reserve

  k,n for Nat,
  x,y,X,Y,Z for set;

theorem ::LemStarb:
  for k being Element of NAT for X being non empty set st 0 < k & k + 1
c= card X for T being Subset of the Points of G_(k,X) st T is STAR for S being
Subset of X holds S = meet T implies card S = k - 1 & T = {A where A is Subset
  of X: card A = k & S c= A}
proof
  let k be Element of NAT;
  let X be non empty set such that
A1: 0 < k & k + 1 c= card X;
  let T be Subset of the Points of G_(k,X);
  assume T is STAR;
  then consider S1 being Subset of X such that
A2: card S1 = k - 1 & T = {A where A is Subset of X: card A = k & S1 c=
  A};
  let S be Subset of X;
  assume
A3: S = meet T;
  S1 = meet T by A1,A2,Th26;
  hence thesis by A3,A2;
end;
