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 Th34:
  len f > 4
proof
  assume
A1: len f <= 4;
A2: len f > 1 by Lm2;
  then
A3: 1 in dom f by FINSEQ_3:25;
A4: len f >= 1+1 by A2,NAT_1:13;
  then
A5: 2 in dom f by FINSEQ_3:25;
  consider i2 such that
A6: i2 in dom f and
A7: (f/.i2)`2 <> (f/.1)`2 by Th31;
  consider i1 such that
A8: i1 in dom f and
A9: (f/.i1)`1 <> (f/.1)`1 by Th30;
  per cases by A4,TOPREAL1:def 5;
  suppose
A10: (f/.(1+1))`1 = (f/.1)`1;
A11: i1 <= len f by A8,FINSEQ_3:25;
A12: f/.len f = f/.1 by FINSEQ_6:def 1;
A13: i1 <> 0 by A8,FINSEQ_3:25;
    now
      i1 <= 4 by A1,A11,XXREAL_0:2;
      then i1 = 0 or ... or i1 = 4;
      then per cases by A9,A10,A13;
      suppose
A14:    i1 = 3;
A15:    now
          assume (f/.(1+1))`2 = (f/.1)`2;
          then f/.(1+1) = |[(f/.1)`1,(f/.1)`2]| by A10,EUCLID:53
            .= f/.1 by EUCLID:53;
          hence contradiction by A3,A5,Th29;
        end;
A16:    len f >= 3 by A8,A14,FINSEQ_3:25;
        then len f > 3 by A9,A12,A14,XXREAL_0:1;
        then
A17:    len f >= 3+1 by NAT_1:13;
        then
A18:    (f/.3)`1 = (f/.(3+1))`1 or (f/.3)`2 = (f/.(3+1))`2 by TOPREAL1:def 5;
A19:    len f = 4 by A1,A17,XXREAL_0:1;
        (f/.2)`2 = (f/.(2+1))`2 by A9,A10,A14,A16,TOPREAL1:def 5;
        hence contradiction by A9,A14,A19,A15,A18,FINSEQ_6:def 1;
      end;
      suppose
        i1 = 4;
        hence contradiction by A1,A9,A11,A12,XXREAL_0:1;
      end;
    end;
    hence contradiction;
  end;
  suppose
A20: (f/.(1+1))`2 = (f/.1)`2;
A21: i2 <= len f by A6,FINSEQ_3:25;
A22: f/.len f = f/.1 by FINSEQ_6:def 1;
A23: i2 <> 0 by A6,FINSEQ_3:25;
    now
      i2 <= 4 by A1,A21,XXREAL_0:2;
      then i2 = 0 or ... or i2 = 4;
      then per cases by A7,A20,A23;
      suppose
A24:    i2 = 3;
A25:    now
          assume (f/.(1+1))`1 = (f/.1)`1;
          then f/.(1+1) = |[(f/.1)`1,(f/.1)`2]| by A20,EUCLID:53
            .= f/.1 by EUCLID:53;
          hence contradiction by A3,A5,Th29;
        end;
A26:    len f >= 3 by A6,A24,FINSEQ_3:25;
        then len f > 3 by A7,A22,A24,XXREAL_0:1;
        then
A27:    len f >= 3+1 by NAT_1:13;
        then
A28:    (f/.3)`2 = (f/.(3+1))`2 or (f/.3)`1 = (f/.(3+1))`1 by TOPREAL1:def 5;
A29:    len f = 4 by A1,A27,XXREAL_0:1;
        (f/.2)`1 = (f/.(2+1))`1 by A7,A20,A24,A26,TOPREAL1:def 5;
        hence contradiction by A7,A24,A29,A25,A28,FINSEQ_6:def 1;
      end;
      suppose
        i2 = 4;
        hence contradiction by A1,A7,A21,A22,XXREAL_0:1;
      end;
    end;
    hence contradiction;
  end;
end;
