theorem
  S-min X = S-max X implies S-most X = {S-min X}
proof
  assume S-min X = S-max X;
  then S-most X c= LSeg(S-min X, S-min X) by Th56;
  then S-most X c= {S-min X} by RLTOPSP1:70;
  hence thesis by ZFMISC_1:33;
end;
