INTERNATIONAL JOURNAL OF ARTIFICIAL INTELLIGENCE
ISSN: 2692-5206, Impact Factor: 12,23
American Academic publishers, volume 05, issue 03,2025
Journal:
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page 1760
USING PREDICATES TO CONSOLIDATE THEORETICAL KNOWLEDGE OF
INEQUALITIES FOR FUTURE PRIMARY SCHOOL TEACHER
Mamadaliyev Kamildjan Bazarbayevich,
Mamadaliyev Baxtiyor Kamildjanovich
Andijan State Pedagogical Institute
Annotation:
This article explores the importance of using predicate algebra formulas in teaching
students how to prove theorems and solve inequalities. The examples and theorems presented in
the article can be used not only for teaching students the applications of predicate algebra, but
also for working with students with creative abilities and organizing group lessons in
mathematics.
Key words:
Inequalities, predicate, range of truth of a predicate, equivalent formulas, theorem,
proof of a theorem, methods of proof.
The science of mathematics is studied and developed based on the laws of mathematical
logic. However, mathematical logic is not taught as a separate subject in secondary schools.
Although the elements of mathematical logic are partially included in mathematics textbooks,
their applications are not sufficiently covered. As a result, students face many difficulties in
studying the theoretical foundations of mathematics in depth, solving equations and inequalities,
and especially in proving theorems. Taking this into account, in this article we will consider the
applications of predicate algebra to solving inequalities and systems of inequalities and proving
theorems.
When studying the applications of predicate algebra, it is important to know its
equivalence formulas. Let us recall the main equivalence formulas of predicate algebra:
P(x)⋀(S x ⋁Q x ) ≡ P(x)⋀S(x)⋁P(x)⋀Q(x)
(1)
P(x)⋁S x ⋀Q x ≡ (P x ⋁S x )⋀(P x ⋁Q x )
(2)
P(x)⋀S(x) ≡ P(x)⋁S(x)
(3)
P(x)⋁S(x) ≡ P(x)⋀S(x)
(4)
P(x) ⟹ S(x) ≡ P(x)⋁S(x)
(5)
P(x) ⟹ S(x) ≡ S(x) ⟹ P(x)
(6)
P(x) ⟺ S(x) ≡ P(x)⋀S(x)⋁P(x)⋀S(x)
(7)
P(x) ⟺ S(x) ≡ (P x ⋁S x )⋀(S x ⋁P x )
(8)
INTERNATIONAL JOURNAL OF ARTIFICIAL INTELLIGENCE
ISSN: 2692-5206, Impact Factor: 12,23
American Academic publishers, volume 05, issue 03,2025
Journal:
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page 1761
Since inequalities consist of predicates, the problem of solving an inequality comes down
to finding the truth domain of the predicate. Let P(x) and S(x) be predicates defined on some set
ℳ
. We denote the truth domains of these predicates by E
p
and E
s
, respectively, the negation of
the predicate P(x) by
P(x)
, and the set
ℳ \ E
p
by
E
p
.
P(x)
,
P(x)⋁S(x)
,
P(x)⋀S(x)
,
P(x) ⟹ S(x)
and
P(x) ⟺ S(x)
We use the following
theorems to find the truth domains of the predicates.
1- theorem.
(∀x ∈ R)(x
2
≤ x ⟹ x ≤ x)
.
Proof. We use the converse method: that is, instead of the given theorem, we prove the
following theorem
(∀x ∈ R)(x ≤ x ⇒ x
2
≤ x)
, which is equally strong. Let
(x ≥ x ⟹
x
2
≥ x) x ≥ x
. Then x≥1. We multiply both sides of this inequality by x (Since the value of x
is positive, the inequality sign does not change).
x ∙ x ≥ x ∙ 1
means that the inequality
x ≥ x
implies the inequality
x
2
≥ x
. Thus, the formula
(∀x ∈ R)(x ≤ x ⇒ x
2
≤ x)
is a theorem.
Therefore, the formula
(∀x ∈ R)(x
2
≤ x ⟹ x ≤ x)
, which is equally strong as this formula, is
also a theorem.
2- theorem.
(E
p
= E
s
) ⟹ (∀x ∈ ℳ)(P x ⟺ S x )
Proof. Let
E
p
= E
s
. Then the element
∀
x belonging to
E
p
also belongs to
E
s
.
At this value of x, P(x) and S(x) are true statements. Therefore, based on the definition of
the equivalence operation, the formula
P x ⟺ S x
is also a true statement. Therefore, from the
equality
E
p
= E
s
it follows that the formula
∀x ∈ ℳ)((P x ⟺ S x )
is a true statement.
The theorem is proved.
Let R be the set of real numbers.
Example 1. Given a predicate x
2
-7x+12<0 in the set R. Find its truth domain.
Solving. We denote the given predicate by P(x) and its truth domain by E
p
. Then,
P(x) ≡
(x
2
− 7x + 12 < 0) ≡ ( x − 3 x − 4 < 0) ≡
≡ (x − 3 < 0) ∧ (x − 4 > 0) ∨ (x − 3 > 0) ∧ (x − 4 < 0) ≡
≡ (x < 3) ∧ (x > 4) ∨ (x > 3) ∧ (x < 4)
.
E
p
= −∞; 3 ∩ 4; ∞ ∪ 3; ∞ ∩ −∞; 4 = ∅ ∪ 3; 4 =
= (3; 4)
. Answer:
E
p
= (3; 4)
.
Example 2. Given a predicate P(x)=(x
2
-x-20>0) defined on a set R. Find its truth domain
E
p
.
Solving.
P x ≡ x
2
− x − 20 > 0 ≡ ( x + 4 ⋅ x − 5 > 0) ≡
≡ (x + 4 < 0) ∧ (x − 5 < 0) ∨ (x + 4 > 0) ∧ (x − 5 > 0) ≡
INTERNATIONAL JOURNAL OF ARTIFICIAL INTELLIGENCE
ISSN: 2692-5206, Impact Factor: 12,23
American Academic publishers, volume 05, issue 03,2025
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page 1762
≡ (x <− 4) ∧ (x < 5) ∨ (x >− 4) ∧ (x > 5)
,
E
p
= x ∈ R x <− 4 ∩ {x ∈ R|x < 5} ∪
∪ x ∈ R x >− 4 ∩ x ∈ R x > 5 = ( − ∞; − 4) ∩ ( − ∞; 5) ∪
∪ −4; ∞ ∩ 5; ∞ = ( − ∞; − 4) ∪ (5; ∞)
.
Answer:
E
p
= ( − ∞; − 4) ∪ (5; ∞)
.
Example 3. Given a predicate
P x =
2x+6
5x−10
≤ 0
in a set R, find its truth domain E
p
.
Solving.
P x =
2x+6
5x−10
≤ 0 ≡ (2x + 6 ≤ 0) ∧ (5x − 10 > 0) ∨
∨ (2x + 6 ≥ 0) ∧ (5x − 10 < 0) ≡ (x ≤− 3) ∧ (x > 2) ∨ (x ≥− 3) ∧ (x < 2)
.
E
p
= x ∈ R x ≤− 3 ∩ x ∈ R x > 2 ∪
∪ x ∈ R x ≥− 3} ∩ {x ∈ R x < 2 = −∞; − 3 ∩ 2; ∞ ∪ −3; ∞ ∩
∩ −∞; 2 = ∅ ∪ −3; 2 = [ − 3; 2)
. Answer:
E
p
= [ − 3; 2)
.
Example 4. Given a predicate P(x)=(|x-2|<3) in the set R. Find its truth domain E
p
.
Solving.
P x = ( x − 2 < 3) ≡ (x − 2 < 3) ∧ (x − 2 >− 3) ≡
≡ (x < 5) ∧ (x >− 1)
.
E
p
= x ∈ R x < 5 ∧ x >− 1 = x ∈ R x < 5} ∩ x ∈ R x >− 1 =
= −∞; 5 ∩ −1; ∞ = ( − 1; 5)
. Answer:
E
p
= (1; 5)
.
Example 5. Given a predicate
P x = (|2x + 6| ≥ 4)
in the set R. Find its truth domain
E
p
.
Solving.
P x = (|2x + 6| ≥ 4) ≡ (2x + 6 ≥ 4) ∨ (2x + 6 ≤− 4) ≡
≡ (2x ≥− 2) ∨ (2x ≤− 10) ≡ (x ≥− 1) ∨ (x ≤− 5)
.
E
p
= x ∈ R x ≥− 1 ∨ x ≤− 5 = x ∈ R x ≥− 1 ∪ x ∈ R x ≤− 5 =
= −1; ∞ ∪ −∞; − 5 = −∞; − 5 ∪ [ − 1; ∞)
.
Answer:
E
p
= −∞; − 5 ∪ [ − 1; ∞)
.
Example 6. Given the predicates
P x = (x
2
− x ≤ 0)
and
S x = (x ≤ x)
defined on a
set R, find
E
p
= ?
,
E
s
= ?
,
E
p∧s
= ?
,
E
p∨s
= ?
,
E
p⟹s
= ?
,
E
s⟹p
= ?
,
E
p⟺s
= ?
INTERNATIONAL JOURNAL OF ARTIFICIAL INTELLIGENCE
ISSN: 2692-5206, Impact Factor: 12,23
American Academic publishers, volume 05, issue 03,2025
Journal:
https://www.academicpublishers.org/journals/index.php/ijai
page 1763
Solving.
E
p
= x ∈ R x
2
− x ≤ 0 = x ∈ R x x − 1 ≤ 0 = {x ∈ R|(x ≤ 0) ∧
∧ (x − 1) ≥ 0} ∨ (x − 1 ≤ 0) ∧ (x ≥ 0)} = {x ∈ R|x ≤ 0} ∩ {x ∈ R|x ≥ 1} ∪
∪ x ∈ R x ≤ 1 ∩ x ∈ R x ≥ 0 = −∞; 0 ∩ 1; ∞ ∪ −∞; 1 ∩ 0; ∞ =
= ∅ ∪ 0; 1 = [0; 1]
;
E
p
= [0; 1]
.
E
s
= x ∈ R x ≤ x = x ∈ R x ≥ 0 ∧ x
2
≤ x = x ∈ R x ≥ 0 ∩
∩ x ∈ R x x − 1 ≤ 0 = 0; ∞ ∩ 0; 1 = [0; 1]
.
E
s
= [0; 1]
.
E
p∧s
= E
p
∩ E
s
= 0; 1 ∩ 0; 1 = [0; 1]
E
p∨s
= E
p
∪ E
s
= 0; 1 ∪ 0; 1 = [0; 1]
E
p⟹s
= E
p
∪ E
s
= −∞; 0 ∪ 1; ∞ ∪ 0; 1 = ( − ∞; ∞)
.
E
s⟹p
= ( − ∞; ∞)
.
E
p⟺s
= E
p⟹s
∩ E
s⟹p
= ( − ∞; ∞)
.
The following theorems can be used to teach students how to solve proof problems using
equivalence formulas of predicate algebra.
3- theorem.
(∀x ∈ R)(x ≤ x ⟹ x
2
≤ x)
.
4- theorem.
(∀x ∈ R)(x ≤ x ⟺ x
2
≤ x)
.
5- theorem.
(∀x ∈ ℳ)(P x ⟹ S x ) ⟹ (E
p
⊂ E
s
)
.
6- theorem.
E
p
⊂ E
s
⟹ (∀x ∈ ℳ)(P x ⟹ S x )
.
7- theorem.
(E
p
= E
s
) ⟹ (∀x ∈ ℳ)(P x ⟺ S x )
[6].
The examples and problems discussed above can be used to teach students the
applications of predicate algebra. When students are taught the laws of mathematical logic, rules
of induction, equivalence formulas, and their applications in depth and in detail, their ability to
solve mathematical problems in the simplest ways, quickly, and without errors, will develop.
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INTERNATIONAL JOURNAL OF ARTIFICIAL INTELLIGENCE
ISSN: 2692-5206, Impact Factor: 12,23
American Academic publishers, volume 05, issue 03,2025
Journal:
https://www.academicpublishers.org/journals/index.php/ijai
page 1764
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