theorem
  x>0 & b < c & a +b|^2 = c|^2 implies a + (b+x)|^2 < (c+x)|^2
  proof assume
    A1: x>0 & b < c & a +b|^2 = c|^2; then
    A1a: b - c < c - c by XREAL_1:9;
    a + (b+x)|^2 - (c+x)|^2 = 2*x*(b-c) by A1,Lm61; then
    a + (b+x)|^2 - (c+x)|^2 + (c+x)|^2 < 0+ (c+x)|^2 by A1,A1a,XREAL_1:6;
    hence thesis;
  end;
