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 Th19:
  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 [x,y] in R holds (SgmX (R,A))/.1 = 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 [x,y] in R and
A4: SgmX (R,A)/.1 <> x;
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: 1 in dom SgmX (R,A) by FINSEQ_5:6;
  then
A9: [x, SgmX (R,A)/.1] in R by A3,A5,PARTFUN2:2;
A10:  field R = X by ORDERS_1:12;
  1 <= i by A6,FINSEQ_3:25;
  then 1 < i by A4,A7,XXREAL_0:1;
  then [SgmX (R,A)/.1, x] in R by A2,A6,A7,A8,Def2;
  hence contradiction by A4,A9,A10,RELAT_2:def 4,def 12;
end;
