reserve X,x,y,z for set;
reserve n,m,k,k9,d9 for Nat;
reserve d for non zero Nat;
reserve i,i0,i1 for Element of Seg d;
reserve l,r,l9,r9,l99,r99,x,x9,l1,r1,l2,r2 for Element of REAL d;
reserve Gi for non trivial finite Subset of REAL;
reserve li,ri,li9,ri9,xi,xi9 for Real;
reserve G for Grating of d;

theorem Th44:
  for X,X9 being Subset of Seg d st cell(l,r) c= cell(l9,r9) &
  (for i holds (i in X & l.i < r.i & [l.i,r.i] is Gap of G.i) or
  (not i in X & l.i = r.i & l.i in G.i)) &
  (for i holds (i in X9 & l9.i < r9.i & [l9.i,r9.i] is Gap of G.i) or
  (not i in X9 & l9.i = r9.i & l9.i in G.i)) holds X c= X9 &
  (for i st i in X or not i in X9 holds l.i = l9.i & r.i = r9.i) &
  for i st not i in X & i in X9 holds
  l.i = l9.i & r.i = l9.i or l.i = r9.i & r.i = r9.i
proof
  let X,X9 be Subset of Seg d;
  assume
A1: cell(l,r) c= cell(l9,r9);
  assume
A2: for i holds i in X & l.i < r.i & [l.i,r.i] is Gap of G.i or
  not i in X & l.i = r.i & l.i in G.i;
  assume
A3: for i holds i in X9 & l9.i < r9.i & [l9.i,r9.i] is Gap of G.i or
  not i in X9 & l9.i = r9.i & l9.i in G.i;
A4: l in cell(l,r) by Th23;
A5: r in cell(l,r) by Th23;
A6: for i holds l9.i <= r9.i by A3;
  thus X c= X9
  proof
    let i be object;
    assume that
A7: i in X and
A8: not i in X9;
    reconsider i as Element of Seg d by A7;
A9: l.i < r.i by A2,A7;
A10: l9.i = r9.i by A3,A8;
A11: l9.i <= l.i by A1,A4,A6,Th21;
    r.i <= r9.i by A1,A5,A6,Th21;
    hence thesis by A9,A10,A11,XXREAL_0:2;
  end;
  set k = card X;
  set k9 = card X9;
A12: card Seg d = d by FINSEQ_1:57;
  then
A13: k <= d by NAT_1:43;
A14: k9 <= d by A12,NAT_1:43;
A15: cell(l,r) in cells(k,G) by A2,A13,Th30;
A16: cell(l9,r9) in cells(k9,G) by A3,A14,Th30;
  thus for i st i in X or not i in X9 holds l.i = l9.i & r.i = r9.i
  proof
    let i;
    assume
A17: i in X or not i in X9;
    l9.i <= r9.i by A3;
    then l.i = l9.i & r.i = r9.i or
    l.i = l9.i & r.i = l9.i or l.i = r9.i & r.i = r9.i by A1,A13,A14,A15,A16
,Th42;
    hence thesis by A2,A3,A17;
  end;
  thus for i st not i in X & i in X9 holds
  l.i = l9.i & r.i = l9.i or l.i = r9.i & r.i = r9.i
  proof
    let i;
    assume that
A18: not i in X and
A19: i in X9;
A20: l.i = r.i by A2,A18;
    l9.i < r9.i by A3,A19;
    hence thesis by A1,A13,A14,A15,A16,A20,Th42;
  end;
end;
