Is $P((A \cap B)\cap C) = P((A \cap B) \cap (B \cap C))$ Is $P((A \cap B)\cap C) = P((A \cap B) \cap (B \cap C))$?
Drawing the Venn diagrams, the above expression seems to hold. But I am not exactly sure. Is there a more rigorous proof?
 A: as long as you can accept the rules of set algebra manipulation
$(A \cap B) \cap (B \cap C)$
$=(A \cap (B \cap C) \cap B \cap (B \cap C)) $
$=(A \cap (B \cap C) \cap (B \cap C)) $
$=A \cap (B \cap C)  $
$=A \cap B \cap C  $
$=(A \cap B) \cap C  $
A: While not a precise proof, you can set up a truth table from set theory.
A | B | C | A∩B | *(A∩B)∩C | B∩C | *(A∩B) ∩ (B∩C)
1 | 1 | 1 |     1    |           *1        |     1    |              *1
1 | 1 | 0 |     1    |           *0        |     0    |              *0
1 | 0 | 1 |     0    |           *0        |     0    |              *0
1 | 0 | 0 |     0    |           *0        |     0    |              *0
0 | 1 | 1 |     0    |           *0        |     1    |              *0
0 | 1 | 0 |     0    |           *0        |     0    |              *0
0 | 0 | 1 |     0    |           *0        |     0    |              *0
0 | 0 | 0 |     0    |           *0        |     0    |              *0
As you can see from the 5th and 7th columns, these sets are equivalent and thus the probabilities will be the same.
A: Let $\Omega$ be the sample space of interest. Let $A$, $B$, $C$ be events of $\Omega$. 
To show that $\mathbb{P}((A\cap B)\cap C)=\mathbb{P}((A\cap B)\cap (B\cap C))$, it suffices to show that $(A\cap B)\cap C=(A\cap B)\cap (B\cap C)$.
We show that $(A\cap B)\cap C\subseteq (A\cap B)\cap (B\cap C)$ and $(A\cap B)\cap (B\cap C)\subseteq (A\cap B)\cap C$. 
Suppose that $\omega\in (A\cap B)\cap C$. By the definition of intersection we have $\omega\in A\cap B$ and $\omega\in C$. Again using the definition of intersection, we have $\omega\in A$, $\omega\in B$ and $\omega\in C$. Since $\omega\in A$, $\omega\in B$ and $\omega\in C$, we have $\omega\in A\cap B$ and $\omega\in B\cap C$. We therefore have $\omega\in A\cap B$ and $\omega\in B\cap C$ whenever $\omega\in (A\cap B)\cap C$. Thus by the definition of a subset, we have $(A\cap B)\cap C\subseteq (A\cap B)\cap (B\cap C)$.
Now suppose $\omega\in (A\cap B)\cap (B\cap C)$. Therefore we have $\omega\in A\cap B$ and $\omega\in B\cap C$, and hence $\omega\in A$, $\omega\in B$ and $\omega\in C$. Thus $\omega\in A\cap B$ and $\omega\in C$. In other words, we have $\omega\in (A\cap B)\cap C$ whenever $\omega\in (A\cap B)\cap (B\cap C)$. We therefore have $(A\cap B)\cap (B\cap C)\subseteq (A\cap B)\cap C$. 
By the above argument, we can conclude that $(A\cap B)\cap C=(A\cap B)\cap (B\cap C)$, and hence $\mathbb{P}((A\cap B)\cap C)=\mathbb{P}((A\cap B)\cap (B\cap C))$.
