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Daca \( M \) este o multime finita vom nota cu \( n(M) \) numarul elementelor sale . Fie \( A,B,C \) trei multimi . Dovediti ca :
\( n(A\cup B\cup C)=n(A)+n(B)+n(C)-[n(A\cap B)+n(A\cap C)+n(B\cap C)]+n(A\cap B\cap C) \)

Desenand cele 3 multimi ca niste diagramele van euler, afirmatia este evidenta.




In desen am notat cu \( M1,M2,M3,M4 \) multimile de puncte situate in cea mai mica suprafata inchisa care le contine.

\( n(A\cup B \cup C)=n(A)+n(B)+n(C)-n(M1)-n(M2)-n(M3)-2n(M4)=n(A)+n(B)+n(C)-[(n(M1)+n(M4)) + (n(M2)+n(M4))+(n(M3)+n(M4))]+n(M4)=n(A)+n(B)+n(C)-[n(A\cap B)+n(A\cap C)+n(B\cap C)]+n(A\cap B\cap C) \)

Eu cred ca merge si astfel :
Folosind proprietatea \( n(A\cup B)=n(A)+n(B)-n(A\cap B) \)
Deci\( n[(A\cup B)\cup C]=n(A\cup B)+n(C)-n[(A\cup B)\cap C]=n(A)+n(B)-n(A\cap B)+n(C)-n[(A\cap C)\cup (A\cap B)]=n(A)+n(B)+n(C)-n(A\cap B)-n(A\cap C)-n(A\cap B)+n(A\cap B\cap C) \) , ceea ce trebuia demonstrat .