theorem
  for P,A,B st 0 < P.B & P.B < 1 & P.|.B.A = P.|.([#] Sigma \ B).A holds
  A,B are_independent_respect_to P
proof
  let P,A,B;
  assume that
A1: 0 < P.B and
A2: P.B < 1 and
A3: P.|.B.A = P.|.([#] Sigma \B).A;
  0 < P.([#] Sigma \ B) & P.(A /\ B)/P.B = P.|.([#] Sigma \B).A by A1,A2,A3
,Def6,Th17;
  then
A4: P.(A /\ B)/P.B = P.(A /\ ([#] Sigma \B))/P.([#] Sigma \ B) by Def6;
A5: B`= ([#] Sigma \ B);
  P.([#] Sigma \ B) <> 0 by A2,Th17;
  then
  P.(A /\ B) * P.([#] Sigma \ B) = P.(A /\ ([#] Sigma \B)) * P.B by A1,A4,Th1;
  then P.(A /\ B) * (1 - P.B) = P.(A /\ ([#] Sigma \B)) * P.B by PROB_1:32;
  then P.(A /\ B) = (P.(A /\ ([#] Sigma \B)) + P.(A /\ B)) * P.B
    .= P.A * P.B by A5,Th14;
  hence thesis;
end;
