Man bestimme G1∩G2{\displaystyle G_{1}\cap G_{2}} und G1∪G2{\displaystyle G_{1}\cup G_{2}}:
G1:V(G1)={1,2,...,8},E(G1)={⟨x,y⟩|x{\displaystyle G_{1}:V(G_{1})=\{1,2,...,8\},E(G_{1})=\{\langle x,y\rangle |x} teilt y,x<y}{\displaystyle y,x<y\}}
G2:V(G2)={1,2,...,5},E(G2)={⟨x,y⟩|x<y<=x+3}{\displaystyle G_{2}:V(G_{2})=\{1,2,...,5\},E(G_{2})=\{\langle x,y\rangle |x<y<=x+3\}}
für V(G1)∩V(G2)={1,2,3,4,5}{\displaystyle V(G_{1})\cap V(G_{2})=\{1,2,3,4,5\}}
für E(G1)∩E(G2)={(1,2),(1,3),(1,4),(2,4)}{\displaystyle E(G_{1})\cap E(G_{2})=\{(1,2),(1,3),(1,4),(2,4)\}}
für V(G1)∪V(G2)={1,2,3,4,5,6,7,8}{\displaystyle V(G_{1})\cup V(_{G}2)=\{1,2,3,4,5,6,7,8\}}
für E(G1)∪E(G2)={(1,2),(1,3),(1,4),(1,5),(1,6),(1,7),(1,8),(2,4),(2,5),(2,6),(2,8),(3,4),(3,5),(3,6)}{\displaystyle E(G_{1})\cup E(G_{2})=\{(1,2),(1,3),(1,4),(1,5),(1,6),(1,7),(1,8),(2,4),(2,5),(2,6),(2,8),(3,4),(3,5),(3,6)\}}