theorem Th25:
  for A,B,P st A,B are_independent_respect_to P holds A,([#] Sigma
  \ B) are_independent_respect_to P
proof
  let A,B,P;
  assume A,B are_independent_respect_to P;
  then
A1: P.(A /\ B) = P.A * P.B;
  P.(A /\ ([#] Sigma \ B)) = P.(A /\ B`) .= P.(A \ B) by SUBSET_1:13
    .= P.(A \ (A /\ B)) by XBOOLE_1:47
    .= P.A * 1 - P.A * P.B by A1,PROB_1:33,XBOOLE_1:17
    .= P.A * (1 - P.B)
    .= P.A * P.([#] Sigma \ B) by PROB_1:32;
  hence thesis;
end;
