reserve i,j,k,n,m for Nat,
  x,y,z,y1,y2 for object, X,Y,D for set,
  p,q for XFinSequence;
reserve k1,k2 for Nat;

theorem Th27:
  for X0,Y0 being finite natural-membered set st
    X0 <N< Y0 & i < (card X0) holds
  rng((Sgm0 (X0\/Y0))|(card X0))=X0 &
  ((Sgm0 (X0\/Y0))|(card X0)).i = (Sgm0 (X0 \/ Y0)).i
proof
  let X0,Y0 be finite natural-membered set;
  assume that
A1: X0 <N< Y0 and
A2: i < card X0;
A3: i in Segm card X0 by A2,NAT_1:44;
  set f=(Sgm0 (X0\/Y0))|(card X0);
  set f0=(Sgm0 (X0\/Y0));
  set Z={ v where v is Element of X0 \/Y0: ex k2 being Nat st v=f.k2 & k2 in
  card X0};
A4: X0 c= X0 \/ Y0 by XBOOLE_1:7;
A5: len (Sgm0 (X0\/Y0))=card (X0\/Y0) by Th20;
  then
A6: len f=card X0 by A4,AFINSQ_1:54,NAT_1:43;
A7: Z c= rng f
  proof
    let y being object;
   assume y in Z;
    then
    ex v0 being Element of X0 \/Y0 st y=v0 & ex k2 being Nat st v0=f.k2
    & k2 in card X0;
    hence thesis by A6,FUNCT_1:def 3;
  end;
  then reconsider Z0=Z as finite set;
  f is one-to-one by FUNCT_1:52;
  then
A8: dom f,(f.:(dom f)) are_equipotent by CARD_1:33;
A9: f.:(dom f)=rng f by RELAT_1:113;
A10: len f0=card (X0 \/Y0) by Th20;
A11: rng f0=X0 \/Y0 by Def4;
A12: rng f c= Z
  proof
    let y being object;
    assume
A13: y in rng f;
    then consider x being object such that
A14: x in dom f and
A15: y=f.x by FUNCT_1:def 3;
    reconsider y0=y as Element of (X0 \/Y0) by Def4,A13;
    ex k2 being Nat st y0=f.k2 & k2 in card X0 by A14,A15;
   hence thesis;
  end;
  then rng f=Z by A7;
  then card Z=card (len f) by A8,A9,CARD_1:5;
  then
A16: card Z= card X0 by A5,A4,AFINSQ_1:54,NAT_1:43;
A17: X0 \/ Y0 <> {} by A2,CARD_1:27,XBOOLE_1:15;
A18: now
    assume that
A19: not Z c= X0 and
A20: not X0 c= Z;
    consider v1 being object such that
A21: v1 in Z and
A22: not v1 in X0 by A19;
    consider v10 being Element of X0 \/Y0 such that
A23: v1=v10 and
A24: ex k2 being Nat st v10=f.k2 & k2 in card X0 by A21;
A25: v10 in Y0 by A17,A22,A23,XBOOLE_0:def 3;
    reconsider nv10 =v10 as Nat;
    consider v2 being object such that
A26: v2 in X0 and
A27: not v2 in Z by A20;
    X0 c= X0\/Y0 by XBOOLE_1:7;
    then consider x2 being object such that
A28: x2 in dom f0 and
A29: v2=f0.x2 by A11,A26,FUNCT_1:def 3;
    reconsider x20=x2 as Nat by A28;
    reconsider nv2 =v2 as Nat by A29;
A30: x20<len f0 by A28,AFINSQ_1:86;
A31: now
      assume x20 < card X0;
      then
A32:  x20 in Segm card X0 by NAT_1:44;
      card X0 <= card (X0 \/Y0) by NAT_1:43,XBOOLE_1:7;
      then card X0 <= len f0 by Th20;
      then f.x20=f0.x20 by A32,AFINSQ_1:53;
      hence contradiction by A4,A26,A27,A29,A32;
    end;
    consider k20 being Nat such that
A33: v10=f.k20 and
A34: k20 in card X0 by A24;
    card X0 <= len f0 by A10,NAT_1:43,XBOOLE_1:7;
    then
A35: f.k20=f0.k20 by A34,AFINSQ_1:53;
    k20<len f by A6,A34,AFINSQ_1:86;
    then k20<x20 by A6,A31,XXREAL_0:2;
    then nv10<nv2 by A33,A29,A35,A30,Def4;
    hence contradiction by A1,A26,A25;
  end;
A36: now
    per cases by A18;
    case
      Z0 c= X0;
      hence Z0=X0 by A16,CARD_2:102;
    end;
    case
      X0 c=Z0;
      hence Z0=X0 by A16,CARD_2:102;
    end;
  end;
  card X0 <= len f0 by A5,NAT_1:43,XBOOLE_1:7;
  hence thesis by A12,A7,A36,A3,AFINSQ_1:53;
end;
