Mathematical-Reasoning · Unit 17 · Video 3 · Interactive Practice

Distributive Law: A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

IKey Formulas

Statement	Meaning	Name
$x \in A \cup B$	$x \in A$ or $x \in B$	Definition of union
$x \in A \cap B$	$x \in A$ and $x \in B$	Definition of intersection
$S = T$	$S \subseteq T$ and $T \subseteq S$	Set equality: two inclusions
$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$	for all subsets $A$ , $B$ , $C$ of $U$	Union distributes over intersection

Key Insight: Both inclusions need cases, for different reasons. In $\subseteq$ the hypothesis $x \in A \cup (B \cap C)$ is an "or", so it splits by itself. In $\supseteq$ the hypothesis is an "and" but the goal $y \in A$ or $y \in B \cap C$ is an "or", so the proof splits on whether $y \in A$ .

IIShade Both Sides

Built region by region, $A \cup (B \cap C)$ and $(A \cup B) \cap (A \cup C)$ shade the same five of the eight regions.

Step 1

IIIOne Region, Both Sides

Each region is one membership pattern for $x$ , and every pattern gives both sides the same answer.

IVWhere Can the Element Be?

Each proof step narrows where the chosen element can lie, and the two cases together cover all five shaded regions.

Inclusion Step 1

1Let $x \in A \cup (B \cap C)$ .

2By the definition of union, $x \in A$ or $x \in B \cap C$ .

Case 1 $x \in A$ : then $x \in A \cup B$ and $x \in A \cup C$ , so $x \in (A \cup B) \cap (A \cup C)$ .

Case 2 $x \in B \cap C$ : then $x \in B$ and $x \in C$ , so $x \in A \cup B$ and $x \in A \cup C$ , hence $x \in (A \cup B) \cap (A \cup C)$ .

5In both cases $x \in (A \cup B) \cap (A \cup C)$ , so $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$ .

1Let $y \in (A \cup B) \cap (A \cup C)$ .

2By the definition of intersection, $y \in A \cup B$ and $y \in A \cup C$ . Goal: $y \in A$ or $y \in B \cap C$ .

Case 1 $y \in A$ : then $y \in A \cup (B \cap C)$ .

Case 2 $y \notin A$ : since $y \in A \cup B$ , $y \in B$ ; since $y \in A \cup C$ , $y \in C$ . So $y \in B \cap C$ , hence $y \in A \cup (B \cap C)$ .

5In both cases $y \in A \cup (B \cap C)$ , so $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$ . With Part 1, the two sets are equal. ∎

VQuiz Questions

Problem 1 · Compute the Right Side

Given: $A = \{1, 2, 3\}$ , $B = \{3, 4, 5, 6\}$ , $C = \{1, 5, 6, 7\}$ — find $(A \cup B) \cap (A \cup C)$ .

✅ Correct! It matches

A \cup (B \cap C) = \{1, 2, 3\} \cup \{5, 6\} = \{1, 2, 3, 5, 6\}

, as the distributive law says it must.

❌ Not quite.

\{5, 6\}

B \cap C

alone. Every element of

A

lies in both

A \cup B

and

A \cup C

, so all of

A

belongs to the answer.

❌ Not quite.

\{1, 5, 6\}

(A \cup B) \cap C

: the second factor must be

A \cup C

, not

C

, so

2

and

3

survive too.

❌ Not quite. That is

(A \cup B) \cup (A \cup C)

. An intersection keeps only elements in both sets:

4 \notin A \cup C

and

7 \notin A \cup B

❌ Not quite. Compute

A \cup B

and

A \cup C

first, then keep the common elements.

Show solution

A \cup B = \{1, 2, 3, 4, 5, 6\}, \qquad A \cup C = \{1, 2, 3, 5, 6, 7\}

(A \cup B) \cap (A \cup C) = \{1, 2, 3, 5, 6\}

Check with the left side: $B \cap C = \{5, 6\}$ , so $A \cup (B \cap C) = \{1, 2, 3, 5, 6\}$ . The two sides agree. The elements $4$ (in $B$ only) and $7$ (in $C$ only) drop out, just as the regions "B only" and "C only" drop out of the Venn diagram.

Problem 2 · Which Identity Is the Law?

Which set equals $A \cup (B \cap C)$ for all subsets $A$ , $B$ , $C$ of a universal set $U$ ?

✅ Correct! Union distributes over intersection:

A

is joined to each of

B

and

C

, and the results are intersected.

❌ Not quite. Here

A

was distributed to only one factor. An element of

A

outside

C

lies in

A \cup (B \cap C)

but not in

(A \cup B) \cap C

❌ Not quite. That is

A \cap (B \cup C)

, the other distributive law. An element of

A

outside both

B

and

C

is in

A \cup (B \cap C)

but not in

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

❌ Not quite. The set being distributed is

A

, so

A

must appear in both factors. An element of

A

outside

B

and

C

is in

A \cup (B \cap C)

but not in

B \cup C

❌ Not quite. Test each option on the region "A only".

Show solution

Write $a$ , $b$ , $c$ for the statements $x \in A$ , $x \in B$ , $x \in C$ . Then $x \in A \cup (B \cap C)$ means $a \vee (b \wedge c)$ , which is equivalent to $(a \vee b) \wedge (a \vee c)$ , that is, $x \in (A \cup B) \cap (A \cup C)$ . So (c) is the law proved in the video.

Each other option fails on the region "A only" ( $a$ true, $b$ and $c$ false): there $A \cup (B \cap C)$ contains $x$ , while $(A \cup B) \cap C$ , $(A \cap B) \cup (A \cap C)$ and $(A \cup B) \cap (B \cup C)$ do not.

Problem 3 · Why Split on y ∈ A?

In Part 2 the proof starts from $y \in (A \cup B) \cap (A \cup C)$ and splits into the cases $y \in A$ and $y \notin A$ . What makes that split the right one?

✅ Correct! The cases

y \in A

and

y \notin A

cover every element, and each one reaches the "or" goal through a different disjunct.

❌ Not quite. By the definition of intersection the hypothesis is an "and":

y \in A \cup B

and

y \in A \cup C

. It does not split by itself; that is the situation of Part 1.

❌ Not quite. Many subset proofs need no cases at all. Here the cases come from the shape of the goal.

❌ Not quite. "

y \in B

y \notin B

" also covers every element. What makes

y \in A

useful is that

y \notin A

, together with

y \in A \cup B

and

y \in A \cup C

, yields

y \in B

and

y \in C

❌ Not quite. Look at whether the hypothesis and the goal are "and" or "or" statements.

Show solution

The hypothesis gives two facts at once: $y \in A \cup B$ and $y \in A \cup C$ . The goal, $y \in A \cup (B \cap C)$ , means $y \in A$ or $y \in B \cap C$ .

Case $y \in A$ : the first disjunct holds, so $y \in A \cup (B \cap C)$ .

Case $y \notin A$ : $y \in A \cup B$ means $y \in A$ or $y \in B$ ; since $y \notin A$ , $y \in B$ . Likewise $y \in A \cup C$ gives $y \in C$ . So $y \in B \cap C$ , the second disjunct.

Every element is in $A$ or not in $A$ , so the two cases cover all possibilities.

Problem 4 · Break a False Law

The claim $A \cup (B \cap C) = (A \cup B) \cap C$ is false. An element $x$ in which of these regions is a counterexample?

✅ Correct! Since

x \in A

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

; since

x \notin C

x \notin (A \cup B) \cap C

. The two sets differ.

❌ Not quite. The centre lies in both sets:

x \in A

puts it on the left, and

x \in A \cup B

with

x \in C

puts it on the right.

❌ Not quite. Here

x \in B \cap C

gives the left side, and

x \in B \subseteq A \cup B

with

x \in C

gives the right side, so both contain

x

❌ Not quite. Here

x

is in neither set:

x \notin A

and

x \notin B \cap C

on the left, and

x \notin A \cup B

on the right.

❌ Not quite. A counterexample must lie in exactly one of the two sets.

Show solution

A counterexample to a set equality is an element in one set but not the other. Test each region:

(a) $x \in A$ , $x \notin B$ , $x \notin C$ : left yes ( $x \in A$ ); right no ( $x \notin C$ ). Counterexample.
(b) centre: left yes, right yes.
(c) $x \in B \cap C$ , $x \notin A$ : left yes, right yes.
(d) $C$ only: left no, right no ( $x \notin A \cup B$ ).

The region "A and B, not C" is a second counterexample. The false claim distributes $A$ into only one factor; the true law is $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$ .

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