If $|A \cup B \cup C | = |A| + |B| + |C|$, then $A, B$, and $C$ must be pairwise disjoint Let A, B, and C be finite sets. Prove or disprove:
If $|A \cup B \cup C | = |A| + |B| + |C|$, then $A, B$, and $C$ must be pairwise disjoint.

I started with inclusion-exclusion formula
$$|A \cup B \cup C | = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$$
Hypothesis says if $|A \cup B \cup C | = |A| + |B| + |C|$, then expression $|A \cap B \cap C|  - |A \cap B| - |A \cap C| - |B \cap C| $, must be equal to zero. So, we have
$$ |A \cap B \cap C|  - |A \cap B| - |A \cap C| - |B \cap C|  = 0$$
$$ |A \cap B \cap C|  = |A \cap B| + |A \cap C| + |B \cap C| \tag{1}$$
Now, I am not sure how finish this proof. I found two ways, but I do not know if are correct.
First way
If  $x \in |A \cap B \cap C|$, then $x \in |A \cap B| \wedge x\in |A \cap C| \wedge x \in |B \cap C|$.
So if we count the cardinality of $|A \cap B \cap C|  = |A \cap B| + |A \cap C| + |B \cap C|$, then if $x \in |A \cap B \cap C|$ then $x$ is counted once but on RHS $x$ is counted three times. Thus the equation is true if $A = B = C = \emptyset$.
Therefore $A, B,$ and $C$ are pairwise disjoint.
Second way
From inclusion-exclusion formula, I know $$|A \cap B \cap C| = - |A| - |B| - |C| + |A \cap B| + |A \cap C| + |B \cap C| - |A \cup B \cup C |$$
I substitute in equation (1)
$$ -|A| - |B| - |C| + |A \cap B| + |A \cap C| + |B \cap C| - |A \cup B \cup C | = |A \cap B| + |A \cap C| + |B \cap C|$$
The expression $|A \cap B| + |A \cap C| + |B \cap C|$ cancels out.
$$ |A| + |B| + |C| + |A \cup B \cup C | = 0$$
And this equation is true if $A = B = C = \emptyset$.
Therefore $A, B,$ and $C$ are pairwise disjoint.

My question is which way is correct and better or if there is a different method to prove the proposition. Thank you. (I am self-studying student, who is reading a  textbook about discrete mathematics).
 A: Suppose $A$, $B$, $C$ are not pairwise disjoint.  Then the intersection of two of them, wlog $A$ and $B$, is nonempty.  Now $|A \cup B| = |A| + |B| - |A \cap B| < |A| + |B|$,
and $|A \cup B \cup C| \le |A \cup B| + |C| <  |A| + |B| + |C|$.
A: *

*Your first way is correct unless $A \cap B \cap C=\emptyset.$ 

*The equation in your second way has to be corrected as $$|A \cap B \cap C| = - |A| - |B| - |C| + |A \cap B| + |A \cap C| + |B \cap C| + |A \cup B \cup C |$$ and substituting this leads to a nothing (as you use the same equation twice).


This is how I would answer this question (using inclusion-exclusion principal):
$A \cap B \cap C\subset A \cap B$ and therefore $|A \cap B \cap C|  \le |A \cap B|, |A \cap C|, |B \cap C|.$
Then we have 
$$3|A \cap B \cap C|  \le |A \cap B| + |A \cap C| + |B \cap C|.$$ Now combining this with what you have already found $$|A \cap B \cap C|  = |A \cap B| + |A \cap C| + |B \cap C|,$$ we can say $$|A \cap B| =|A \cap C| = |B \cap C|=|A \cap B \cap C|=0 .$$
