( 2м ∨ 2с)∨ 1в 2м 4о)∨ 3м(( 1и ∨ 1в 4н) 2с ∨ 1д ∨ 1т ∨ 1п)). c) it is obvious that only the second multiplier depends on 2. It has a complex structure, therefore we should split it into addends: • ( 1и 2ж ∨ 1р( 2м ∨ 2с)∨ 1в 2м 4о); 3е( 1и 2ж ∨ 1р( 2м ∨ 2с)∨ 1в 2м 4о) 3м(( 1и ∨ 1в 4н) 2с ∨ 1д ∨ 1т ∨ 1п) d) split every addend into multipliers: • 3е; • 3м; • (( 1и ∨ 1в 4н) 2с ∨ 1д ∨ 1т ∨ 1п). e) the first and third multipliers do not depend on 2. Split the rest into addends to get: • 1и 2ж; • 1р( 2м ∨ 2с); • 1в 2м 4о; • ( 1и ∨ 1в 4н) 2с; • 1д ∨ 1т ∨ 1п. f) the first and third addends are disjunctive normal forms. Apply the operation ∃ 2( ) to get: ∃ 2( 1и 2ж) = 1и; ∃ 2( 1в 2м 4о) = 1в 4о. g) the second and fourth addends should be split into multipliers to get: • 1р; • ( 2м ∨ 2с); • ( 1и ∨ 1в 4н); • 2с. h) apply the operation ∃ 2:
∃ 2( 2м ∨ 2с) = 1; ∃ 2( 2с) = 1. i) taking into consideration the fact that all the other multipliers have not changed, we get the result in the following form: ∃ 2( ) = ( 8м( 7б ∨ 7в ∨ 7д ∨ 7з ∨ 7л ∨ 7м ∨ 7н ∨ 7п ∨ ∨ 7р ∨ 7с ∨ 7т ∨ 7ф)∨ 7а ∨ 7е ∨ 7и ∨ 7й ∨ 7о ∨ ∨ 7 ∨ 7ы ∨ 7ю ∨ 7я)∧ 3е( 1и ∨ 1р ∨ 1в 4о)∨ у
∨ 3м(( 1и ∨ 1в 4н)∨ 1д ∨ 1т ∨ 1п)).
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:
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: 3 = е and 1 = в.
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.
1ы = ( 8т( 7б ∨ 7в ∨ 7д ∨ 7з ∨ 7л ∨ 7м ∨ 7н ∨ 7п ∨ 7р ∨ 7с ∨ 7т ∨ 7ф)∨ 7ц)∧
∧ ( 3е 1р 2ж ∨ 3м( 1и ∨ 1в 4н)( 2м ∨ 2ж)) , ∨ 7р ∨ 7с ∨ 7т ∨ 7ф)∨ 7а ∨ 7е ∨ 7и ∨ 7й ∨ 7о ∨ 7у ∨ 7ы ∨ 7ю ∨ 7я)∧ ( 2ж)) = 1. ∃ 7((( 8м( 7б ∨ 7в ∨ 7д ∨ 7з ∨ 7л ∨ 7м ∨ 7н ∨ 7п ∨ 7р ∨ 7с ∨ 7т ∨ 7ф)∨ 7а ∨ 7е ∨ 7и ∨ у ∨ 7й ∨ 7о ∨ 7 ∨ 7ы ∨ 7ю ∨ 7я)∧ ( 2ж))) = (( 8м)∧ ( 2ж)) = 1.
∃ 8(( 8м)∧ ( 2ж)) = ( 2ж) = 1.
Thus, the result is 2ж = 1. Hence, for the given initial values the variable 2 takes on the value {ж}.
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.
[1] Y. P. Shabanov-Kushnarenko, Teoria intellekta. Matematicheskie sredstva [The theory of intelligence. Mathematical tools]. Visha Shkola Publs., Kharkiv,1984. [2] Y. P. Shabanov-Kushnarenko Teoria intellekta. Tehnicheskie sredstva [The theory of intelligence.
[3] Y. P. Shabanov-Kushnarenko, Teoria intellekta. Problemy i perspectivy, [The theory of intelligence. Problems and prospects]. Visha Shkola Publs., Kharkiv,1987. [4] D.E. Sitnikov, B. D’cruz, P.E. Sitnikova, Extracting Salient Discrete Object Features Based On Composing And Manipulating Logical Equations. Vestnik kharkivskogo gоsydarstvennogo politechnicheckogo yniversiteta [Bulletin of Kharkiv State Polytechnic University] 51 (1999) 186– 192. [5] M. Kracht, G. Pullum, The Mathematics of Language. The Mathematical Intelligencer 28 (2006) 74–78. [6] S. Gebus, E. Juuso, K. Leiviskä, Knowledge-based linguistic equations for defect detection through functional testing of printed circuit boards. Expert Systems with Applications 36.1 (2009) 292–302. [7] A. Isokangas, E. Juuso, Fuzzy Modelling with Linguistic Equations. University of Oulu: Control
[8] K. J. Chen, P. H. Dedhia, J. R. Imbus, F. David, Schneider Thyroid Ultrasound Reports: Will the Thyroid Imaging, Reporting, and Data System Improve Natural Language Processing Capture of
[9] A. Okhotin, O. Yakimova, Language equations with complementation: Expressive power.
Theoretical Computer Science 416 (2012) 71–86. [10] A. Okhotin, Decision problems for language equations. Journal of Computer and System Sciences 76, Issues 3–4 (2010) 251–266. [11] S. Ganji, K. Dhawan, R. Sinha, Novel textual features for language modelling of intra-sentential code-switching data. Computer Speech & Language 64 (2020) 101099. [12] C.A. MacArthur, A. Jennings, Z.A. Philippakos, Which linguistic features predict quality of argumentative writing for college basic writers, and how do those features change with instruction? Reading and Writing 32 (2019) 1553–1574. [13] A. Ionescu, E. L. Leiss, On the role of complementation in implicit language equations and relations. Journal of Computer and System Sciences 80.2 (2014) 457–467. [14] G. Weikum et al, Big Data Methods for Computational Linguistics. IEEE Data Eng. Bull. 35 (2012) 46–64.