reserve A for set, x,y,z for object,
  k for Element of NAT;
reserve n for Nat,
  x for object;
reserve V, C for set;

theorem Th20:
  for X being non empty set, A being non empty finite Subset of X,
R being Order of X, x being Element of X st x in A & R linearly_orders A & for
y being Element of X st y in A holds [y,x] in R holds SgmX (R,A)/.len SgmX (R,A
  ) = x
proof
  let X be non empty set, A be non empty finite Subset of X, R be Order of X,
  x be Element of X;
  assume that
A1: x in A and
A2: R linearly_orders A and
A3: for y being Element of X st y in A holds [y,x] in R and
A4: SgmX (R,A)/.len SgmX (R,A) <> x;
  set L = len SgmX (R,A);
A5: A = rng SgmX (R,A) by A2,Def2;
  then consider i being Element of NAT such that
A6: i in dom SgmX (R,A) and
A7: SgmX (R,A)/.i = x by A1,PARTFUN2:2;
  SgmX (R,A) is non empty by A2,Def2,RELAT_1:38;
  then
A8: L in dom SgmX (R,A) by FINSEQ_5:6;
  then
A9: [SgmX (R,A)/.L,x] in R by A3,A5,PARTFUN2:2;
A10: field R = X by ORDERS_1:12;
  i <= L by A6,FINSEQ_3:25;
  then i < L by A4,A7,XXREAL_0:1;
  then [x, SgmX (R,A)/.L] in R by A2,A6,A7,A8,Def2;
  hence contradiction by A4,A9,A10,RELAT_2:def 4,def 12;
end;
