theorem
  for G1,G2 being strict Graph holds
  G1 c= G2 implies G1 \/ G2 = G2 & G2 \/ G1 = G2
proof
  let G1,G2 be strict Graph;
  assume
A1: G1 c= G2;
then  (G1 \/ G2) c= G2 & G2 c= (G1 \/ G2) by Th20,Th22;
  thus
then  G1 \/ G2 = G2 by Th16;
  hence thesis by A1,Th12;
end;
