
theorem Th6:
  for X being Ordinal, a, b be finite Subset of X st a <> b holds (
  a,1)-bag <> (b,1)-bag
proof
  let X be Ordinal, a,b be finite Subset of X such that
A1: a <> b;
  assume
A2: (a,1)-bag = (b,1)-bag;
  now
    let x be object;
    x in a iff (b,1)-bag.x = 1 by A2,UPROOTS:6,7;
    hence x in a iff x in b by UPROOTS:6,7;
  end;
  hence contradiction by A1,TARSKI:2;
end;
