Karakteristic fonsiyonlari kullanirsak:
1A∩B=1A⋅1Band1AΔB=1A+1B−1A∩B
O halde islemlerimiz:
1AΔ(BΔC)=1A+1BΔC−1A∩(BΔC)
=1A+1B+1C−1B∩C−1A⋅(1B+1C−1B∩C)
=1A+1B+1C−1A∩B−1B∩C−1C∩A+1A∩(B∩C).
Burdan sonrasi bitti aslinda. Sunu da ispatlayalim: A∩(B∩C)=A∩B∩C:
1A∩(B∩C)=1A1B∩C=1A1B1C=1A∩B1C=1(A∩B)∩C.