reserve x,y,X for set,
  i,j,k,m,n for Nat,
  p for FinSequence of X,
  ii for Integer;
reserve G for Graph,
  pe,qe for FinSequence of the carrier' of G,
  p,q for oriented Chain of G,
  W for Function,
  U,V,e,ee for set,
  v1,v2,v3,v4 for Vertex of G;
reserve G for finite Graph,
  P,Q for oriented Chain of G,
  v1,v2,v3 for Vertex of G;
reserve G for finite oriented Graph,
  P,Q for oriented Chain of G,
  W for Function of (the carrier' of G), Real>=0,
  v1,v2,v3,v4 for Vertex of G;
reserve f,g,h for Element of REAL*,
  r for Real;
reserve G for oriented Graph,
  v1,v2 for Vertex of G,
  W for Function of (the carrier' of G), Real>=0;

theorem
  for p,q being FinSequence of NAT,f be Element of REAL*,i,n be Element
  of NAT st p is_simple_vertex_seq_at f,i,n & q is_simple_vertex_seq_at f,i,n
  holds p = q
proof
  let p,q be FinSequence of NAT,f be Element of REAL*,i,n be Element of NAT;
  assume that
A1: p is_simple_vertex_seq_at f,i,n and
A2: q is_simple_vertex_seq_at f,i,n;
A3: p.1=1 by A1;
A4: q.1=1 by A2;
A5: p is_vertex_seq_at f,i,n by A1;
  then
A6: p.(len p)=i & for k st 1<=k & k < len p holds p.(len p-k)=f.(p/.(len p-k
  +1)+ n);
A7: q is_vertex_seq_at f,i,n by A2;
  then
A8: q.(len q)=i;
A9: for k st 1<=k & k < len q holds q.(len q-k)=f.(q/.(len q-k+1)+n) by A7;
A10: len p > 1 by A1;
A11: now
    assume
A12: len p <> len q;
    per cases;
    suppose
A13:  len p < len q;
A14:  len p -1 > 1-1 by A10,XREAL_1:14;
      then reconsider k=len p -1 as Element of NAT by INT_1:3;
A15:  len p - k =0+1;
      then
A16:  len q - k > 1 by A13,XREAL_1:14;
      then reconsider m=len q -k as Element of NAT by INT_1:3;
A17:  k < len p by XREAL_1:146;
      k >= 0+1 by A14,INT_1:7;
      then
A18:  1=q.(len q-k) by A3,A6,A8,A9,A13,A15,A17,Lm10;
      q is one-to-one & m <= len q by A2,Lm9;
      hence contradiction by A4,A18,A16,GRAPH_5:6;
    end;
    suppose
A19:  len p >= len q;
A20:  p is one-to-one by A1;
A21:  len p > len q by A12,A19,XXREAL_0:1;
      hereby
        per cases;
        suppose
          len q <=1;
          hence contradiction by A2;
        end;
        suppose
          len q > 1;
          then
A22:      len q -1 > 1-1 by XREAL_1:14;
          then reconsider k=len q -1 as Element of NAT by INT_1:3;
A23:      k < len q by XREAL_1:146;
A24:      len q - k =0+1;
          then
A25:      len p - k > 1 by A21,XREAL_1:14;
          then reconsider m=len p -k as Element of NAT by INT_1:3;
          k >= 0+1 by A22,INT_1:7;
          then
A26:      1=p.(len p-k) by A6,A4,A8,A9,A19,A24,A23,Lm10;
          m <= len p by Lm9;
          hence contradiction by A3,A20,A26,A25,GRAPH_5:6;
        end;
      end;
    end;
  end;
  now
    let k be Nat;
    assume that
A27: 1 <=k and
A28: k <= len p;
    per cases;
    suppose
      k=len p;
      hence p.k=q.k by A5,A8,A11;
    end;
    suppose
      k <> len p;
      then k < len p by A28,XXREAL_0:1;
      then
A29:  len p-k > k-k by XREAL_1:14;
      then reconsider m=len p-k as Element of NAT by INT_1:3;
A30:  len q - m =0+k by A11;
      len p - k <= len p -1 by A27,XREAL_1:13;
      then
A31:  m < len p by XREAL_1:146,XXREAL_0:2;
      m >= 0+1 by A29,INT_1:7;
      hence p.k=q.k by A6,A8,A9,A30,A31,Lm10;
    end;
  end;
  hence thesis by A11,FINSEQ_1:14;
end;
