To’plamlarning birlashmasi va uning xossalari. Ma’ruza matni: To’plamlarning kesishmasi Ta’rif

Theorem. Let A, B, and C be subsets of some universal set U . Then we have two “distributive laws:”
A∩(B ∪C) = (A∩B)∪(A∩C), and A∪(B ∩C) = (A∪B)∩(A∪C).
Proof. As you might expect the above can be easily demonstrated through Venn diagrams (see Exercise 1 below). Here, I’ll give a formal proof of the first result (viz., that “intersection distributes over union”).
Let x ∈ A ∩ (B ∪ C) and so x ∈ A and x ∈ B ∪ C. From this we see that either x ∈ A and x ∈ B or that x ∈ A and x ∈ C, which means, of course, that x ∈ (A ∩ B) ∪ (A ∩ C), proving that A∩(B∪C) ⊆ (A∩B)∪(A∩C). Conversely, if x ∈ (A∩B)∪(A∩C), then x ∈ A∩ B or x ∈ A∩ C. In either case x ∈ A, but also x ∈ B ∪ C, which means that x ∈ A∩(B∪C), proving that A∩(B∪C) ⊆ (A∩B)∪(A∩C). It follows that A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C). The motivated student
will have no difficulty in likewise providing a formal proof of the second distributive law.²


7°. A∪(B∩C) = (A∪B)∩(A∪C) (birlashmaning kesishmaga nisbatan distributivlik xossasi).
To‘plamlar soni ikkitadan ortiq bo‘lganda ham yig‘indi uchun chiqarilgan xulosalar to‘g‘ri bo‘ladi.
Kommutativlik va kesishmaning birlashmaga nisbatan distributivlik xossalarining to’g’riligini ko’rsatamiz
1) (kommutativlik xossasi)

a) b)

8.2 – chizma
8.2- a) b) chizmalardagi shtrixlangan sohalar bir xil bo’lgani uchun kesishmalar teng.
2) (kesishmaning birlashmaga nisbatan distributivlik xossasi)

8.3 – chizma 8.4 – chizma

8.3-chizmada tenglikning chap qismi birlashma vertical va garizantal shtrixlangan.
8.4-chizmada va kesishma gorizantal shtrizlangan. esa vertikal shtrixlangan. 8.3 va 8.4 chizmalardagi ikki marta shtrixlangan sohalar bir xil bo’lganligidan tenglikning to’g’riligi ko’rinadi.