reserve

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

theorem Th30:
  for k being Element of NAT for X being non empty set st k + 1 c=
card X for A being finite Subset of X st card A = k-1 holds meet ^^(A,X,k) = A
proof
  let k be Element of NAT;
  let X be non empty set such that
A1: k + 1 c= card X;
  let A be finite Subset of X such that
A2: card A = k - 1;
  ^^(A,X,k) = ^^(A,X) by A1,A2,Def13;
  hence thesis by A1,A2,Th26;
end;
