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,Th198
    .= (A///a) /\ (B///a) by A2;
end;
