reserve x for set,
  D for non empty set,
  k, n for Element of NAT,
  z for Nat;
reserve N for with_zero set,
  S for
    IC-Ins-separated non empty with_non-empty_values AMI-Struct over N,
  i for Element of the InstructionsF of S,
  l, l1, l2, l3 for Element of NAT,
  s for State of S;
reserve ss for Element of product the_Values_of S;

theorem Th2:
  for f1, f2 being sequence of NAT st f1 is bijective & (for
m, n being Element of NAT holds m <= n iff f1.m <= f1.n,S) & f2 is bijective &
 (for m, n being Element of NAT holds m <= n iff f2.m <= f2.n, S) holds f1 = f2
proof
  let f1, f2 be sequence of NAT such that
A1: f1 is bijective and
A2: for m, n being Element of NAT holds m <= n iff f1.m <= f1.n,S and
A3: f2 is bijective and
A4: for m, n being Element of NAT holds m <= n iff f2.m <= f2.n,S;
A5: dom f1 = NAT by FUNCT_2:def 1;
A6: dom f2 = NAT by FUNCT_2:def 1;
  defpred P[Nat] means f1.$1 <> f2.$1;
  assume f1 <> f2;
  then ex c being Element of NAT st P[c] by FUNCT_2:63;
  then
A7: ex c being Nat st P[c];
  consider d being Nat such that
A8: P[d] and
A9: for n being Nat st P[n] holds d <= n from NAT_1:sch 5(A7);
  reconsider d as Element of NAT by ORDINAL1:def 12;
A10: rng f1 = NAT by A1,FUNCT_2:def 3;
A11:  rng f2 = NAT by A3,FUNCT_2:def 3;
  consider d1 being object such that
A12: d1 in dom f1 and
A13: f2.d = f1.d1 by A10,FUNCT_1:def 3;
  reconsider d1 as Element of NAT by A12;
  consider d2 being object such that
A14: d2 in dom f2 and
A15: f1.d = f2.d2 by A11,FUNCT_1:def 3;
  reconsider d2 as Element of NAT by A14;
  per cases;
  suppose
A16: d1 <= d & d2 <= d;
    then f2.d2 <= f2.d, S by A4;
    then d <= d1 by A2,A15,A13;
    hence contradiction by A8,A13,A16,XXREAL_0:1;
  end;
  suppose
A17: d <= d1 & d2 <= d;
    f2.d2 = f1.d2
    proof
      assume not thesis;
      then d <= d2 by A9;
      hence contradiction by A8,A15,A17,XXREAL_0:1;
    end;
    hence contradiction by A1,A8,A15,A5,FUNCT_1:def 4;
  end;
  suppose
A18: d1 <= d & d <= d2;
    f1.d1 = f2.d1
    proof
      assume not thesis;
      then d <= d1 by A9;
      hence contradiction by A8,A13,A18,XXREAL_0:1;
    end;
    hence contradiction by A3,A8,A13,A6,FUNCT_1:def 4;
  end;
  suppose
A19: d <= d1 & d <= d2;
    then f2.d <= f2.d2,S by A4;
    then d1 <= d by A2,A15,A13;
    hence contradiction by A8,A13,A19,XXREAL_0:1;
  end;
end;
