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Fie \( T \) o multime si \( A, B, X \subset T \). Daca \( (X \cup \overline{A}) \cap (\overline {X} \cup A) = B \) atunci demonstrati ca \( A \Delta B = \overline{X} \).
( unde, pentru doua multimi \( Y, Z \subset T \) folosim notatiile \( \overline {Y}=T-Y \) si \( Y\Delta Z=(Y-Z)\cup(Z-Y) \) )

Avem \( B = (\overline{A} \cap \overline{X}) \cup (A \cap X) \). Deci \( A \Delta B = (A \cup (\overline{A} \cap \overline{X}) \cup (A \cap X)) \backslash (A \cap ((\overline{A} \cap \overline{X}) \cup (A \cap X))) \) \( = (A \cup (\overline{A} \cap \overline{X})) \backslash ((A \cap (\overline{A} \cap \overline{X})) \cup (A \cap (A \cap X))) \) \( = (A \cup \overline{X}) \backslash (\emptyset \cup (A \cap X)) \) \( = (A \cup \overline{X}) \cap \overline{A \cap X} \) \( = (A \cup \overline{X}) \cap (\overline{A} \cup \overline{X}) \) \( = \overline{X} \cup (A \cap \overline{X}) \cup (\overline{A} \cap \overline{X}) = \overline{X} \).