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,Th199
    .= (A///a) \ (B///a) by A2;
end;
