Mathematical-Reasoning · Unit 16 · Video 3 · Interactive Practice

Proving Set Equality by Two Inclusions: A − (A − B) = A ∩ B

IKey Formulas

Statement	Name	Role in a proof
$A = B \iff A \subseteq B \text{ and } B \subseteq A$	Set equality	An equality proof is two subset proofs
$x \notin A - B \iff x \notin A \text{ or } x \in B$	Not in a difference	Negation of " $x \in A$ and $x \notin B$ "
$A - (A - B) = A \cap B$	Proposition 1	Proved by choosing elements in both directions
$A - B = A \cap B^c$	Proposition 2	Rewrites any difference as an intersection

Key Insight: Every step in both proofs applies one of three definitions: $x \in A - B$ means $x \in A$ and $x \notin B$ ; $x \in A \cap B$ means $x \in A$ and $x \in B$ ; $x \in B^c$ means $x \notin B$ . A Venn diagram suggests an equality, but only the two inclusions prove it for every pair of sets.

IITwo Ways Out of A − B

An element fails to be in $A - B$ for one of two reasons: it is outside $A$ , or it is inside $B$ .

Place x in

IIIName the Rule Behind Each Line

Each line of the proof of $A - (A - B) = A \cap B$ follows from exactly one rule.

Part 1 · $A - (A - B) \subseteq A \cap B$

1Let

x \in A - (A - B)

x \in A

and

x \notin A - B

x \notin A

x \in B

4Since

x \in A

, we have

x \in B

x \in A \cap B

A - (A - B) \subseteq A \cap B

Part 2 · $A \cap B \subseteq A - (A - B)$

7Let

y \in A \cap B

y \in A

and

y \in B

y \notin A

y \in B

y \notin A - B

y \in A - (A - B)

A \cap B \subseteq A - (A - B)

Conclusion

A - (A - B) = A \cap B

IVOne Instance Versus Every Pair of Sets

One element in one side but not the other refutes an equality; agreement for one choice of $A$ and $B$ proves nothing.

Compare L and R

💡 Challenge: for $L = A - (A - B)$ and $R = A$ , the inclusion $L \subseteq R$ never fails; find the one region where an element must sit to make $R \subseteq L$ fail.

VQuiz Questions

Problem 1 · Not in a Difference

Given: $A$ and $B$ are subsets of a universal set $U$ and $x \in U$ . Which statement is equivalent to $x \notin A - B$ ?

✅ Correct! By De Morgan's law the negation of "

x \in A

and

x \notin B

" is "

x \notin A

x \in B

❌ The connective must change. Negating an "and" statement gives an "or" statement. With "and", an element of

A^c \cap B^c

would count as being in

A - B

, which it is not.

❌ That is not a negation. Neither part was negated; this statement is true for every element of

A - B

, so it cannot mean "not in

A - B

❌ Only one part was negated correctly. The negation of

x \notin B

x \in B

. This option describes

x \notin A \cap B

, and it is true for every element of

A - B

Show solution

By definition, $x \in A - B$ if and only if $x \in A$ and $x \notin B$ . Negating both sides,

x \notin A - B \iff \neg\,(x \in A \text{ and } x \notin B).

De Morgan's law $\neg(P \wedge Q) \equiv \neg P \vee \neg Q$ turns the right side into " $x \notin A$ or $x \in B$ ". So $x \notin A - B$ exactly when $x$ is outside $A$ or inside $B$ .

Problem 2 · One Direction Only

Given: a student writes: "Let $x \in A \cap B$ . Then $x \in A$ and $x \in B$ . Since $x \in B$ , the statement ' $x \notin A$ or $x \in B$ ' is true, so $x \notin A - B$ . Hence $x \in A - (A - B)$ . Therefore $A - (A - B) = A \cap B$ ." What is wrong with this proof?

✅ Correct! Every step is valid, but choosing an element of

A \cap B

proves only that

A \cap B

is a subset of the other set. Equality needs the second inclusion too.

❌ Not quite. The proof starts with

x \in A \cap B

and ends with

x \in A - (A - B)

, so it shows one inclusion. Set equality requires both

A - (A - B) \subseteq A \cap B

and

A \cap B \subseteq A - (A - B)

❌ The reverse is true. A subset proof must choose an arbitrary element and use only its membership in the chosen set; a specific element would prove nothing about the others.

❌ That step is valid.

x \notin A - B

is equivalent to "

x \notin A

x \in B

", and an "or" statement is true as soon as one part, here

x \in B

, is true.

Show solution

The argument correctly proves $A \cap B \subseteq A - (A - B)$ : each line follows from the previous ones and $x$ is arbitrary.

By the test for set equality, $A - (A - B) = A \cap B$ also needs $A - (A - B) \subseteq A \cap B$ . The missing half: let $x \in A - (A - B)$ . Then $x \in A$ and $x \notin A - B$ , so $x \notin A$ or $x \in B$ . Since $x \in A$ , we get $x \in B$ , so $x \in A \cap B$ .

Only with both inclusions does the conclusion $A - (A - B) = A \cap B$ follow.

Problem 3 · A New Difference

Given: to prove $A - (A \cap B) \subseteq A - B$ , let $x \in A - (A \cap B)$ . By the definition of set difference, $x \in A$ and $x \notin A \cap B$ . Complete the next two steps.

Step 1 · Negate membership in the intersection

Step 2 · Use the membership in A

✅ Correct! Now

x \in A

and

x \notin B

, so

x \in A - B

, which proves

A - (A \cap B) \subseteq A - B

❌ Check the negation.

x \in A \cap B

means

x \in A

and

x \in B

; negate that "and" statement.

❌ The connective must change. De Morgan's law turns the negation of an "and" statement into an "or" statement. An element of

A - B

is not in

A \cap B

, yet it is in

A

❌ That is the negation of $x \in A - B$ . Here the set is

A \cap B

, whose membership condition is "

x \in A

and

x \in B

", so both parts become "

\notin

❌ That is not a negation of $x \in A \cap B$ . Every element of

A \cap B

satisfies "

x \in A

x \notin B

", so this statement cannot mean "not in

A \cap B

❌ Use the "or" statement. Knowing

x \in A

rules out one of the two options in "

x \notin A

x \notin B

❌ Not quite. The surviving option of "

x \notin A

x \notin B

" is

x \notin B

, not

x \in B

❌ That option is the one ruled out. We already know

x \in A

, so "

x \notin A

" is false and the other option must hold.

❌ That contradicts the given. The definition of set difference already gave

x \notin A \cap B

Show solution

Let $x \in A - (A \cap B)$ . Then $x \in A$ and $x \notin A \cap B$ .

Since $x \in A \cap B$ means " $x \in A$ and $x \in B$ ", De Morgan's law gives

x \notin A \cap B \iff x \notin A \text{ or } x \notin B.

We know $x \in A$ , so the first option is ruled out and $x \notin B$ . Hence $x \in A$ and $x \notin B$ , that is, $x \in A - B$ . Since $x$ was arbitrary, $A - (A \cap B) \subseteq A - B$ .

(The reverse inclusion also holds, so in fact $A - (A \cap B) = A - B$ .)

Problem 4 · Which Inclusions Hold?

Given: $L = (A - B) \cup B$ and $R = A$ , where $A$ and $B$ are arbitrary subsets of a universal set $U$ . Which statement is true?

✅ Correct! Every element of

A

is in

A - B

or in

B

, but an element of

B

outside

A

lies in

L

and not in

R

❌ One inclusion fails. Take

A = \emptyset

and

B = \{1\}

: then

L = \{1\}

and

R = \emptyset

❌ The directions are reversed. With

A = \emptyset

and

B = \{1\}

1 \in L

but

1 \notin R

, so

L \subseteq R

fails;

R \subseteq L

never does.

❌ One inclusion always holds. If

x \in A

, then either

x \in B

, so

x \in L

, or

x \notin B

, so

x \in A - B \subseteq L

. Hence

R \subseteq L

for every

A

and

B

Show solution

$R \subseteq L$ always. Let $x \in A$ . Case 1: $x \in B$ ; then $x \in (A - B) \cup B$ . Case 2: $x \notin B$ ; then $x \in A$ and $x \notin B$ , so $x \in A - B$ and again $x \in (A - B) \cup B$ . Since $x$ was arbitrary, $A \subseteq (A - B) \cup B$ .

$L \subseteq R$ can fail. Let $U = \{1\}$ , $A = \emptyset$ and $B = \{1\}$ . Then $L = \emptyset \cup \{1\} = \{1\}$ and $R = \emptyset$ , so $1 \in L$ but $1 \notin R$ .

One counterexample is enough to show that $L = R$ is not an identity, while the inclusion $R \subseteq L$ needs a proof for every $A$ and $B$ .

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