Mathematical-Reasoning · Unit 17 · Video 4 · Interactive Practice

Complement Laws and Proofs by the Algebra of Sets

IKey Formulas

Property	Statement (for $A, B \subseteq U$ )
Basic, $\emptyset$ and $U$	$(A^c)^c = A$ , $\ A - B = A \cap B^c$ , $\ A - \emptyset = A$ , $\ A - U = \emptyset$ , $\ \emptyset^c = U$ , $\ U^c = \emptyset$
De Morgan's Laws	$(A \cap B)^c = A^c \cup B^c$ and $(A \cup B)^c = A^c \cap B^c$
Subsets and complements	$A \subseteq B \iff B^c \subseteq A^c$ (the order reverses)
Three equivalent conditions	$A \subseteq B \iff A \cap B^c = \emptyset \iff A^c \cup B = U$

Key Insight: An element-chasing proof follows one $x$ through the definitions; a proof by the algebra of sets chooses no element and rewrites one side, one named law per line, until it becomes the other side.

IIDe Morgan's Law, One Element at a Time

For every position of $x$ , the statements $x \in (A \cup B)^c$ and $x \in A^c \cap B^c$ have the same truth value.

Law x lies in

IIIThe Algebra of Sets: $(A \cup B) - C$

Each rewrite is an equality justified by a named law, so the shaded set never changes.

Step 0

$(A \cup B) - C$

$= (A \cup B) \cap C^c$ · basic property: $X - Y = X \cap Y^c$

$= C^c \cap (A \cup B)$ · commutative law

$= (C^c \cap A) \cup (C^c \cap B)$ · distributive law, $X = C^c$ , $Y = A$ , $Z = B$

$= (A \cap C^c) \cup (B \cap C^c)$ · commutative law, inside each piece

$= (A - C) \cup (B - C)$ · basic property, read right to left

IVThree Ways to Say $A \subseteq B$

Conditions (1), (2) and (3) are equivalent: wherever $A$ sits, they are all true or all false.

Shade A: horizontal position 190

(1) $A \subseteq B$

(2) $A \cap B^c = \emptyset$

(3) $A^c \cup B = U$

💡 The part of $U$ that $A^c \cup B$ misses is $(A^c \cup B)^c = A \cap B^c$ by De Morgan's Law and $(A^c)^c = A$ . So $A^c \cup B = U$ exactly when $A \cap B^c = \emptyset$ , which is why (2) and (3) always agree.

VQuiz Questions

Problem 1 · De Morgan with Actual Elements

Given: $U = \{1, 2, 3, 4, 5, 6, 7, 8\}$ , $A = \{1, 2, 3, 4\}$ and $B = \{3, 4, 5, 6\}$ . Find $A^c \cup B^c$ .

✅ Correct! By De Morgan,

A^c \cup B^c = (A \cap B)^c = \{3, 4\}^c = \{1, 2, 5, 6, 7, 8\}

❌ Not quite.

\{7, 8\}

A^c \cap B^c = (A \cup B)^c

: you intersected the complements instead of taking their union.

❌ Not quite.

\{3, 4\}

A \cap B

itself;

A^c \cup B^c

is its complement.

❌ Close, but…

7

and

8

are in neither

A

nor

B

, so they lie in

A^c

(and in

B^c

); complements are taken in

U

, not in

A \cup B

❌ Not quite. Compute

A^c

and

B^c

inside

U

first.

Show solution

Directly: $A^c = \{5, 6, 7, 8\}$ and $B^c = \{1, 2, 7, 8\}$ , so

A^c \cup B^c = \{1, 2, 5, 6, 7, 8\}.

Check with De Morgan's Law: $A \cap B = \{3, 4\}$ , and $(A \cap B)^c = \{1, 2, 5, 6, 7, 8\}$ , the same set.

Problem 2 · Complements Reverse Inclusion

Given: $A$ and $B$ are subsets of a universal set $U$ with $A \subseteq B$ . Which statement must be true?

✅ Correct!

A \subseteq B \iff B^c \subseteq A^c

: taking complements reverses the order of the two sets.

❌ Close, but… the order reverses. With

U = \{1, 2, 3\}

A = \{1\}

B = \{1, 2\}

A^c = \{2, 3\} \not\subseteq \{3\} = B^c

❌ Not quite. By De Morgan,

A^c \cup B^c = (A \cap B)^c

, so this says

A \cap B = \emptyset

, which fails whenever

A \neq \emptyset

❌ Not quite. By De Morgan,

A^c \cap B^c = (A \cup B)^c

, so this says

A \cup B = U

, which

A \subseteq B

does not guarantee.

❌ Not quite. Test each option on

U = \{1, 2, 3\}

A = \{1\}

B = \{1, 2\}

Show solution

Let $x \in B^c$ , so $x \notin B$ . If $x$ were in $A$ , then $x \in B$ because $A \subseteq B$ , a contradiction. So $x \notin A$ , that is $x \in A^c$ . Hence $B^c \subseteq A^c$ .

The example $U = \{1, 2, 3\}$ , $A = \{1\}$ , $B = \{1, 2\}$ satisfies $A \subseteq B$ and refutes the other three: $A^c = \{2, 3\} \not\subseteq \{3\} = B^c$ ; $A^c \cup B^c = \{2, 3\} \neq U$ ; $A^c \cap B^c = \{3\} \neq \emptyset$ .

Problem 3 · Read an Algebra-of-Sets Chain

Given: the chain $\begin{aligned} A - (B \cup C) &= A \cap (B \cup C)^c \\ &= A \cap (B^c \cap C^c) \\ &= (A \cap B^c) \cap C^c \\ &= (A - B) \cap C^c \\ &= \ ? \end{aligned}$

Which law justifies the second line, $A \cap (B \cup C)^c = A \cap (B^c \cap C^c)$ ?

What is the final line?

✅ Correct! The chain proves

A - (B \cup C) = (A - B) - C

without choosing an element.

❌ Check the law. Only the complement of a union changed:

(B \cup C)^c

became

B^c \cap C^c

❌ Not quite.

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

; removing

B \cup C

means removing both, which is an intersection of the pieces, not a union.

❌ Check the last step. Apply

X \cap Y^c = X - Y

with

X = A - B

and

Y = C

Show solution

The justifications, line by line:

\begin{aligned} A - (B \cup C) &= A \cap (B \cup C)^c && \text{basic property } X - Y = X \cap Y^c \\ &= A \cap (B^c \cap C^c) && \text{De Morgan's Law} \\ &= (A \cap B^c) \cap C^c && \text{associative law} \\ &= (A - B) \cap C^c && \text{basic property, right to left} \\ &= (A - B) - C && \text{basic property with } X = A - B,\ Y = C \end{aligned}

So $A - (B \cup C) = (A - B) - C$ : removing the elements of $B$ or $C$ is the same as removing $B$ first and then $C$ .

Problem 4 · Equivalent Conditions with $C^c$ in Place of $B$

Given: $A, C \subseteq U$ . Apply the theorem on three equivalent conditions with $B = C^c$ . Which condition is equivalent to $A \subseteq C^c$ ?

✅ Correct! Condition (2) with

B = C^c

reads

A \cap (C^c)^c = \emptyset

, and

(C^c)^c = C

: the set

A

lies outside

C

exactly when

A

and

C

are disjoint.

❌ Close, but… condition (2) needs the complement of

B = C^c

, which is

(C^c)^c = C

, not

C^c

. Your option says

A \subseteq C

❌ Not quite.

A \cup C = U

means

C^c \subseteq A

: the inclusion points the other way.

❌ Not quite.

A^c \cap C = \emptyset

means

C \subseteq A

, a different condition.

❌ Not quite. Substitute

B = C^c

into

A \cap B^c = \emptyset

Show solution

The theorem gives $A \subseteq B \iff A \cap B^c = \emptyset \iff A^c \cup B = U$ . With $B = C^c$ and $(C^c)^c = C$ :

A \subseteq C^c \iff A \cap C = \emptyset \iff A^c \cup C^c = U.

The example $U = \{1, 2, 3\}$ , $A = \{1\}$ , $C = \{2\}$ has $A \subseteq C^c = \{1, 3\}$ , yet $A \cap C^c = \{1\} \neq \emptyset$ , $A \cup C = \{1, 2\} \neq U$ and $A^c \cap C = \{2\} \neq \emptyset$ , so none of the other options is equivalent.

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Complement Laws and Proofs by the Algebra of Sets

IKey Formulas

IIDe Morgan's Law, One Element at a Time

IIIThe Algebra of Sets: (A∪B)−C(A \cup B) - C(A∪B)−C

IVThree Ways to Say A⊆BA \subseteq BA⊆B

VQuiz Questions

IIIThe Algebra of Sets: $(A \cup B) - C$

IVThree Ways to Say $A \subseteq B$