theorem Th2:
  r <> 0 implies 1/r * r|^(i+1) = r|^i
proof
  assume
A1: r <> 0;
  thus 1/r * r|^(i+1) = 1/r * (r|^i * r) by NEWTON:6
    .= 1/r * r * r|^i
    .= 1*r|^i by A1,XCMPLX_1:106
    .= r|^i;
end;
