
theorem Th52:
  for K be Relation st rng K is empty-membered holds
    union rng K = {}
proof
   let K be Relation;
   assume A2: rng K is empty-membered;
   now let x be object;
    assume x in union rng K; then
    ex A be set st x in A & A in rng K by TARSKI:def 4;
    hence x in {} by A2;
   end; then
   union rng K c= {} by TARSKI:def 3;
   hence union rng K = {};
end;
