theorem
  rng seq c= dom h implies Re ( (h/*seq)*Ns ) = Re (h/*(seq*Ns))
proof
  assume
A1: rng seq c= dom h;
  now
    let n be Element of NAT;
    (seq * Ns) is subsequence of seq by VALUED_0:def 17;
    then
A2: rng (seq*Ns) c= rng seq by VALUED_0:21;
    thus (Re ( (h/*seq)*Ns )).n = Re( ((h/*seq)*Ns).n ) by COMSEQ_3:def 5
      .= Re( (h/*seq).(Ns.n) ) by FUNCT_2:15
      .= Re( h/.(seq.(Ns.n)) ) by A1,FUNCT_2:109
      .=Re( h/.((seq*Ns).n) ) by FUNCT_2:15
      .=Re( (h/*(seq*Ns)).n ) by A1,A2,FUNCT_2:109,XBOOLE_1:1
      .=(Re (h/*(seq*Ns)) ).n by COMSEQ_3:def 5;
  end;
  hence thesis by FUNCT_2:63;
end;
