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