theorem
  a <> 0 implies (A\+\B) /// a = (A///a) \+\ (B///a)
proof
  assume
A1: a <> 0;
A2: {a}"" = {a"} by Th37;
  thus (A\+\B)///a = a"**(A\+\B) by Th37
    .= (a"**A)\+\(a"**B) by A1,Th200
    .= (A///a) \+\ (B///a) by A2;
end;
