Himpunan

Jika S adalah himpunan semesta, A dan B adalah himpunan bagian dari S, maka: $$n(S)=n(A)+n(B)-n(A\cap B)+n(A\cup B)'$$ Bukti:

b=n(A\cap B)

\begin{aligned} a+b&=n(A)\\ a+n(A\cap B)&=n(A)\\ a&=n(A)-n(A\cap B) \end{aligned}

\begin{aligned} b+c&=n(B)\\ n(A\cap B)+c&=n(B)\\ c&=n(B)-n(A\cap B) \end{aligned}

d=n(A\cup B)'

\begin{aligned} n(S)&=a+b+c+d\\ &=n(A)-n(A\cap B)+n(A\cap B)+n(B)-n(A\cap B)+n(A\cup B)'\\ &=n(A)+n(B)-n(A\cap B)+n(A\cup B)' \end{aligned}

Jika S adalah himpunan semesta, A, B dan C adalah himpunan bagian dari S, maka: $$n(S)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)+n(A\cup B\cup C)'$$ Bukti:

g=n(A\cap B\cap C)

\begin{aligned} d+g&=n(A\cap B)\\ d+n(A\cap B\cap C)&=n(A\cap B)\\ d&=n(A\cap B)-n(A\cap B\cap C) \end{aligned}

\begin{aligned} e+g&=n(A\cap C)\\ e+n(A\cap B\cap C)&=n(A\cap C)\\ e&=n(A\cap C)-n(A\cap B\cap C) \end{aligned}

\begin{aligned} f+g&=n(B\cap C)\\ f+n(A\cap B\cap C)&=n(B\cap C)\\ f&=n(B\cap C)-n(A\cap B\cap C) \end{aligned}

\begin{aligned} a+d+e+g&=n(A)\\ a+n(A\cap B)-n(A\cap B\cap C)+n(A\cap C)-n(A\cap B\cap C)+n(A\cap B\cap C)&=n(A)\\ a+n(A\cap B)+n(A\cap C)-n(A\cap B\cap C)&=n(A)\\ a&=n(A)-n(A\cap B)-n(A\cap C)+n(A\cap B\cap C) \end{aligned}

\begin{aligned} b+d+f+g&=n(B)\\ b+n(A\cap B)-n(A\cap B\cap C)+n(B\cap C)-n(A\cap B\cap C)+n(A\cap B\cap C)&=n(B)\\ b+n(A\cap B)+n(B\cap C)-n(A\cap B\cap C)&=n(B)\\ b&=n(B)-n(A\cap B)-n(B\cap C)+n(A\cap B\cap C) \end{aligned}

\begin{aligned} c+e+f+g&=n(C)\\ c+n(A\cap C)-n(A\cap B\cap C)+n(B\cap C)-n(A\cap B\cap C)+n(A\cap B\cap C)&=n(C)\\ c+n(A\cap C)+n(B\cap C)-n(A\cap B\cap C)&=n(C)\\ c&=n(C)-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C) \end{aligned}

h=n(A\cup B\cup C)'

\begin{aligned} n(S)&=a+b+c+d+e+f+g+h\\ &=n(A)-n(A\cap B)-n(A\cap C)+n(A\cap B\cap C)+n(B)-n(A\cap B)-n(B\cap C)+n(A\cap B\cap C)+n(C)-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C)+n(A\cap B)-n(A\cap B\cap C)+n(A\cap C)-n(A\cap B\cap C)+n(B\cap C)-n(A\cap B\cap C)+n(A\cap B\cap C)+n(A\cup B\cup C)'\\ &=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)+n(A\cup B\cup C)' \end{aligned}