SOLUTION: Theorem 1.4.3 is P(AuB)=P(AnB')+P(B) = [P(A) - P(A n B)] + P(B) = P(A) + P(B) - P(A n B) Theorem 1.4.4 is P(A u B u C) = P(A) + P(B) + P(C) - P(A n B) - P(A n C) - P(B n C)

In order to understand the formula of inclusion-exclusion principle,
it is very useful to keep in mind this simple reasoning.


When we calculate the sum 

    P(A) + P(B) + P(C) for P(A U B U C),     (1)

it seems very natural and does not arouse suspicion - so, it looks as a good first approximation.


But thinking longer, you understand that every part  P(A n B),  P(A n C)  and  P(B n C)
you count twice in this sum  P(A) + P(B) + P(C).


Therefore, next step is to subract  P(A n B) + P(A n C) + P(B n C)  from the sum   P(A) + P(B) + P(C).


So, you get then  P(A) + P(B) + P(C) - P(A n B) - P(A n C) - P(B n C).     (2)


It is good as the next, second approximation.



But thinking further, you understand that in expression  P(A) + P(B) + P(C) - P(A n B) - P(A n C) - P(B n C)

the part P(A n B n C) is added three times in the first three addends and subtracted three times

in the next three terms. So, now this part  P(A n B n C)  simply ABSENTS in the second approximation (2).


THEREFORE, you MUST add  P(A n B n C)  to (2),  and after doing it, you get  
final formula of the Inclusion-Exclusion principle


    +-------------------------------------------------------------------------+
    |   P(A U B U C) = P(A) + P(B) + P(C) - P(AB) - P(AC) - P(BC) + P(ABC).   |
    +-------------------------------------------------------------------------+


You may consider it as a formal or informal proof of the formula.