theorem Th35:
  s ^\k^\m = s^\(k+m)
proof
  now
    let n be Element of NAT;
    thus (s ^\k^\m).n=(s ^\k).(n+m) by Def2
      .=s.(n+m+k) by Def2
      .=s.(n+(k+m))
      .=(s ^\(k+m)).n by Def2;
  end;
  hence thesis by FUNCT_2:63;
end;
