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 Th29:
  for X,Y being finite natural-membered set st X <N<
  Y & i< len (Sgm0 X) holds (Sgm0 X).i = (Sgm0 (X \/ Y)).i
proof
  let X,Y be finite natural-membered set;
  assume that
A1: X <N< Y and
A2: i< len (Sgm0 X);
  reconsider h=(Sgm0 (X \/ Y))|(len (Sgm0 X)) as XFinSequence of NAT;
A3: len (Sgm0 X)=card X by Th20;
  then
A4: h.i=(Sgm0 (X \/ Y)).i by A1,A2,Th27;
  Segm card X c= Segm card (X \/ Y) by CARD_1:11,XBOOLE_1:7;
  then
A5: card X <= card (X \/ Y) by NAT_1:39;
  then card X <= len (Sgm0 (X \/ Y)) by Th20;
  then
A6: len (Sgm0 X) <= len (Sgm0 (X \/ Y)) by Th20;
A7: len (Sgm0 (X \/ Y))=card (X \/Y) by Th20;
  then
A8: len h=len (Sgm0 X) by A5,A3,AFINSQ_1:54;
A9: len h=card X by A5,A3,A7,AFINSQ_1:54;
A10: for l,m,k1,k2 being Nat st l < m & m < len h & k1=h.l & k2=h.m holds k1
  < k2
  proof
    let l,m,k1,k2 be Nat;
    assume that
A11: l < m and
A12: m < len h and
A13: k1=h.l and
A14: k2=h.m;
A15: m<len (Sgm0 (X \/ Y)) by A8,A6,A12,XXREAL_0:2;
    l < card X by A9,A11,A12,XXREAL_0:2;
    then
A16: h.l= (Sgm0 (X \/ Y)).l by A1,A3,Th27;
    h.m=(Sgm0 (X \/ Y)).m by A1,A3,A8,A12,Th27;
    hence thesis by A11,A13,A14,A16,A15,Def4;
  end;
  rng h=X by A1,A2,A3,Th27;
  hence thesis by A10,A4,Def4;
end;
