Answer: The step-by-step proof is shown below.
Step-by-step explanation:
To prove this, we need to first show that
|A U B| = |A| + |B| - |A n B|
Suppose A' and B' are the compliments of A and B respectively.
In a Venn diagram where set A and B intersects, we can see that
B = A' n B
So
|A U B| = |A U (A' n B)|
Where A and A' n B are mutually disjoint.
|A U B| = |A| U |A' n B|
= |A| + |A' U B|
= |A| + |B| - |A n B| (equation 1)
Now, to the main business.
Rewrite A U B U C to look like what we have proven above.
A U B U C = A U (B U C)
All we now have to do is prove for |A U (B U C)|
Easily from (equation 1)
|A U (B U C)| = |A| + |B U C| - |A n (B U C)|
= |A| + [|B| + |C| - |B n C|] - [(A n B) U (A n C)]
= |A| + [|B| + |C| - |B n C|] - [|A n B| + |A n C| - |(A n B) n (A n C)]|
= |A| + |B| + |C| - |B n C| - |A n B| - |A n C| + |A n B n C|
Because
(A n B) n (A n C) = A n B n C
Hence, the proof.
|A U (B U C)| = |A| + |B| + |C| - |B n C| - |A n B| - |A n C| + |A n B n C|
