theorem
  (s (#) s9)*' = (s*') (#) (s9*')
proof
  now
    let n be Element of NAT;
    thus (s (#) s9)*'.n = ((s (#) s9).n)*' by Def2
      .= (s.n * s9.n)*' by VALUED_1:5
      .= (s.n)*' * (s9.n)*' by COMPLEX1:35
      .= (s*'.n) * (s9.n)*' by Def2
      .= (s*'.n) * (s9*'.n) by Def2
      .= (s*' (#) s9*').n by VALUED_1:5;
  end;
  hence thesis by FUNCT_2:63;
end;
