theorem Th22:
  Re (seq*Nseq)=(Re seq)*Nseq & Im (seq*Nseq)=(Im seq)*Nseq
proof
  now
    let n be Element of NAT;
    thus (Re (seq*Nseq)).n =Re((seq*Nseq).n) by Def5
      .=Re(seq.(Nseq.n)) by FUNCT_2:15
      .=Re seq.(Nseq.n) by Def5
      .=(Re seq*Nseq).n by FUNCT_2:15;
  end;
  hence Re (seq*Nseq)=Re seq*Nseq;
  now
    let n be Element of NAT;
    thus Im (seq*Nseq).n =Im((seq*Nseq).n) by Def6
      .=Im(seq.(Nseq.n)) by FUNCT_2:15
      .=Im seq.(Nseq.n) by Def6
      .=(Im seq*Nseq).n by FUNCT_2:15;
  end;
  hence thesis;
end;
