theorem Th6:
  h + g = h + f or g + h = f + h implies g = f
proof
  assume h + g = h + f or g + h = f + h;
  then -h + (h + g) = -h + h + f or g + h + -h = f + (h + -h)
    by RLVECT_1:def 3;
  then -h + (h + g) = 0_G + f or g + (h + -h) = f + (h + -h)
    by RLVECT_1:def 3,Def5;
  then -h + (h + g) = f or g + 0_G = f + (h + -h) by Def4,Def5;
  then -h + h + g = f or g = f + (h + -h) by RLVECT_1:def 3,Def4;
  then -h + h + g = f or g = f + 0_G by Def5;
  then 0_G + g = f or g = f by Def4,Def5;
  hence thesis by Def4;
end;
