theorem Th44:
  q < 0 & b < c & a|^2 +b|^2 = c|^2 implies
    a|^2 + (b+q)|^2 > (c+q)|^2
  proof assume
    A1: q < 0 & b < c & a|^2 +b|^2 = c|^2; then
    A1a: b - c < c - c by XREAL_1:9;
    a|^2 + (b+q)|^2 - (c+q)|^2 = 2*q*(b-c) by A1,Lm61; then
    a|^2 + (b+q)|^2 - (c+q)|^2 + (c+q)|^2 > 0+ (c+q)|^2 by A1,A1a,XREAL_1:6;
    hence thesis;
  end;
