theorem Th2:
  b <= a & a <= 0 implies |.a.| <= |.b.|
proof
  assume that
A1: b <= a and
A2: a <= 0;
  per cases by A2;
  suppose
    a = 0;
    then |.a.| = 0 by ABSVALUE:2;
    hence thesis by COMPLEX1:46;
  end;
  suppose
A3: a < 0;
    then
A4: |.a.| = -a by ABSVALUE:def 1;
    |.b.| = -b by A1,A3,ABSVALUE:def 1;
    hence thesis by A1,A4,XREAL_1:24;
  end;
end;
