theorem Th2:
  x in field R or R-Seg(x) = {}
proof
  assume
A1: not x in field R;
  set y = the Element of R-Seg(x);
  assume R-Seg(x) <> {};
  then [y,x] in R by Th1;
  hence contradiction by A1,RELAT_1:15;
end;
