Question

Daca se pricepe cineva, am nevoie de ajutor pentru a-mi putea da seama cum se rezolvă.​

Answer 1

Îți voi arăta cum să procedezi la primele 2, iar ultimele le faci singur:

Definiții și idei folosite:

1) $\text{ \wedge - inseamna cuvantul "si", iar \lor - sau}.$

2) $x \in A \cap B \iff (x \in A) \wedge (x \in B)$

3) $x \in (A \cup B) \iff (x \in A) \lor (x \in B)$

4) $(a \wedge b) \lor c = (a \lor c) \wedge (b \lor c)$

5) $(a \lor b) \wedge c = (a \wedge c) \lor (b \wedge c)$

6) $(A \subseteq B) \wedge (B \subseteq A) \iff A = B$

7)  $\text{Daca } a \in A \implies a \in A \cup B, \text{ pentru oricare multime B }.$

a) $(A \cup B) \cap A = A$

$\text{Fie } x \in (A \cup B) \cap A \implies x \in A \cup B \text{ }\wedge\text{ } x \in A\\((x \in A) \lor (x \in B)) \wedge (x \in A) \implies ((x \in A) \wedge (x \in A)) \lor ((x \in B) \wedge (x \in A)) \implies (x \in A) \lor (x \in A \cap B) \implies x \in A \implies (A \cup B) \cap A \subseteq A.$

$\text{Fie }x \in A \implies (x \in A \cup B) \wedge (x \in A) => x \in (A \cup B) \cap A \implies (A \cup B) \cap A \subseteq A\\\boxed{A = (A \cup B) \cap A}$

b) $(A \cap B) \cup A = A$

$\text{Fie }x \in (A \cap B) \cup A \implies (x \in A \cap B) \lor (x \in A) \implies ((x \in A) \wedge (x \in B)) \lor (x \in A) \implies ((x \in A) \lor (x \in A)) \wedge ((x \in B) \lor (x \in A)) \implies (x \in A) \wedge ((x \in B) \lor (x \in A)) \implies x \in A \implies (A \cap B) \cup A \subseteq A.$

$\text{Fie }x \in A \implies (x \in (A \cap B)) \lor (x \in A) \implies A \subseteq (A \cap B) \cup A\\\boxed{A = (A \cap B) \cup A}$