Prove the distributive property for sets:

$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$

I'm not good with proofs but my understanding is that I have to prove 2 things:

(1) $A \cup (B \cap C) \subset (A \cup B) \cap (A \cap C)$

(2) $A \cap (B \cap C) \supset (A \cup B) \cap (A \cup C)$

This is what I have done so far:

Part (1)

If $x\in A$, then $x \in (A \cup B)$ and $x \in (A \cup C)$.

$\therefore x \in (A \cup B) \cap (A \cap C)$

If $x \in (B \cap C)$ then $x \in (A \cup B)$ and $x \in (A \cup C)$ because $x \in B$ and $x \in C$.

$\therefore x \in (A \cup B) \cap (A \cap C)$

$\therefore A \cup (B \cap C) \subset (A \cup B) \cap (A \cup C)$

Part (2)

Now we have to prove the reverse inequality: $(A \cup B) \cap (A \cap C)$. Then $x \in A \cup B$ and $x \in (A \cup C)$

If $x \in A$, then $x \in A \cup (B \cap C)$

This is where I am up to. I wanted to know whether my approach is correct and if I did part (1) correctly. I'm stuck on part (2) and don't know how to proceed. I'd appreciate any help.

Thank you!!

  • $\begingroup$ Alright I'm no math major but I put some stuff together below. If anyone could verify I would appreciate it. I just learned about the symbols $\land \lor$ now... $\endgroup$ May 12, 2013 at 19:00

3 Answers 3


You must first prove 2 cases:

(1) $A \cap (B \cup C) \subset (A \cap B) \cup (A \cap C)$

(2) $(A \cap B) \cup (A \cap C) \subset A \cap (B \cup C)$

Note that in mathematics we use the following symbols:

$\cap=$ AND = $\land$

$\cup=$ OR = $\lor$

Case 1: $A \cap (B \cup C) \subset (A \cap B) \cup (A \cap C)$

Let $x \in A \cap (B \cup C) \implies x \in A \land x \in (B \cup C)$

$\implies x \in A \land \{ x \in B \lor x \in C \}$

$\implies \{ x \in A \land x \in B \} \lor\{ x \in A \land x \in C \} $

$\implies x \in (A \cap B) \lor x \in (A \cap C)$

$\implies x \in (A \cap B) \cup (A \cap C)$

$\therefore x \in A \cap (B \cup C) \implies x \in (A \cap B) \cup (A \cap C)$

$\therefore A \cap (B \cup C) \subset (A \cap B) \cup (A \cap C)$

Case 2: $(A \cap B) \cup (A \cap C) \subset A \cap (B \cup C)$

Let $x \in (A \cap B) \cup (A \cap C) \implies x \in (A \cap B) \lor x \in (A \cap C)$

$\implies \{x \in A \land x \in B \} \lor \{ x \in A \land x \in C \}$

$\implies x \in A \land \{ x \in B \lor x \in C\}$

$\implies x \in A \land \{B \cup C \}$

$\implies x \in A \cap (B \cup C)$

$\therefore x \in (A \cap B) \cup (A \cap C) \implies x \in A \cap (B \cup C)$

$\therefore (A \cap B) \cup (A \cap C) \subset A \cap (B \cup C)$

$\therefore A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$

  • $\begingroup$ Why must we prove the case A∩(B∪C)⊂(A∩B)∪(A∩C) instead of proving the case A∪(B∩C)⊂(A∪B)∩(A∪C) ? $\endgroup$
    – quantif
    Sep 3, 2017 at 18:03

Let $X = A \cap (B \cup C)$ and $Y = (A \cap B) \cup (A \cap C)$

To show that $X=Y$, we must show that:

  1. $X \subseteq Y$
  2. $Y \subseteq X$

Case 1: $X \subseteq Y$

If $x \in X$, then $x \in A$ and $x \in (B \cup C)$

The latter implies that $x$ is a member of at least one of $B$ or $C$.

We will proceed from here in 3 cases:

Case A: $x \in B$ and $x \notin C$

We know from above that $x \in A$.

If $x \in B$ and $x \in A$, then $x \in (A \cap B)$.

If $x \in (A \cap B)$, then $x \in (A \cap B) \cup (A \cap C)$.

$(A \cap B) \cup (A \cap C) = Y$, therefore $x \in Y$.

Case B: $x \notin B$ and $x \in C$

This is symmetric to Case A.

Case C: $x \in B$ and $x \in C$

This is simply an extension of Case A and Case B.

Therefore, if $x \in X$, then $x \in Y$, which implies $X \subseteq Y$.

Case 2: $Y \subseteq X$

If $y \in Y$, then $y$ is a member of at least one of $(A \cap B)$ or $(A \cap C)$.

Again, we have 3 cases:

Case A: $y \in (A \cap B)$ and $y \notin (A \cap C)$.

The former implies $y \in A$ and $y \in B$ by the definition of intersection.

If $y \in B$, then $y \in (B \cup C)$.

If $y \in A$ and $y \in (B \cup C)$, then $y \in (A \cap (B \cup C))$.

$(A \cap (B \cup C)) = X$, therefore $y \in X$.

Case B: $y \notin (A \cap B)$ and $y \in (A \cap C)$.

This is symmetric to Case A.

Case C: $y \in (A \cap B)$ and $y \in (A \cap C)$.

This is simply an extension of Case A and Case B.

Therefore, if $y \in Y$, then $y \in X$, which implies $Y \subseteq X$.

From Case 1 and Case 2, we have:

$X \subseteq Y$ and $Y \subseteq X$, therefore $X = Y$, which implies $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$. This completes the proof.


A simple way is to do a calculational proof, starting at the most complex side, to calculate which elements are in $(A \cup B) \cap (A \cup C)$: for all $x$, $$ \begin{align} & x \in (A \cup B) \cap (A \cup C) \\ \equiv & \;\;\;\;\;\text{"definition of $\cap$; definition of $\cup$, twice"} \\ & (x \in A \lor x \in B) \land (x \in A \lor x \in C) \\ (*) \; \equiv & \;\;\;\;\;\text{"logic: simplify by 'factoring out' $x \in A$, using the fact that $\lor$ distributes over $\land$"} \\ & x \in A \lor (x \in B \land x \in C) \\ \equiv & \;\;\;\;\;\text{"definition of $\cap$; definition of $\cup$"} \\ & x \in A \cup (B \cap C) \\ \end{align} $$ Now by extensionality (i.e., equal sets have the same elements) it follows that $$(0) \;\;\; A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$$

Note how we didn't need to prove two separate $\subseteq$ cases. Also note that the key step $(*)$ uses the logical law $$(1) \;\;\; P \lor (Q \land R) \;\equiv\; (P \lor Q) \land (P \lor R)$$ We see that $(0)$ and $(1)$ have the same structure. Since $(1)$ is in the logic domain, it is more generally useful than $(0)$, which is only applicable when dealing with sets.

  • $\begingroup$ that's a trick, not a real proof $\endgroup$
    – LJNG
    Jan 30, 2020 at 21:42

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