Principle of inclusion-exclusion proof

[mottov2]

There are 3 events A,B and C prove that

P(A\cupB\cupC) = P(A)+P(B)+P(C)-P(A\capB)-P(A\capC)-P(B\capC)+P(A\capB\capC)
each event is disjoint so by the additivity rule...

My attempt:
A\cupB\cupC = (A\capB\capC)\cup(A\cap[A\capB\capC]^c)\cup([B\capC]\cap[A\capB\capC]^c)\cup(C\capA^c\cap([B\capC]\cap[A\capB\capC]^c)^c)\cup(B\capA^c\cap([B\capC]\cap[A\capB\capC]^c)^c)

each event is disjoint so by the additivity rule...
P(A\cupB\cupC) = P(A\capB\capC)+P(A\cap[A\capB\capC]^c)+P([B\capC]\cap[A\capB\capC]^c)+P(C\capA^c\cap([B\capC]\cap[A\capB\capC]^c)^c)+P(B\capA^c\cap([B\capC]\cap[A\capB\capC]^c)^c)

P(A\cap(A\capB\capC)^c) = P(A)-P(A\capB\capC)
P((B\capC)\cap(A\capB\capC)^c) = P(B\capC)-P(A\capB\capC)
P(C\capA^c\cap([B\capC]\cap[A\capB\capC]^c)^c) = P(C\cap(A\capC)^c\cap([B\capC]\cap[A\capB\capC]^c)^c) = P(C)-P(A\capC)-P((B\capC)\cap(A\capB\capC)^c) = P(C)-P(A\capC)-P(B\capC)+P(A\capB\capC)
P(B\capA^c\cap([B\capC]\cap[A\capB\capC]^c)^c) = P(B\cap(A\capB)^c\cap([B\capC]\cap[A\capB\capC]^c)^c) = P(B)-P(A\capB)-P((B\capC)\cap(A\capB\capC)^c) = P(B)-P(A\capB)-P(B\capC)+P(A\capB\capC)

then by substitution...

P(A\cupB\cupC) = P(A\capB\capC)+P(A)-P(A\capB\capC)+P(B\capC)-P(A\capB\capC)+P(C)-P(A\capC)-P(B\capC)+P(A\capB\capC)+P(B)-P(A\capB)-P(B\capC)+P(A\capB\capC)
= P(A)+P(B)+P(C)-P(A\capC)-P(A\capB)-P(B\capC)+P(A\capB\capC)

did i do this right? I feel like i may have overcomplicated it..

 

[Stephen Tashi]

You haven't explained what laws of probability can be assumed in order to prove the result and you didn't say which laws you used. Putting a 'P' in front of each set in an expression is not "substitution".

 

[M-quest]

Proof By Induction is a lot more elegant,but you have to be careful of your notation