theorem Th11:
  (for n holds B.n = A) implies Intersection B = A
proof
  assume
A1: for n holds B.n = A;
  now
    let x be object;
    assume x in rng B;
    then ex n st x = B.n by Th4;
    hence x = A by A1;
  end;
  then rng B = {A} by ZFMISC_1:35;
  then meet rng B = A by SETFAM_1:10;
  hence thesis by Th8;
end;
