
theorem
  for L being non empty reflexive antisymmetric RelStr, R being
auxiliary(i) (Relation of L), C being strict_chain of R st (for c being Element
  of L st c in C holds ex_sup_of SetBelow (R,C,c),L & c = sup SetBelow (R,C,c))
  holds R satisfies_SIC_on C
proof
  let L be non empty reflexive antisymmetric RelStr, R be auxiliary(i) (
  Relation of L), C be strict_chain of R;
  assume
A1: for c being Element of L st c in C holds ex_sup_of SetBelow (R,C,c),
  L & c = sup SetBelow (R,C,c);
  let x, z be Element of L;
  assume that
A2: x in C and
A3: z in C and
A4: [x,z] in R and
A5: x <> z;
A6: z = sup SetBelow (R,C,z) by A1,A3;
  per cases;
  suppose
A7: not ex y being Element of L st y in SetBelow (R,C,z) & x < y;
    reconsider x as Element of L;
A8: SetBelow (R,C,z) is_<=_than x
    proof
      let b be Element of L;
      assume
A9:   b in SetBelow (R,C,z);
      then
A10:  not x < b by A7;
      per cases;
      suppose
A11:    x <> b;
        b in C by A9,Th15;
        then
A12:    [x,b] in R or x = b or [b,x] in R by A2,Def3;
        not x <= b by A11,A10,ORDERS_2:def 6;
        hence b <= x by A12,WAYBEL_4:def 3;
      end;
      suppose
        x = b;
        hence b <= x;
      end;
    end;
A13: for a being Element of L st SetBelow (R,C,z) is_<=_than a holds x <= a
    by A2,A4,Th15;
    ex_sup_of SetBelow (R,C,z),L by A1,A3;
    hence thesis by A13,A5,A6,A8,YELLOW_0:def 9;
  end;
  suppose
    ex y being Element of L st y in SetBelow (R,C,z) & x < y;
    then consider y being Element of L such that
A14: y in SetBelow (R,C,z) and
A15: x < y;
    take y;
    thus y in C by A14,Th15;
    hence [x,y] in R by A2,A15,Th2;
    thus [y,z] in R by A14,Th15;
    thus thesis by A15;
  end;
end;
