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;

theorem
  f is special & i <= len GoB f & j <= width GoB f implies Int cell(GoB
  f,i,j) misses L~f
proof
  assume that
A1: f is special and
A2: i <= len GoB f and
A3: j <= width GoB f;
A4: Int cell(GoB f,i,j) = Int(v_strip(GoB f,i) /\ h_strip(GoB f,j)) by
GOBOARD5:def 3
    .= Int v_strip(GoB f,i) /\ Int h_strip(GoB f,j) by TOPS_1:17;
  assume Int cell(GoB f,i,j) meets L~f;
  then consider x be object such that
A5: x in Int cell(GoB f,i,j) and
A6: x in L~f by XBOOLE_0:3;
  L~f = union { LSeg(f,k) : 1 <= k & k+1 <= len f } by TOPREAL1:def 4;
  then consider X being set such that
A7: x in X and
A8: X in { LSeg(f,k) : 1 <= k & k+1 <= len f } by A6,TARSKI:def 4;
  consider k such that
A9: X = LSeg(f,k) and
  1 <= k and
  k+1 <= len f by A8;
  reconsider p = x as Point of TOP-REAL 2 by A7,A9;
  per cases by A1,SPPOL_1:19;
  suppose
    LSeg(f,k) is horizontal;
    then consider j0 being Nat such that
A10: 1 <= j0 and
A11: j0 <= width GoB f and
A12: for p st p in LSeg(f,k) holds p`2 = (GoB f)*(1,j0)`2 by Th10;
    now
A13:  j0 > j implies j0 >= j+1 by NAT_1:13;
      assume
A14:  p in Int h_strip(GoB f,j);
      per cases by A13,XXREAL_0:1;
      suppose
A15:    j0 < j;
        0 <> len GoB f by MATRIX_0:def 10;
        then 1 <= len GoB f by NAT_1:14;
        then
A16:    (GoB f)*(1,j)`2 > (GoB f)*(1,j0)`2 by A3,A10,A15,GOBOARD5:4;
        j >= 1 by A10,A15,XXREAL_0:2;
        then p`2 > (GoB f)*(1,j)`2 by A3,A14,GOBOARD6:27;
        hence contradiction by A7,A9,A12,A16;
      end;
      suppose
        j0 = j;
        then p`2 > (GoB f)*(1,j0)`2 by A10,A11,A14,GOBOARD6:27;
        hence contradiction by A7,A9,A12;
      end;
      suppose
A17:    j0 > j+1;
        then j + 1 <= width GoB f by A11,XXREAL_0:2;
        then j < width GoB f by NAT_1:13;
        then
A18:    p`2 < (GoB f)*(1,j+1)`2 by A14,GOBOARD6:28;
        0 <> len GoB f by MATRIX_0:def 10;
        then
A19:    1 <= len GoB f by NAT_1:14;
        j+1 >= 1 by NAT_1:11;
        then (GoB f)*(1,j+1)`2 < (GoB f)*(1,j0)`2 by A11,A17,A19,GOBOARD5:4;
        hence contradiction by A7,A9,A12,A18;
      end;
      suppose
A20:    j0 = j+1;
        then j < width GoB f by A11,NAT_1:13;
        then p`2 < (GoB f)*(1,j0)`2 by A14,A20,GOBOARD6:28;
        hence contradiction by A7,A9,A12;
      end;
    end;
    hence contradiction by A5,A4,XBOOLE_0:def 4;
  end;
  suppose
    LSeg(f,k) is vertical;
    then consider i0 being Nat such that
A21: 1 <= i0 and
A22: i0 <= len GoB f and
A23: for p st p in LSeg(f,k) holds p`1 = (GoB f)*(i0,1)`1 by Th11;
    now
A24:  i0 > i implies i0 >= i+1 by NAT_1:13;
      assume
A25:  p in Int v_strip(GoB f,i);
      per cases by A24,XXREAL_0:1;
      suppose
A26:    i0 < i;
        0 <> width GoB f by MATRIX_0:def 10;
        then 1 <= width GoB f by NAT_1:14;
        then
A27:    (GoB f)*(i,1)`1 > (GoB f)*(i0,1)`1 by A2,A21,A26,GOBOARD5:3;
        i >= 1 by A21,A26,XXREAL_0:2;
        then p`1 > (GoB f)*(i,1)`1 by A2,A25,GOBOARD6:29;
        hence contradiction by A7,A9,A23,A27;
      end;
      suppose
        i0 = i;
        then p`1 > (GoB f)*(i0,1)`1 by A21,A22,A25,GOBOARD6:29;
        hence contradiction by A7,A9,A23;
      end;
      suppose
A28:    i0 > i+1;
        then i + 1 <= len GoB f by A22,XXREAL_0:2;
        then i < len GoB f by NAT_1:13;
        then
A29:    p`1 < (GoB f)*(i+1,1)`1 by A25,GOBOARD6:30;
        0 <> width GoB f by MATRIX_0:def 10;
        then
A30:    1 <= width GoB f by NAT_1:14;
        i+1 >= 1 by NAT_1:11;
        then (GoB f)*(i+1,1)`1 < (GoB f)*(i0,1)`1 by A22,A28,A30,GOBOARD5:3;
        hence contradiction by A7,A9,A23,A29;
      end;
      suppose
A31:    i0 = i+1;
        then i < len GoB f by A22,NAT_1:13;
        then p`1 < (GoB f)*(i0,1)`1 by A25,A31,GOBOARD6:30;
        hence contradiction by A7,A9,A23;
      end;
    end;
    hence contradiction by A5,A4,XBOOLE_0:def 4;
  end;
end;
