theorem
  r(#)Re seq = Re (r (#) seq) & r(#)Im seq = Im (r (#) seq)
proof
  now
    let n be Element of NAT;
    thus (r(#)Re seq).n= r*(Re seq.n) by SEQ_1:9
      .=r*Re(seq.n) by Def5
      .=Re(r*(seq.n)) by Th17
      .=Re((r (#) seq).n) by VALUED_1:6
      .=Re (r (#) seq).n by Def5;
  end;
  hence r(#)Re seq=Re (r (#) seq);
  now
    let n be Element of NAT;
    thus (r(#)Im seq).n= r*(Im seq.n) by SEQ_1:9
      .=r*Im(seq.n) by Def6
      .=Im(r*(seq.n)) by Th17
      .=Im((r(#) seq).n) by VALUED_1:6
      .=Im(r(#)seq).n by Def6;
  end;
  hence thesis;
end;
