theorem ThA37:
  g + h = h + g implies i * (g + h) = i * g + ( i  *h)
proof
  assume
A1: g + h = h + g;
  per cases;
  suppose
A2: i >= 0;
    then
A3:  i * h = |.i.| * h by Def8;
    i * (g + h) = |.i.| * (g + h) & i * g = |.i.| * g by A2,Def8;
    hence thesis by A1,A3,Lm11;
  end;
  suppose
A4: i < 0;
    hence i * (g + h) = -(|.i.| * (h + g)) by A1,Def8
      .= -( |.i.| * h + ( |.i.| * g)) by A1,Lm11
      .= -( |.i.| * g) + -( |.i.| * h) by Th16
      .= i * g + -( |.i.| * h) by A4,Def8
      .= i * g + ( i * h) by A4,Def8;
  end;
end;
