theorem Th23:
  (Re seq)^\k =Re (seq^\k) & (Im seq)^\k =Im (seq^\k)
proof
  now
    let n be Element of NAT;
    thus (Re seq^\k).n =Re seq.(n+k) by NAT_1:def 3
      .=Re(seq.(n+k)) by Def5
      .=Re((seq^\k).n) by NAT_1:def 3
      .=Re ((seq)^\k).n by Def5;
    thus (Im seq^\k).n =Im seq.(n+k) by NAT_1:def 3
      .=Im(seq.(n+k)) by Def6
      .=Im((seq^\k).n) by NAT_1:def 3
      .=Im ((seq)^\k).n by Def6;
  end;
  hence thesis;
end;
