reserve x,y,z,X,Y for set;
reserve X,Y for non empty set,
  f for Function of X,Y;

theorem Th10:
  for A being Element of Fin X holds f.:A = {} implies A = {}
proof
  let A be Element of Fin X;
  assume
A1: f.:A = {};
  assume A <> {};
  then consider x being Element of X such that
A2: x in A by Th9;
  f.x in f.:A by A2,FUNCT_2:35;
  hence contradiction by A1;
end;
