reserve f for non empty FinSequence of TOP-REAL 2,
  i,j,k,k1,k2,n,i1,i2,j1,j2 for Nat,
  r,s,r1,r2 for Real,
  p,q,p1,q1 for Point of TOP-REAL 2,
  G for Go-board;
reserve f for non constant standard special_circular_sequence;

theorem Th30:
  ex i st i in dom f & (f/.i)`1 <> (f/.1)`1
proof
  assume
A1: for i st i in dom f holds (f/.i)`1 = (f/.1)`1;
A2: len f > 1 by Lm2;
  then
A3: len f >= 1+1 by NAT_1:13;
  then
A4: 1+1 in dom f by FINSEQ_3:25;
A5: now
    assume
A6: (f/.2)`2 = (f/.1)`2;
    (f/.2)`1 = (f/.1)`1 by A1,A4;
    then f/.2 = |[(f/.1)`1,(f/.1)`2]| by A6,EUCLID:53
      .= f/.1 by EUCLID:53;
    hence contradiction by A4,Th29,FINSEQ_5:6;
  end;
  len f = 2 implies f/.2 = f/.1 by FINSEQ_6:def 1;
  then
A7: 2 < len f by A3,A5,XXREAL_0:1;
  per cases by A5,XXREAL_0:1;
  suppose
A8: (f/.2)`2 < (f/.1)`2;
    defpred P[Nat] means
2 <= $1 & $1 < len f implies (f/.$1)`2 <=
    (f/.2)`2 & (f/.($1+1))`2 < (f/.$1)`2;
A9: for j st P[j] holds P[j+1]
    proof
      let j such that
A10:  2 <= j & j < len f implies (f/.j)`2 <= (f/.2)`2 & (f/.(j+1))`2
      < (f/.j)`2 and
A11:  2 <= j+1 and
A12:  j+1 < len f;
      1+1 <= j+1 by A11;
      then
A13:  1 <= j by XREAL_1:6;
      thus (f/.(j+1))`2 <= (f/.2)`2
      proof
        per cases by A11,XXREAL_0:1;
        suppose
          2 = j+1;
          hence thesis;
        end;
        suppose
          2 < j+1;
          hence thesis by A10,A12,NAT_1:13,XXREAL_0:2;
        end;
      end;
A14:  j+1+1 <= len f by A12,NAT_1:13;
A15:  now
        per cases by A11,XXREAL_0:1;
        suppose
          1+1 = j+1;
          hence (f/.(j+1))`2 < (f/.j)`2 by A8;
        end;
        suppose
          2 < j+1;
          hence (f/.(j+1))`2 < (f/.j)`2 by A10,A12,NAT_1:13;
        end;
      end;
A16:  1 <= j+1 by NAT_1:11;
      then
A17:  j+1 in dom f by A12,FINSEQ_3:25;
      then
A18:  (f/.(j+1))`1 = (f/.1)`1 by A1;
      j < len f by A12,NAT_1:13;
      then
A19:  j in dom f by A13,FINSEQ_3:25;
      then
A20:  (f/.j)`1 = (f/.1)`1 by A1;
      1 <= j+1+1 by NAT_1:11;
      then
A21:  j+1+1 in dom f by A14,FINSEQ_3:25;
      then
A22:  (f/.(j+1+1))`1 = (f/.1)`1 by A1;
      assume
A23:  (f/.(j+1+1))`2 >= (f/.(j+1))`2;
      per cases by A23,XXREAL_0:1;
      suppose
A24:    (f/.(j+1+1))`2 > (f/.(j+1))`2;
        now
          per cases;
          suppose
            (f/.j)`2 <= (f/.(j+1+1))`2;
            then f/.j in LSeg(f/.(j+1),f/.(j+1+1)) by A15,A20,A18,A22,Th7;
            then
A25:        f/.j in LSeg(f,j+1) by A14,A16,TOPREAL1:def 3;
            j+1+1 = j+(1+1);
            then
A26:        LSeg(f,j) /\ LSeg(f,j+1) = {f/.(j+1)} by A14,A13,TOPREAL1:def 6;
            f/.j in LSeg(f,j) by A12,A13,TOPREAL1:21;
            then f/.j in LSeg(f,j) /\ LSeg(f,j+1) by A25,XBOOLE_0:def 4;
            then f/.j = f/.(j+1) by A26,TARSKI:def 1;
            hence contradiction by A19,A17,Th29;
          end;
          suppose
            (f/.j)`2 >= (f/.(j+1+1))`2;
            then f/.(j+1+1) in LSeg(f/.j,f/.(j+1)) by A20,A18,A22,A24,Th7;
            then
A27:        f/.(j+1+1) in LSeg(f,j) by A12,A13,TOPREAL1:def 3;
            j+1+1 = j+(1+1);
            then
A28:        LSeg(f,j) /\ LSeg(f,j+1) = {f/.(j+1)} by A14,A13,TOPREAL1:def 6;
            f/.(j+1+1) in LSeg(f,j+1) by A14,A16,TOPREAL1:21;
            then f/.(j+1+1) in LSeg(f,j) /\ LSeg(f,j+1) by A27,XBOOLE_0:def 4;
            then f/.(j+1+1) = f/.(j+1) by A28,TARSKI:def 1;
            hence contradiction by A17,A21,Th29;
          end;
        end;
        hence contradiction;
      end;
      suppose
A29:    (f/.(j+1+1))`2 = (f/.(j+1))`2;
        (f/.(j+1+1))`1 = (f/.1)`1 by A1,A21
          .= (f/.(j+1))`1 by A1,A17;
        then f/.(j+1+1) = |[(f/.(j+1))`1,(f/.(j+1))`2]| by A29,EUCLID:53
          .= f/.(j+1) by EUCLID:53;
        hence contradiction by A17,A21,Th29;
      end;
    end;
A30: len f -'1+1 = len f by A2,XREAL_1:235;
    then
A31: 2 <= len f -'1 & len f-'1 < len f by A7,NAT_1:13;
A32: P[0];
A33: for j holds P[j] from NAT_1:sch 2(A32,A9);
    then
A34: (f/.(len f-'1))`2 <= (f/.2)`2 by A31;
    (f/.len f)`2 < (f/.(len f -'1))`2 by A33,A30,A31;
    then (f/.len f)`2 < (f/.2)`2 by A34,XXREAL_0:2;
    hence contradiction by A8,FINSEQ_6:def 1;
  end;
  suppose
A35: (f/.2)`2 > (f/.1)`2;
    defpred P[Nat] means
2 <= $1 & $1 < len f implies (f/.$1)`2 >=
    (f/.2)`2 & (f/.($1+1))`2 > (f/.$1)`2;
A36: for j st P[j] holds P[j+1]
    proof
      let j such that
A37:  2 <= j & j < len f implies (f/.j)`2 >= (f/.2)`2 & (f/.(j+1))`2
      > (f/.j)`2 and
A38:  2 <= j+1 and
A39:  j+1 < len f;
      1+1 <= j+1 by A38;
      then
A40:  1 <= j by XREAL_1:6;
      thus (f/.(j+1))`2 >= (f/.2)`2
      proof
        per cases by A38,XXREAL_0:1;
        suppose
          2 = j+1;
          hence thesis;
        end;
        suppose
          2 < j+1;
          hence thesis by A37,A39,NAT_1:13,XXREAL_0:2;
        end;
      end;
A41:  j+1+1 <= len f by A39,NAT_1:13;
A42:  now
        per cases by A38,XXREAL_0:1;
        suppose
          1+1 = j+1;
          hence (f/.(j+1))`2 > (f/.j)`2 by A35;
        end;
        suppose
          2 < j+1;
          hence (f/.(j+1))`2 > (f/.j)`2 by A37,A39,NAT_1:13;
        end;
      end;
A43:  1 <= j+1 by NAT_1:11;
      then
A44:  j+1 in dom f by A39,FINSEQ_3:25;
      then
A45:  (f/.(j+1))`1 = (f/.1)`1 by A1;
      j < len f by A39,NAT_1:13;
      then
A46:  j in dom f by A40,FINSEQ_3:25;
      then
A47:  (f/.j)`1 = (f/.1)`1 by A1;
      1 <= j+1+1 by NAT_1:11;
      then
A48:  j+1+1 in dom f by A41,FINSEQ_3:25;
      then
A49:  (f/.(j+1+1))`1 = (f/.1)`1 by A1;
      assume
A50:  (f/.(j+1+1))`2 <= (f/.(j+1))`2;
      per cases by A50,XXREAL_0:1;
      suppose
A51:    (f/.(j+1+1))`2 < (f/.(j+1))`2;
        now
          per cases;
          suppose
            (f/.j)`2 >= (f/.(j+1+1))`2;
            then f/.j in LSeg(f/.(j+1),f/.(j+1+1)) by A42,A47,A45,A49,Th7;
            then
A52:        f/.j in LSeg(f,j+1) by A41,A43,TOPREAL1:def 3;
            j+1+1 = j+(1+1);
            then
A53:        LSeg(f,j) /\ LSeg(f,j+1) = {f/.(j+1)} by A41,A40,TOPREAL1:def 6;
            f/.j in LSeg(f,j) by A39,A40,TOPREAL1:21;
            then f/.j in LSeg(f,j) /\ LSeg(f,j+1) by A52,XBOOLE_0:def 4;
            then f/.j = f/.(j+1) by A53,TARSKI:def 1;
            hence contradiction by A46,A44,Th29;
          end;
          suppose
            (f/.j)`2 <= (f/.(j+1+1))`2;
            then f/.(j+1+1) in LSeg(f/.j,f/.(j+1)) by A47,A45,A49,A51,Th7;
            then
A54:        f/.(j+1+1) in LSeg(f,j) by A39,A40,TOPREAL1:def 3;
            j+1+1 = j+(1+1);
            then
A55:        LSeg(f,j) /\ LSeg(f,j+1) = {f/.(j+1)} by A41,A40,TOPREAL1:def 6;
            f/.(j+1+1) in LSeg(f,j+1) by A41,A43,TOPREAL1:21;
            then f/.(j+1+1) in LSeg(f,j) /\ LSeg(f,j+1) by A54,XBOOLE_0:def 4;
            then f/.(j+1+1) = f/.(j+1) by A55,TARSKI:def 1;
            hence contradiction by A44,A48,Th29;
          end;
        end;
        hence contradiction;
      end;
      suppose
A56:    (f/.(j+1+1))`2 = (f/.(j+1))`2;
        (f/.(j+1+1))`1 = (f/.1)`1 by A1,A48
          .= (f/.(j+1))`1 by A1,A44;
        then f/.(j+1+1) = |[(f/.(j+1))`1,(f/.(j+1))`2]| by A56,EUCLID:53
          .= f/.(j+1) by EUCLID:53;
        hence contradiction by A44,A48,Th29;
      end;
    end;
A57: len f -'1+1 = len f by A2,XREAL_1:235;
    then
A58: 2 <= len f -'1 & len f-'1 < len f by A7,NAT_1:13;
A59: P[0];
A60: for j holds P[j] from NAT_1:sch 2(A59,A36);
    then
A61: (f/.(len f-'1))`2 >= (f/.2)`2 by A58;
    (f/.len f)`2 > (f/.(len f -'1))`2 by A60,A57,A58;
    then (f/.len f)`2 > (f/.2)`2 by A61,XXREAL_0:2;
    hence contradiction by A35,FINSEQ_6:def 1;
  end;
end;
