theorem
  E-min X = E-max X implies E-most X = {E-min X}
proof
  assume E-min X = E-max X;
  then E-most X c= LSeg(E-min X, E-min X) by Th48;
  then E-most X c= {E-min X} by RLTOPSP1:70;
  hence thesis by ZFMISC_1:33;
end;
