theorem Th16:
  for A be non empty Subset of X, B be set holds B meets dom chi(A
  ,A) implies rng (chi(A,A)|B) = {1}
proof
  let A be non empty Subset of X;
  let B be set;
A1: dom (chi(A,A)|B) = B /\ dom (chi(A,A)) by RELAT_1:61;
  rng (chi(A,A)|B) c= rng (chi(A,A)) by RELAT_1:70;
  then
A2: rng (chi(A,A)|B) c= {1} by Th15;
  assume B /\ dom (chi(A,A)) <> {};
  then rng (chi(A,A)|B) <> {} by A1,RELAT_1:42;
  hence thesis by A2,ZFMISC_1:33;
end;
