De Morgan's Laws for Events

Why Complements Change Unions and Intersections

Compound events often describe several conditions at once: at least one condition occurs, both conditions occur, neither occurs, or it is not the case that both occur. To translate these descriptions reliably, it helps to know how taking a complement interacts with union and intersection. De Morgan’s laws give the exact identities. The key idea is that negating “at least one” means “neither,” while negating “both” means “at least one fails.”

Definition: Let $\Omega$ be a sample space and let $A,B\subseteq\Omega$ be events. The complement $A^c$ is the event $\Omega\setminus A$, containing exactly those outcomes in $\Omega$ that are not in $A$. The union $A\cup B$ contains outcomes in at least one of $A$ and $B$; the intersection $A\cap B$ contains outcomes in both.

Theorem (De Morgan’s laws): For any events $A,B\subseteq\Omega$, $$ (A\cup B)^c=A^c\cap B^c \qquad\text{and}\qquad (A\cap B)^c=A^c\cup B^c. $$

Each identity says that complementing a compound event reverses the operation: a union becomes an intersection, and an intersection becomes a union. Both complements on the right are taken relative to the same sample space $\Omega$.

Proof by Checking an Arbitrary Outcome

Two sets are equal precisely when they have the same members. To prove each identity, take an arbitrary outcome $\omega\in\Omega$ and show that membership in the left-hand event is equivalent to membership in the right-hand event.

Complement of a union.
$\omega\in(A\cup B)^c$ means $\omega\notin A\cup B$. By the definition of union, this means $\omega\notin A$ and $\omega\notin B$. Equivalently, $\omega\in A^c$ and $\omega\in B^c$, so $\omega\in A^c\cap B^c$.

Complement of an intersection.
$\omega\in(A\cap B)^c$ means $\omega\notin A\cap B$. By the definition of intersection, this means it is not the case that $\omega\in A$ and $\omega\in B$. Thus $\omega\notin A$ or $\omega\notin B$, so $\omega\in A^c$ or $\omega\in B^c$, which means $\omega\in A^c\cup B^c$.

The membership equivalences hold for every $\omega\in\Omega$, so each pair of events has exactly the same outcomes. Therefore,

(A\cup B)^c=A^c\cap B^c, \qquad (A\cap B)^c=A^c\cup B^c.

This proof uses only the definitions of complement, union, and intersection; it does not depend on probabilities or on whether the sample space is finite.

Translate the Event Description First

A useful way to remember the laws is to translate the event into words before manipulating the symbols. “$A$ or $B$” means at least one occurs; its complement is “neither $A$ nor $B$.” “$A$ and $B$” means both occur; its complement is “$A$ fails or $B$ fails,” where “or” includes the possibility that both fail.

Event	Equivalent description
$(A\cup B)^c$	Neither $A$ nor $B$ occurs
$(A\cap B)^c$	At least one of $A$ and $B$ does not occur
$A^c\cap B^c$	Both $A$ and $B$ do not occur
$A^c\cup B^c$	At least one of $A$ and $B$ does not occur

Worked Example: “Neither Test Is Positive”

Let $A$ be the event that test 1 is positive, and let $B$ be the event that test 2 is positive. Write the event that neither test is positive in two equivalent ways.

“Neither test is positive” says that test 1 is not positive and test 2 is not positive. In symbols, the event is $A^c\cap B^c$.

It also says that it is not the case that at least one test is positive. “At least one is positive” is $A\cup B$, so the same event is $(A\cup B)^c$. De Morgan’s law confirms the equivalence:

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

The two descriptions are useful in different contexts: one names the failures separately, while the other negates a single compound event.

Worked Example: “Not Both” Is Not “Neither”

Let $A$ mean that a component passes a stress test, and let $B$ mean that it passes a heat test. Express the event that the component does not pass both tests.

“Passes both tests” is $A\cap B$. Negating that entire event gives $(A\cap B)^c$. Applying De Morgan’s law,

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

Thus the component fails at least one of the two tests. It may fail the stress test only, the heat test only, or both. By contrast, “neither test is passed” is $A^c\cap B^c$, which requires failure of both tests. The phrase “not both” must not be mistaken for “neither.”

Finite Collections of Events

The same rule applies to any finite collection of events. For events $A_1,\ldots,A_m$, the complement of their union is the event that none occurs. The complement of their intersection is the event that at least one fails.

Theorem (finite De Morgan’s laws): For a positive integer $m$ and events $A_1,\ldots,A_m\subseteq\Omega$, $$ \left(\bigcup_{i=1}^{m}A_i\right)^c=\bigcap_{i=1}^{m}A_i^c \qquad\text{and}\qquad \left(\bigcap_{i=1}^{m}A_i\right)^c=\bigcup_{i=1}^{m}A_i^c. $$

For the first identity, take any $\omega\in\Omega$. It belongs to $\left(\bigcup_{i=1}^{m}A_i\right)^c$ exactly when it belongs to none of $A_1,\ldots,A_m$. That is, $\omega\notin A_i$ for every $i$, which is equivalent to $\omega\in A_i^c$ for every $i$. This is precisely the condition $\omega\in\bigcap_{i=1}^{m}A_i^c$.

For the second identity, $\omega\in\left(\bigcap_{i=1}^{m}A_i\right)^c$ exactly when it is not true that $\omega$ belongs to every $A_i$. This means there is at least one index $i$ for which $\omega\notin A_i$. Equivalently, $\omega\in A_i^c$ for at least one $i$, or $\omega\in\bigcup_{i=1}^{m}A_i^c$. The equivalences hold for each outcome, proving both identities. $\square$

Worked Example: Simplifying a Three-Condition Event

Let $A_1,A_2,A_3$ be the events that three required checks are passed. A system is rejected if it does not pass all three checks. Express the rejection event without a complement around an intersection.

Passing all three checks is $A_1\cap A_2\cap A_3$. Rejection is its complement. By the finite De Morgan identity,

(A_1\cap A_2\cap A_3)^c =A_1^c\cup A_2^c\cup A_3^c.

So rejection occurs if at least one check is failed. This description includes outcomes where one, two, or all three checks are failed; it does not require all three to fail.

Simplifying a Probability Calculation

De Morgan’s laws can make a probability calculation easier by replacing a complicated verbal event with an equivalent set of outcomes. In a finite model where each outcome $\omega$ has an assigned probability $p(\omega)$, the probability of an event $E$ is calculated by summing $p(\omega)$ over the outcomes in $E$. The identities let us identify $E$ in a different but equivalent form before carrying out that sum.

Worked Example: Calculating the Probability of Neither Event

Consider $\Omega=\{1,2,3,4,5,6\}$, with the following assigned probabilities:

Outcome $\omega$	1	2	3	4	5	6
$p(\omega)$	0.05	0.10	0.15	0.20	0.25	0.25

The weights sum to $0.05+0.10+0.15+0.20+0.25+0.25=1$. Let $A=\{2,4,6\}$ and $B=\{4,5,6\}$. Find the probability that neither event occurs.

The event is $(A\cup B)^c$. First, $A\cup B=\{2,4,5,6\}$, so $(A\cup B)^c=\{1,3\}$. Equivalently, De Morgan’s law gives $A^c\cap B^c$. Indeed, $A^c=\{1,3,5\}$, $B^c=\{1,2,3\}$, and their intersection is $\{1,3\}$.

Summing the assigned probabilities of these two outcomes gives

P\big((A\cup B)^c\big)=p(1)+p(3)=0.05+0.15=0.20.

The set identity provides two routes to the same event; the outcome list verifies that both routes give exactly the outcomes 1 and 3.

Common Mistake: Forgetting to Switch the Operation

A frequent error is to write $(A\cup B)^c=A^c\cup B^c$, or $(A\cap B)^c=A^c\cap B^c$. These expressions do not generally describe the same events as the original complements. Taking a complement changes “or” to “and,” and changes “and” to “or.”

Incorrect: “Not $A$ or $B$” means $A^c\cup B^c$. Correction: The complement of $A\cup B$ is $A^c\cap B^c$: neither event occurs.
Incorrect: “Not $A$ and $B$” means $A^c\cap B^c$. Correction: The complement of $A\cap B$ is $A^c\cup B^c$: at least one event fails.

A quick check is to test an outcome that belongs to one event but not the other. Such an outcome is in $A\cup B$, so it cannot be in $(A\cup B)^c$. But it does belong to at least one of $A^c$ or $B^c$, which shows why $A^c\cup B^c$ cannot be the complement of the union.

Key takeaway: Complementing a union produces an intersection of complements; complementing an intersection produces a union of complements. Translate the event into words, switch the operation, and then check which outcomes the resulting event contains.

Check Your Understanding

Use event descriptions as well as symbols when explaining your answers.

Rewrite $(A\cup B)^c$ using only $A^c$, $B^c$, and an intersection. What does this event mean in words?
A device works only if both $A$ and $B$ occur. Express the event that the device does not work using complements and a union.
Explain the difference between $(A\cap B)^c$ and $A^c\cap B^c$.
Write the complement of $A_1\cup A_2\cup A_3$ as an intersection of complements.
If $\Omega=\{1,2,3,4,5\}$, $A=\{1,3\}$, and $B=\{3,4\}$, list the outcomes in $(A\cup B)^c$ and in $A^c\cap B^c$.

Event	Equivalent description
\((A\cup B)^c\)	Neither \(A\) nor \(B\) occurs
\((A\cap B)^c\)	At least one of \(A\) and \(B\) does not occur
\(A^c\cap B^c\)	Both \(A\) and \(B\) do not occur
\(A^c\cup B^c\)	At least one of \(A\) and \(B\) does not occur

What You'll Learn