USING PREDICATES TO CONSOLIDATE THEORETICAL KNOWLEDGE OF INEQUALITIES FOR FUTURE PRIMARY SCHOOL TEACHER

Аннотация

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, proofofa theorem, methods ofproof. 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 themain equivalenceformulas ofpredicatealgebra: (1) P(x)⋀(S x ⋁Q x ) ≡ P(x)⋀S(x)⋁P(x)⋀Q(x) (2) P(x)⋁S x ⋀Q x ≡ (P x ⋁S x )⋀(P x ⋁Q x ) (3) P(x)⋀S(x) ≡ P(x)⋁S(x) (4) P(x)⋁S(x) ≡ P(x)⋀S(x) (5) P(x) ⟹ S(x) ≡ P(x)⋁S(x) (6) P(x) ⟹ S(x) ≡ S(x) ⟹ P(x) (7) P(x) ⟺ S(x) ≡ P(x)⋀S(x)⋁P(x)⋀S(x) (8) P(x) ⟺ S(x) ≡ (P x ⋁S x )⋀(S x ⋁P x ) page1760 INTERNATIONAL JOURNAL OF ARTIFICIAL INTELLIGENCE ISSN:2692-5206,ImpactFactor:12,23 AmericanAcademicpublishers,volume05,issue03,2025 Journal: https://www.academicpublishers.org/journals/index.php/ijai 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 . We denote the truth domains of these predicates by E and E, respectively, the negation of p s thepredicateP(x)by ,and theset by . ℳ , P(x), ,ℳ\Ep Ep and We use the following theorems tofind thetruth domains ofthepredicates. P(x) P(x)⋁S(x) P(x)⋀S(x) P(x) ⟹ S(x) P(x) ⟺ S(x) 1-theorem. . 2 Proof. We us(e∀xth∈e cRo)n(vxer≤sexm⟹ethoxd≤: thaxt)is, instead of the given theorem, we prove the following theorem , which is equally strong. Let . Then x≥1. We multiply bo2th sides of this inequality by x (Since the value of x (∀x ∈ R)(x ≤ x ⇒ x ≤ x) (x ≥ x ⟹ is2positive, the inequality sign does not change). means that the inequality x ≥ x) x ≥ x implies the inequality . Thus, the formula is a theorem. x∙x ≥ x∙1 x ≥ x Therefore, the formula 2 , which is equally stro2ng as this formula, is also atheorem. x ≥ x 2 (∀x ∈ R)(x ≤ x ⇒ x ≤ x) (∀x ∈ R)(x ≤ x ⟹ x ≤ x) 2-theorem. Proof. Let (Ep =.ET s)he⟹n th(∀exel∈emℳen)t(∀P xxb⟺elonSgixng)to also belongs to . At thisvaluEepo=f xE,sP(x) and S(x) are true statements. TheErpefore, based onthEesdefinition of the equivalence operation, the formula is also a true statement. Therefore, from the equality it follows that the formula is a true statement. P x ⟺ S x Thetheorem isproved. Ep = Es ∀x ∈ ℳ)((P x ⟺ S x ) Let Rbe theset of real numbers. Example1. Given a predicatex2-7x+12 0)∨(x−3 >.0)∧(x−4 4)∨(x > 3)∧(x 0) defined on a set R. Find its truth domain E . p Solving. 2 P x ≡ x −x−20 > 0 ≡ ( x+4 ⋅ x−5 > 0) ≡ ≡ (x+4 0)∧(x−5 > 0) ≡ page1761 INTERNATIONAL JOURNAL OF ARTIFICIAL INTELLIGENCE ISSN:2692-5206,ImpactFactor:12,23 AmericanAcademicpublishers,volume05,issue03,2025 Journal: https://www.academicpublishers.org/journals/index.php/ijai , ≡ (x −4)∧(x > 5) Ep = x ∈ R x −4 ∩ x ∈ R x > 5 = (−∞;.−4)∩(−∞;5)∪ ∪Ans−w4e;r∞: ∩ 5;∞ = (−∞;−4)∪. (5;∞) Ep = (−∞;−4)∪(5;∞) Example3.Given a predicate inaset R,find itstruth domain E . p 2x+6 P x = 5x−10 ≤ 0 Solving. 2x+6 P x = 5x−10 ≤ 0 ≡ (2x+6 ≤ 0)∧(5x−10 > 0)∨ . ∨(2x+6 ≥ 0)∧(5x−10 2)∨(x ≥−3)∧(x 2 ∪ ∪ x ∈ R x ≥−3}∩{x ∈ R x −3) ≡ ≡ (x −1) Ep = x ∈ R x sw−er1: = x ∈ R.x −1 = = −∞E;5xam∩pl−e15;.∞Giv=en(a−pr1e;d5ic)ate Ep = (1;5) in the set R. Find its truth domain E . p P x = (|2x+6| ≥ 4) Solving. P x = (|2x+6| ≥ 4) ≡ (2x+6 ≥.4)∨(2x+6 ≤−4) ≡ ≡ (2x ≥−2)∨(2x ≤−10) ≡ (x ≥−1)∨(x ≤−5) Ep = x ∈ R x ≥−1 ∨ x