theorem
  N-min X = N-max X implies N-most X = {N-min X}
proof
  assume N-min X = N-max X;
  then N-most X c= LSeg(N-min X, N-min X) by Th40;
  then N-most X c= {N-min X} by RLTOPSP1:70;
  hence thesis by ZFMISC_1:33;
end;
