theorem
  for E being finite non empty set, A,B,C being Event of E st A misses B
& A misses C & B misses C holds prob(A \/ B \/ C) = prob(A) + prob(B) + prob(C)
proof
  let E be finite non empty set, A,B,C be Event of E;
  assume that
A1: A misses B and
A2: A misses C and
A3: B misses C;
A4: prob(A /\ (B /\ C)) = 0 by A1,Th16,XBOOLE_1:74;
  prob(A \/ B \/ C) = ( prob(A) + prob(B) + prob(C) ) - ( prob(A /\ B) +
  prob(A /\ C) + prob(B /\ C) ) + prob(A /\ B /\ C) by Th29
    .= ( prob(A) + prob(B) + prob(C) ) - ( prob(A /\ B) + prob(A /\ C) +
  prob(B /\ C) ) + 0 by A4,XBOOLE_1:16
    .= ( prob(A) + prob(B) + prob(C) ) - ( prob(A /\ B) + prob(A /\ C) + 0 )
  by A3,Th16
    .= ( prob(A) + prob(B) + prob(C) ) - ( prob(A /\ B) + 0 ) by A2,Th16
    .= ( prob(A) + prob(B) + prob(C) ) - 0 by A1,Th16;
  hence thesis;
end;
