reserve i,i1,i2,i9,i19,j,j1,j2,j9,j19,k,l,m,n for Nat;
reserve r,s,r9,s9 for Real;
reserve D for set,
  f for FinSequence of D,
  G for Matrix of D;
reserve G for Go-board,
  p for Point of TOP-REAL 2;
reserve T for non empty Subset of TOP-REAL 2;
reserve C for
  compact non vertical non horizontal non empty Subset of TOP-REAL 2;
reserve i, j, n for Nat;

theorem
  i <= len Gauge(C,n) implies cell(Gauge(C,n),i,0) misses C
proof
  set G = Gauge(C,n);
  assume
A1: i <= len G;
A2: len G = width G by Def1;
  assume cell(G,i,0) /\ C <> {};
  then consider p such that
A3: p in cell(G,i,0) /\ C by SUBSET_1:4;
A4: p in cell(G,i,0) by A3,XBOOLE_0:def 4;
A5: p in C by A3,XBOOLE_0:def 4;
  4 <= len G by Th10;
  then
A6: 1 <= len G by XXREAL_0:2;
  set W = W-bound C, S = S-bound C, E = E-bound C, N = N-bound C;
  set NS = (N-S)/(2|^n);
  [1,1] in Indices G by A2,A6,MATRIX_0:30;
  then G*(1,1) = |[W+((E-W)/(2|^n))*(1-2), S+NS*(1-2)]| by Def1;
  then
A7: G*(1,1)`2 = S+NS*(-1) by EUCLID:52;
A8: 2|^n > 0 by NEWTON:83;
  N > S by Th9;
  then N-S > 0 by XREAL_1:50;
  then NS > 0 by A8,XREAL_1:139;
  then NS*(-1) < 0 * (-1) by XREAL_1:69;
  then
A9: G*(1,1)`2 < S+0 by A7,XREAL_1:6;
A10: i = 0 or i >= 1+0 by NAT_1:9;
  per cases by A1,A10,XXREAL_0:1;
  suppose i = 0;
    then cell(G,i,0) ={ |[r,s]| : r <= G*(1,1)`1 & s <= G*(1,1)`2 }
    by GOBRD11:24;
    then consider r,s such that
A11: p = |[r,s]| and
    r <= G*(1,1)`1 and
A12: s <= G*(1,1)`2 by A4;
    p`2 = s by A11,EUCLID:52;
    then S > p`2 by A9,A12,XXREAL_0:2;
    hence contradiction by A5,PSCOMP_1:24;
  end;
  suppose i = len G;
    then cell(G,i,0)={ |[r,s]| : G*(len G,1)`1 <= r & s <= G*(1,1)`2 }
    by GOBRD11:27;
    then consider r,s such that
A13: p = |[r,s]| and
    G*(len G,1)`1 <= r and
A14: s <= G*(1,1)`2 by A4;
    p`2 = s by A13,EUCLID:52;
    then S > p`2 by A9,A14,XXREAL_0:2;
    hence contradiction by A5,PSCOMP_1:24;
  end;
  suppose 1 <= i & i < len G;
    then cell(G,i,0) = { |[r,s]|: G*(i,1)`1 <= r & r <= G*(i+1,1)`1 & s <= G*
    (1,1)`2 } by GOBRD11:30;
    then consider r,s such that
A15: p = |[r,s]| and
    G*(i,1)`1 <= r and r <= G*(i+1,1)`1 and
A16: s <= G*(1,1)`2 by A4;
    p`2 = s by A15,EUCLID:52;
    then S > p`2 by A9,A16,XXREAL_0:2;
    hence contradiction by A5,PSCOMP_1:24;
  end;
end;
