theorem
  x>0 & a|^2 +b|^2 = (b+1)|^2 implies a|^2 + (b-x)|^2 > (b+1-x)|^2
  proof
    A0: b+1 > b+0 by XREAL_1:6;
    assume
    A1: x>0 & a|^2 +b|^2 = (b+1)|^2;
    consider q such that
    A2: q = -x;
    a|^2 + (b+q)|^2 > (b+1+q)|^2 by A0,A1,A2,Th44;
    hence thesis by A2;
  end;
