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Predicates and Quantifiers

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Quick Reference

Field Value
Textbook Rosen, Discrete Mathematics and Its Applications, 8th edition
Chapter 1 (The Foundations: Logic and Proofs)
Section 1.4 (Predicates and Quantifiers)
Subsections 1.4.2 Predicates; 1.4.3 Quantifiers; 1.4.6 Negating Quantified Expressions
Pages 40-59
Tables referenced Table 1 (quantifier truth conditions); Table 2 (De Morgan’s Laws for Quantifiers, p.55)

Before You Start: Prerequisite Check

📋 Can you do these? (Click to reveal self-test)

Test yourself on these prerequisite skills:

  1. Truth value of a proposition: Is the statement “$5$ is greater than $3$” true or false?

    Check

    True. A proposition is a declarative statement that is either true or false, and this one is true.

  2. Negation of a proposition: Write the negation of “$7$ is even”.

    Check

    “$7$ is not even” (equivalently, “$7$ is odd”). The negation flips the truth value.

  3. Membership in a set: If the domain is the set $\{1, 2, 3, 4\}$, which elements are greater than $2$?

    Check

    $3$ and $4$. The elements $1$ and $2$ are not greater than $2$.

If you struggled:


Try This First

Read each row of the table out loud and decide, before any formula appears, whether the sentence is true or false. The variable $x$ ranges over the whole numbers $1, 2, 3, 4, 5$.

Sentence True or false?
“$x$ is greater than $3$” when $x = 5$ ?
“$x$ is greater than $3$” when $x = 2$ ?
“Every value of $x$ is greater than $3$” ?
“Some value of $x$ is greater than $3$” ?
What did you notice?

The first two rows have different answers (true, then false) even though the sentence is identical. A value had to be chosen before the sentence could be judged. The last two rows judge the sentence across the entire collection at once: “every” is false (because $2$ fails), and “some” is true (because $5$ succeeds).

That difference is the whole section. The bare sentence “$x$ is greater than $3$” is a predicate. The words “every” and “some” are quantifiers, and they turn a predicate into a statement that has a single truth value over a whole domain.


Why Predicates and Quantifiers Matter

A proposition such as “$5$ is greater than $3$” already has a fixed truth value, so propositional logic cannot say anything about a variable. Mathematics is full of statements about variables: “every integer has a successor”, “some real number squares to $2$”, “for each input there is an output”. These statements cannot be written with propositional connectives alone.

A predicate is a process: it takes an input and gives back a proposition. Feed it the value $5$ and the process returns the true proposition “$5 > 3$”; feed it the value $2$ and the process returns the false proposition “$2 > 3$”. A quantifier then collapses all of those individual propositions into one claim about the whole domain. The universal quantifier asks whether the predicate holds for every input. The existential quantifier asks whether it holds for at least one input.

This pairing (a predicate that responds to inputs, and a quantifier that sweeps the domain) is the language used to state every definition, theorem, and proof in the rest of the course.


Prerequisite Hub

Predicates and quantifiers belong to the logic strand of discrete mathematics. The map shows what to master first and what this skill opens up next.

graph LR
    subgraph Builds On
        A["Propositional<br/>Logic"]
        B["Propositional<br/>Equivalences"]
        C["Set-Builder and<br/>Roster Notation"]
    end

    subgraph ThisSkill["This Skill"]
        D["Predicates and<br/>Quantifiers"]
    end

    subgraph Unlocks
        E["Nested<br/>Quantifiers"]
        F["Rules of<br/>Inference"]
    end

    A --> D
    B --> D
    C --> D
    D --> E
    D --> F

    style D fill:#d1fae5,stroke:#a565f0,stroke-width:3px

    click E "../ch1-sec5/nested-quantifiers.html"

Builds on (master these first):

Skill Skill ID Why it is needed
Propositional Logic dm-propositional-logic A predicate becomes a proposition once a value is supplied, so propositions and their truth values come first
Propositional Equivalences dm-propositional-equivalences The De Morgan laws for quantifiers extend the propositional De Morgan equivalences
Set-Builder and Roster Notation dm-pre-set-builder-and-roster-notation Every quantifier needs a domain of discourse, which is a set

Unlocks (these come next):

Skill Skill ID What it adds
Nested Quantifiers dm-nested-quantifiers Two or more quantifiers in one statement, where order changes meaning
Rules of Inference dm-rules-of-inference Universal and existential instantiation and generalization in formal proofs

Cross-course prerequisites: none.


Official Definitions

The definitions below are quoted from Rosen 8th edition, Section 1.4, on the pages noted.

Predicate (propositional function)

A statement such as “$x$ is greater than $3$” has two parts: the variable $x$, the subject of the statement, and the predicate “is greater than $3$”, which refers to a property the subject can have. The statement is denoted $P(x)$, where $P$ is the predicate and $x$ is the variable. The statement $P(x)$ is the value of the propositional function $P$ at $x$; once a value is assigned to $x$, $P(x)$ becomes a proposition with a truth value.

Cite: Rosen 8e, Section 1.4, subsection 1.4.2 Predicates, p.41.

Domain of discourse (domain)

A statement that is true for all values of a variable in a particular domain, called the domain of discourse (or the universe of discourse), is expressed using universal quantification. The domain specifies the possible values of the variable and must always be specified when a quantifier is used.

Cite: Rosen 8e, Section 1.4, p.44.

Universal quantifier (universal quantification)

The universal quantification of $P(x)$ is the statement “$P(x)$ for all values of $x$ in the domain.” The notation $\forall x\, P(x)$ denotes the universal quantification of $P(x)$. Here $\forall$ is called the universal quantifier. An element for which $P(x)$ is false is called a counterexample to $\forall x\, P(x)$.

Cite: Rosen 8e, Section 1.4, Definition 1, p.44.

Existential quantifier (existential quantification)

The existential quantification of $P(x)$ is the proposition “There exists an element $x$ in the domain such that $P(x)$.” The notation $\exists x\, P(x)$ denotes the existential quantification of $P(x)$. Here $\exists$ is called the existential quantifier.

Cite: Rosen 8e, Section 1.4, Definition 2, p.45.

Truth conditions at a glance

Statement True when False when
$\forall x\, P(x)$ $P(x)$ is true for every $x$ in the domain there is at least one $x$ for which $P(x)$ is false (a counterexample)
$\exists x\, P(x)$ there is at least one $x$ for which $P(x)$ is true $P(x)$ is false for every $x$ in the domain

Truth conditions follow Table 1, Rosen 8e, Section 1.4.

A reminder that quantification depends on the domain: $\forall x\, (x > 3)$ is false over the integers (take $x = 0$) and true over the domain $\{4, 5, 6\}$. State the domain before judging a quantified statement.


Key Theorem: De Morgan’s Laws for Quantifiers

De Morgan’s Laws for Quantifiers. Negating a quantified statement swaps the quantifier and pushes the negation onto the predicate:

$$\boxed{\;\lnot\,\forall x\, P(x) \;\equiv\; \exists x\, \lnot P(x) \qquad\text{and}\qquad \lnot\,\exists x\, Q(x) \;\equiv\; \forall x\, \lnot Q(x)\;}$$

Statement: $\lnot\,\forall x\, P(x)$ is logically equivalent to $\exists x\, \lnot P(x)$, and $\lnot\,\exists x\, Q(x)$ is logically equivalent to $\forall x\, \lnot Q(x)$.

Cite: Rosen 8e, Section 1.4, Table 2 (De Morgan’s Laws for Quantifiers), p.55.

Reading the laws in words. The statement “not everything has the property” means the same as “something fails to have the property.” The statement “nothing has the property” means the same as “everything fails to have the property.” Each law turns a $\forall$ into an $\exists$ (or the reverse) and moves the negation inside.


Worked Examples

Example 1: A predicate is a process, not yet a proposition

Let the domain be the integers, and let $P(x)$ be the predicate “$x^2 \geq x$.”

Predict first. Before computing, guess whether $P(x)$ has a single truth value as written. Then test it at two inputs.

Check.

The predicate $P(x)$ on its own has no fixed truth value; it returns a different proposition for each input. Only after an input is supplied does it become a proposition. The prediction that “$P(x)$ already has one truth value” is the misconception this example targets.

Example 2: Reading $\forall$ and finding a counterexample

Let the domain be the integers $\{1, 2, 3, 4, 5\}$, and let $Q(x)$ be “$x$ is greater than $3$.” Decide the truth value of $\forall x\, Q(x)$.

Predict first. The claim says every listed integer exceeds $3$. Scan the list and predict true or false.

Check. Evaluate $Q$ at each element:

$x$ $Q(x):\ x > 3$ Truth
$1$ $1 > 3$ false
$2$ $2 > 3$ false
$3$ $3 > 3$ false
$4$ $4 > 3$ true
$5$ $5 > 3$ true

The value $x = 1$ makes $Q(x)$ false, so $x = 1$ is a counterexample. A single counterexample is enough to make $\forall x\, Q(x)$ false. Notice that $x = 2$ and $x = 3$ are also counterexamples; the universal statement fails the moment the first one appears.

Example 3: Reading $\exists$

Keep the same domain $\{1, 2, 3, 4, 5\}$ and the same predicate $Q(x): x > 3$. Decide the truth value of $\exists x\, Q(x)$.

Predict first. This claim asks only for one success. Predict before checking.

Check. From the table in Example 2, $Q(4)$ is true. One witness is enough, so $\exists x\, Q(x)$ is true. The existential statement needs only a single value that works, while the universal statement needed every value to work.

Example 4: Negating a universal statement with De Morgan

Let the domain be all students in a class, and let $S(x)$ be “$x$ has taken a calculus course.” Consider the statement $\forall x\, S(x)$ (“every student has taken a calculus course”). Write its negation as an existential statement.

Predict first. In plain words, what would make “every student has taken calculus” false? Predict the English sentence, then convert it to symbols.

Check. The statement is false exactly when at least one student has not taken calculus. By De Morgan for quantifiers:

$$\lnot\,\forall x\, S(x) \;\equiv\; \exists x\, \lnot S(x).$$

In words: “There is a student who has not taken a calculus course.” The negation turned $\forall$ into $\exists$ and pushed the negation onto the predicate $S(x)$, giving $\lnot S(x)$.

Example 5: Negating an existential statement, two ways

Let the domain be the real numbers, and let $R(x)$ be “$x^2 = -1$.” Consider $\exists x\, R(x)$ (“some real number squares to $-1$”). Negate it.

One way to see this. Apply De Morgan for quantifiers directly:

$$\lnot\,\exists x\, R(x) \;\equiv\; \forall x\, \lnot R(x),$$

which reads “for every real number $x$, it is not the case that $x^2 = -1$”, that is “every real number has $x^2 \neq -1$.”

Another way to see this. Reason about the original statement first. Over the reals, $x^2 \geq 0$ for every $x$, so no real number squares to $-1$; the existential statement $\exists x\, R(x)$ is false, and its negation is therefore true. The De Morgan form $\forall x\, (x^2 \neq -1)$ agrees: it is a true statement over the reals. Which path feels clearer to you, the symbolic rewrite or the reasoning about values? Both reach the same equivalent statement.


Common Misconceptions

Common misconception

a predicate $P(x)$ already has a truth value before any input is supplied. A predicate is a process that returns a proposition once given a value, not a proposition itself. The expression “$x > 3$” is true at $x = 5$ and false at $x = 2$, so it cannot have a single truth value as written. Asking “is $P(x)$ true?” without naming the input, or naming the domain and quantifier, has no answer. This is the input-output confusion: treating the rule as if it were already one of its outputs.

Common misconception

a universal statement and an existential statement carry the same evidence. The two quantifiers demand opposite amounts of evidence. To prove $\forall x\, P(x)$ true, every value in the domain must satisfy $P$; one counterexample destroys it. To prove $\exists x\, P(x)$ true, one witness is enough; showing that some particular value fails proves nothing about the existential claim. Treating a quantifier as a fixed all-or-nothing label, rather than a condition that sweeps the whole domain, leads to the wrong evidence being gathered.

Common misconception

to negate $\forall x\, P(x)$, keep the $\forall$ and just negate the predicate. The negation $\lnot\,\forall x\, P(x)$ is not $\forall x\, \lnot P(x)$. Test it on the domain $\{1, 2, 3\}$ with $P(x): x > 0$. The original $\forall x\, P(x)$ is true, so its negation is false. The wrong form $\forall x\, \lnot P(x)$ says “every element is not positive”, which is also false, so the two agree here by accident. Now take $P(x): x > 2$ on the same domain. The original is false (so its negation is true), but $\forall x\, \lnot P(x)$ says “every element is at most $2$”, which is false. The wrong form gives false where the correct negation $\exists x\, \lnot P(x)$ gives true. The quantifier must flip.


Practice Problems

Level 1 Evaluating a Predicate at a Value

Let $P(x)$ be the predicate “$x + 1 = 4$” with the integers as the domain. Find the truth value of $P(3)$ and of $P(5)$.

Thought Process

Substitute each value into the predicate and decide whether the resulting proposition is true.

Show Answer

$P(3)$ is “$3 + 1 = 4$”, that is “$4 = 4$”, which is true.

$P(5)$ is “$5 + 1 = 4$”, that is “$6 = 4$”, which is false.

Level 2 Truth Value of a Universal Statement

The domain is $\{2, 4, 6, 8\}$. Let $E(x)$ be “$x$ is even.” Find the truth value of $\forall x\, E(x)$, and give a counterexample if one exists.

Thought Process

A universal statement is true only when every element satisfies the predicate. Check each element; a single failure is a counterexample.

Show Answer

Every element of $\{2, 4, 6, 8\}$ is even, so $\forall x\, E(x)$ is true. There is no counterexample, because no element fails the predicate.

Level 3 Truth Value of an Existential Statement

The domain is all real numbers. Let $T(x)$ be “$x^2 = 2$.” Find the truth value of $\exists x\, T(x)$, and name a witness if the statement is true.

Thought Process

An existential statement is true when at least one value in the domain satisfies the predicate. Ask whether any real number squares to $2$.

Show Answer

The statement is true. A witness is $x = \sqrt{2}$, because $(\sqrt{2})^2 = 2$. (The value $x = -\sqrt{2}$ is a second witness, but one is enough.)

Level 4 Negating a Quantified Statement

The domain is all integers. Let $D(x)$ be “$x$ is divisible by $5$.” Write the negation of $\forall x\, D(x)$ as an existential statement, both in symbols and in words.

Thought Process

Apply De Morgan for quantifiers: the negation of a universal is an existential with the predicate negated. Then translate $\lnot D(x)$ into plain words.

Show Answer

By De Morgan for quantifiers,

$$\lnot\,\forall x\, D(x) \;\equiv\; \exists x\, \lnot D(x).$$

In words: “There exists an integer that is not divisible by $5$.” (This negation is true; for example, $7$ is not divisible by $5$.)

Level 5 Negate, Simplify, and Judge

The domain is all real numbers. Consider the statement $\exists x\, (x^2 < 0)$. Write its negation using De Morgan for quantifiers, simplify the negated predicate, and state the truth value of both the original statement and its negation.

Thought Process

First negate with De Morgan, turning $\exists$ into $\forall$ and negating the predicate. Then simplify $\lnot(x^2 < 0)$ to a clean inequality. Finally, reason about whether each statement holds over the reals.

Show Answer

Let $P(x)$ be “$x^2 < 0$.” By De Morgan for quantifiers,

$$\lnot\,\exists x\, P(x) \;\equiv\; \forall x\, \lnot P(x).$$

Simplify the negated predicate: $\lnot(x^2 < 0)$ is “$x^2 \geq 0$.” So the negation is

$$\forall x\, (x^2 \geq 0).$$

Truth values. For every real number $x$, the square $x^2$ is at least $0$, so no real number satisfies $x^2 < 0$. The original statement $\exists x\, (x^2 < 0)$ is false, and its negation $\forall x\, (x^2 \geq 0)$ is true. The two truth values are opposite, as a statement and its negation must be.


Mastery Checklist

Novice (Level 1-2):

Competent (Level 3-4):

Proficient (Level 5):


Mental Model

The domain sweep. Picture the domain as a row of boxes, one box per element. A predicate $P$ is a stamp that prints “true” or “false” on a box once a value is dropped in. The universal quantifier $\forall$ walks the whole row and reports “true” only if no box came back false. The existential quantifier $\exists$ walks the row and reports “true” the moment one box comes back true. Negation flips the report and switches which quantifier is doing the walking: “not every box is true” is the same as “some box is false.”


Connections

Looking back:

Looking ahead:

Real-world connections:


Resources

Resource What it covers Link
Rosen, Discrete Mathematics and Its Applications, 8th ed., Section 1.4 Predicates and quantifiers, including Table 1 (quantifier truth conditions) and Table 2 (De Morgan for quantifiers), pages 40-59 Course textbook, Section 1.4
Levin, Discrete Mathematics: An Open Introduction Open-access companion treatment of predicates and quantifiers discrete.openmathbooks.org
Tree node: Nested Quantifiers The next section this skill unlocks nested-quantifiers.md


Last updated: 2026-06-16