theorem Th17:
  P.A < 1 iff 0 < P.([#] Sigma \ A)
proof
  thus P.A < 1 implies 0 < P.([#] Sigma \ A)
  proof
    assume P.A < 1;
    then 1 - P.([#] Sigma \ A) < 1 by Th16;
    then 1 + - P.([#] Sigma \ A) < 1;
    then - P.([#] Sigma \ A) < 1 - 1 by XREAL_1:20;
    hence thesis;
  end;
  assume 0 < P.([#] Sigma \ A);
  then 0 < 1 - P.A by PROB_1:32;
  then P.A + 0 < 1 by XREAL_1:20;
  hence thesis;
end;
