Principle of inclusion-exclusion proof

[mottov2]

There are 3 events A,B and C prove that

P(A[itex]\cup[/itex]B[itex]\cup[/itex]C) = P(A)+P(B)+P(C)-P(A[itex]\cap[/itex]B)-P(A[itex]\cap[/itex]C)-P(B[itex]\cap[/itex]C)+P(A[itex]\cap[/itex]B[itex]\cap[/itex]C)
each event is disjoint so by the additivity rule...

My attempt:
A[itex]\cup[/itex]B[itex]\cup[/itex]C = (A[itex]\cap[/itex]B[itex]\cap[/itex]C)[itex]\cup[/itex](A[itex]\cap[/itex][A[itex]\cap[/itex]B[itex]\cap[/itex]C]^c)[itex]\cup[/itex]([B[itex]\cap[/itex]C][itex]\cap[/itex][A[itex]\cap[/itex]B[itex]\cap[/itex]C]^c)[itex]\cup[/itex](C[itex]\cap[/itex]A^c[itex]\cap[/itex]([B[itex]\cap[/itex]C][itex]\cap[/itex][A[itex]\cap[/itex]B[itex]\cap[/itex]C]^c)^c)[itex]\cup[/itex](B[itex]\cap[/itex]A^c[itex]\cap[/itex]([B[itex]\cap[/itex]C][itex]\cap[/itex][A[itex]\cap[/itex]B[itex]\cap[/itex]C]^c)^c)

each event is disjoint so by the additivity rule...
P(A[itex]\cup[/itex]B[itex]\cup[/itex]C) = P(A[itex]\cap[/itex]B[itex]\cap[/itex]C)+P(A[itex]\cap[/itex][A[itex]\cap[/itex]B[itex]\cap[/itex]C]^c)+P([B[itex]\cap[/itex]C][itex]\cap[/itex][A[itex]\cap[/itex]B[itex]\cap[/itex]C]^c)+P(C[itex]\cap[/itex]A^c[itex]\cap[/itex]([B[itex]\cap[/itex]C][itex]\cap[/itex][A[itex]\cap[/itex]B[itex]\cap[/itex]C]^c)^c)+P(B[itex]\cap[/itex]A^c[itex]\cap[/itex]([B[itex]\cap[/itex]C][itex]\cap[/itex][A[itex]\cap[/itex]B[itex]\cap[/itex]C]^c)^c)

P(A[itex]\cap[/itex](A[itex]\cap[/itex]B[itex]\cap[/itex]C)^c) = P(A)-P(A[itex]\cap[/itex]B[itex]\cap[/itex]C)
P((B[itex]\cap[/itex]C)[itex]\cap[/itex](A[itex]\cap[/itex]B[itex]\cap[/itex]C)^c) = P(B[itex]\cap[/itex]C)-P(A[itex]\cap[/itex]B[itex]\cap[/itex]C)
P(C[itex]\cap[/itex]A^c[itex]\cap[/itex]([B[itex]\cap[/itex]C][itex]\cap[/itex][A[itex]\cap[/itex]B[itex]\cap[/itex]C]^c)^c) = P(C[itex]\cap[/itex](A[itex]\cap[/itex]C)^c[itex]\cap[/itex]([B[itex]\cap[/itex]C][itex]\cap[/itex][A[itex]\cap[/itex]B[itex]\cap[/itex]C]^c)^c) = P(C)-P(A[itex]\cap[/itex]C)-P((B[itex]\cap[/itex]C)[itex]\cap[/itex](A[itex]\cap[/itex]B[itex]\cap[/itex]C)^c) = P(C)-P(A[itex]\cap[/itex]C)-P(B[itex]\cap[/itex]C)+P(A[itex]\cap[/itex]B[itex]\cap[/itex]C)
P(B[itex]\cap[/itex]A^c[itex]\cap[/itex]([B[itex]\cap[/itex]C][itex]\cap[/itex][A[itex]\cap[/itex]B[itex]\cap[/itex]C]^c)^c) = P(B[itex]\cap[/itex](A[itex]\cap[/itex]B)^c[itex]\cap[/itex]([B[itex]\cap[/itex]C][itex]\cap[/itex][A[itex]\cap[/itex]B[itex]\cap[/itex]C]^c)^c) = P(B)-P(A[itex]\cap[/itex]B)-P((B[itex]\cap[/itex]C)[itex]\cap[/itex](A[itex]\cap[/itex]B[itex]\cap[/itex]C)^c) = P(B)-P(A[itex]\cap[/itex]B)-P(B[itex]\cap[/itex]C)+P(A[itex]\cap[/itex]B[itex]\cap[/itex]C)

then by substitution...

P(A[itex]\cup[/itex]B[itex]\cup[/itex]C) = P(A[itex]\cap[/itex]B[itex]\cap[/itex]C)+P(A)-P(A[itex]\cap[/itex]B[itex]\cap[/itex]C)+P(B[itex]\cap[/itex]C)-P(A[itex]\cap[/itex]B[itex]\cap[/itex]C)+P(C)-P(A[itex]\cap[/itex]C)-P(B[itex]\cap[/itex]C)+P(A[itex]\cap[/itex]B[itex]\cap[/itex]C)+P(B)-P(A[itex]\cap[/itex]B)-P(B[itex]\cap[/itex]C)+P(A[itex]\cap[/itex]B[itex]\cap[/itex]C)
= P(A)+P(B)+P(C)-P(A[itex]\cap[/itex]C)-P(A[itex]\cap[/itex]B)-P(B[itex]\cap[/itex]C)+P(A[itex]\cap[/itex]B[itex]\cap[/itex]C)

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