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Rules of Inference

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Textbook: Rosen, Discrete Mathematics and Its Applications, 8th ed.  •  Chapter: 1  •  Section: 6

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Textbook Rosen, Discrete Mathematics and Its Applications, 8th ed.
Chapter 1 (The Foundations: Logic and Proofs)
Section 1.6 Rules of Inference
Subsections 1.6.2 Valid Arguments in Propositional Logic; 1.6.3 Rules of Inference for Propositional Logic; 1.6.4 Using Rules of Inference; 1.6.6 Rules of Inference for Quantified Statements
Pages p. 73-83
Course MATH301 (Discrete Mathematics)
Difficulty Intermediate
Time ~35 minutes

Before You Start: Prerequisite Check

đź“‹ Can you do these? (Click to reveal self-test)

Test yourself on these warm-up skills:

  1. Tautology check: Build a truth table for $((p \rightarrow q) \wedge p) \rightarrow q$. Is it true in every row?

    Check

    Yes. The conditional is true in all four rows, so it is a tautology. This is the fact that makes modus ponens a valid rule.

  2. Conditional and its parts: In the statement $p \rightarrow q$, which proposition is the hypothesis and which is the conclusion?

    Check

    $p$ is the hypothesis (also called the antecedent). $q$ is the conclusion (also called the consequent).

  3. Reading a quantifier: What does $\forall x\, P(x)$ assert, and what may you conclude about a specific element $c$ in the domain?

    Check

    It asserts that $P(x)$ holds for every element of the domain. For any particular $c$ in the domain, you may conclude $P(c)$.

If you struggled:


Try This First

✏️ A short argument to test before any formula (click to open)

Read these three sentences as a single argument.

If the chemistry homework is graded tonight, then the answer key is finalized. The chemistry homework is graded tonight.

Predict: does the truth of those two sentences force any further sentence to be true? Write down the sentence you think must follow before reading on.

Compare your prediction

The sentence that must follow is “the answer key is finalized.” If the first sentence is true (graded forces finalized) and the second is true (it is graded), then there is no way for “the answer key is finalized” to be false. You just used a rule of inference, the one named modus ponens, without naming it.

A Sentence That Must Follow

Mathematical reasoning is a sequence of small forced steps. At each step you already hold some true statements, and you ask one question: which new statement is forced to be true by what you already have?

A rule of inference is one such forced step, written as a template. The template has a short list of statements above a line (the things you already hold) and one statement below the line (the thing that is forced). Reading from the warm-up argument:

$$ \begin{array}{l} p \rightarrow q \\ p \\ \hline \therefore\ q \end{array} $$

The horizontal line reads “therefore.” The two statements above the line are the premises; the statement below is the conclusion. The claim of the rule is that any time the premises are all true, the conclusion cannot be false.

Each rule earns its place for one reason: the conditional that says “if all the premises are true then the conclusion is true” is a tautology, a statement true in every row of its truth table. So the validity of a rule is checked once, in advance, by a truth table, and then the rule is reused without rechecking. Tautologies are what make the rules trustworthy.

Prerequisite Hub

graph LR
    subgraph Builds_On["Builds On"]
        A["Propositional<br/>Equivalences"]
        B["Predicates and<br/>Quantifiers"]
        G["Nested<br/>Quantifiers"]
    end

    subgraph ThisSkill["This Skill"]
        C["Rules of<br/>Inference"]
    end

    subgraph Unlocks
        D["Introduction<br/>to Proofs"]
    end

    A --> C
    B --> C
    G --> C
    C --> D

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

    click C "rules-of-inference.html"
    click D "../ch1-sec7/introduction-to-proofs.html"

Builds on:

Skill Why it helps
dm-propositional-equivalences Each rule of inference is valid exactly because a matching conditional is a tautology, so checking tautologies and equivalences is the tool that justifies every rule (strong prerequisite).
dm-predicates-and-quantifiers Universal and existential instantiation and generalization act directly on $\forall$ and $\exists$ statements (strong prerequisite).
dm-nested-quantifiers Arguments that combine several quantified premises reuse the reading order from nested quantifiers (helpful, not required).

Leads into:

Skill What it adds
dm-introduction-to-proofs Direct proofs, proof by contraposition, and proof by contradiction are built by chaining these rules into longer arguments about specific mathematical objects.

Cross-course prerequisites: none.


Official Definitions

Each definition below is quoted from Rosen, 8th edition, Section 1.6, with the page citation. These are the canonical statements to put on the board and on a study sheet.

Argument, premises, conclusion, and validity

By an argument, we mean a sequence of statements that end with a conclusion. By valid, we mean that the conclusion, or final statement of the argument, must follow from the truth of the preceding statements, or premises, of the argument. That is, an argument is valid if and only if it is impossible for all the premises to be true and the conclusion to be false.

Rosen 8e, Section 1.6, subsection 1.6.1 Introduction, p. 73.

Two cautions sit inside this definition. First, validity is about the form of the argument, not about whether the premises happen to be true. An argument can be valid while resting on a false premise; in that case validity guarantees nothing about the conclusion in the real world. Second, “must follow” is the strong word: there is to be no row of the combined truth table in which every premise is true and the conclusion is false.

Rule of inference

Rules of inference are templates for constructing valid arguments; they are the basic tools for establishing the truth of statements. Each rule of inference can be established using a tautology.

Rosen 8e, Section 1.6, p. 73-74 (verbatim discussion).

The phrase “established using a tautology” is the working definition for a study sheet. To show that a proposed rule is legitimate, form the conditional whose hypothesis is the conjunction of the premises and whose conclusion is the rule’s conclusion, then confirm by truth table that this conditional is a tautology.

Modus ponens (law of detachment)

The rule of inference modus ponens, or the law of detachment, is based on the tautology $((p \rightarrow q) \wedge p) \rightarrow q$. It is written: premise $p \rightarrow q$, premise $p$, therefore conclusion $q$.

Rosen 8e, Section 1.6, p. 74-75 and Table 1 Rules of Inference.

The name translates as “mode that affirms”: the second premise affirms the hypothesis $p$, which detaches the conclusion $q$ from the conditional.


Table 1: Rules of Inference for Propositional Logic

These are the eight rules of Section 1.6 for propositional logic, each with the tautology that establishes it.

Rule of inference Premises and conclusion Tautology that establishes it
Modus ponens $p \rightarrow q,\ p\ \therefore q$ $((p \rightarrow q) \wedge p) \rightarrow q$
Modus tollens $p \rightarrow q,\ \neg q\ \therefore \neg p$ $((p \rightarrow q) \wedge \neg q) \rightarrow \neg p$
Hypothetical syllogism $p \rightarrow q,\ q \rightarrow r\ \therefore p \rightarrow r$ $((p \rightarrow q) \wedge (q \rightarrow r)) \rightarrow (p \rightarrow r)$
Disjunctive syllogism $p \vee q,\ \neg p\ \therefore q$ $((p \vee q) \wedge \neg p) \rightarrow q$
Addition $p\ \therefore p \vee q$ $p \rightarrow (p \vee q)$
Simplification $p \wedge q\ \therefore p$ $(p \wedge q) \rightarrow p$
Conjunction $p,\ q\ \therefore p \wedge q$ $((p) \wedge (q)) \rightarrow (p \wedge q)$
Resolution $p \vee q,\ \neg p \vee r\ \therefore q \vee r$ $((p \vee q) \wedge (\neg p \vee r)) \rightarrow (q \vee r)$

Rosen 8e, Section 1.6, Table 1 Rules of Inference, p. 74-77.

One way to read each rule: the conclusion is forced. Another way: the rule throws away information. Simplification, for instance, takes the strong premise $p \wedge q$ and keeps only $p$; the rule is safe precisely because it concludes something weaker than what it was given. Asking, for each rule, “is the conclusion stronger or weaker than the premises” is a quick check on whether a step is legitimate.


Table 2: Rules of Inference for Quantified Statements

When arguments involve $\forall$ and $\exists$, four further rules move between a quantified statement and a statement about a particular element. Here $c$ denotes an element of the domain.

Rule From Conclude Condition on $c$
Universal instantiation $\forall x\, P(x)$ $P(c)$ $c$ is any element of the domain
Universal generalization $P(c)$ $\forall x\, P(x)$ $c$ is an arbitrary element, chosen with no special property
Existential instantiation $\exists x\, P(x)$ $P(c)$ $c$ is a new name for some element that makes $P$ true
Existential generalization $P(c)$ $\exists x\, P(x)$ $c$ is some specific element of the domain

Rosen 8e, Section 1.6, Table 2 Rules of Inference for Quantified Statements, subsection 1.6.6, p. 81-83.

The two conditions in the right column are where most errors occur. Universal generalization is legitimate only when $c$ was chosen with no special assumptions, so that whatever is proved about $c$ holds for every element. Existential instantiation must introduce a fresh name, because $\exists x\, P(x)$ promises that some element works without telling you which one.


Worked Examples

Example 1: A single step of modus tollens

Premises: “If the autograder passes, then the build compiled” and “the build did not compile.” What follows?

Predict first. The second premise denies the conclusion of the conditional. Predict whether you can deny its hypothesis.

Set up the symbols. Let $p$ be “the autograder passes” and $q$ be “the build compiled.” The premises are $p \rightarrow q$ and $\neg q$.

Apply the rule. This matches modus tollens:

$$ \begin{array}{l} p \rightarrow q \\ \neg q \\ \hline \therefore\ \neg p \end{array} $$

Conclusion. $\neg p$: “the autograder did not pass.” The prediction is confirmed; denying the conclusion of a conditional lets you deny its hypothesis.

Example 2: A chain of rules

Show that the conclusion $r$ follows from the premises $p \vee q$, $\neg p$, and $q \rightarrow r$.

Step Statement Justification
1 $p \vee q$ Premise
2 $\neg p$ Premise
3 $q$ Disjunctive syllogism, from steps 1 and 2
4 $q \rightarrow r$ Premise
5 $r$ Modus ponens, from steps 4 and 3

Each line is forced by lines already established, so the argument is valid and the conclusion $r$ holds.

Why might this be valid as a whole? Each individual step rests on a tautology, and a chain of forced steps is itself forced: if there were a truth assignment making every premise true and $r$ false, it would have to break one of the five lines, but no line can break. Convince a classmate by asking them to find a single row of the truth table for $p$, $q$, $r$ in which all three premises are true and $r$ is false; there is none.

Example 3: An argument with quantifiers

Domain: all students in the section. Premises: “Every student who submitted the lab earned points” and “Avery submitted the lab.” Conclude that Avery earned points.

Let $S(x)$ be “$x$ submitted the lab” and $E(x)$ be “$x$ earned points.”

Step Statement Justification
1 $\forall x\,(S(x) \rightarrow E(x))$ Premise
2 $S(\text{Avery}) \rightarrow E(\text{Avery})$ Universal instantiation, from step 1
3 $S(\text{Avery})$ Premise
4 $E(\text{Avery})$ Modus ponens, from steps 2 and 3

Universal instantiation brings the general rule down to the one element named in the second premise, and then modus ponens finishes the step. This pattern, universal instantiation followed by modus ponens, is so common that it has its own name, universal modus ponens.


Common Misconceptions

Common misconception

a valid argument has a true conclusion. Validity is about form, not about the real-world truth of the premises. Consider the valid modus ponens argument with premises “if $2$ is odd then $3$ is even” and “$2$ is odd.” The form is correct, so the argument is valid, yet the conclusion “$3$ is even” is false because the second premise is false. Validity guarantees a true conclusion only when every premise is also true. Checking validity and checking the premises are two separate jobs.

Common misconception

affirming the conclusion of a conditional lets you affirm its hypothesis. From $p \rightarrow q$ and $q$, the statement $p$ does not follow. Test it on a small case: let $p$ be “it rained last night” and $q$ be “the sidewalk is wet.” The sidewalk being wet ($q$ true) does not force rain, because a sprinkler could have wet it. The truth table for $((p \rightarrow q) \wedge q) \rightarrow p$ has a row ($p$ false, $q$ true) where the hypothesis is true and the conclusion false, so it is not a tautology and not a rule. This error pattern has the traditional name “affirming the consequent.”

Common misconception

denying the hypothesis of a conditional lets you deny its conclusion. From $p \rightarrow q$ and $\neg p$, the statement $\neg q$ does not follow. With the same rain example, “it did not rain” ($\neg p$) does not force “the sidewalk is dry” ($\neg q$), again because of the sprinkler. The conditional $((p \rightarrow q) \wedge \neg p) \rightarrow \neg q$ fails in the row $p$ false, $q$ true. The legitimate move that uses $\neg q$ is modus tollens, which concludes $\neg p$, not the reverse. This error has the name “denying the antecedent.”


Practice Problems

Level 1 Name the Rule

Which rule of inference has premises $p \rightarrow q$ and $p$, with conclusion $q$?

Thought Process

The second premise affirms the hypothesis of the conditional, and the conclusion is the consequent. Match this against the rows of Table 1.

Show Answer

Modus ponens (the law of detachment), established by the tautology $((p \rightarrow q) \wedge p) \rightarrow q$.

Level 2 Single-Step Application

Premises: $\neg q$ and $p \rightarrow q$. Which rule applies, and what is the conclusion?

Thought Process

The premise $\neg q$ denies the consequent of the conditional. Find the rule in Table 1 that uses a negated consequent.

Show Answer

Modus tollens. From $p \rightarrow q$ and $\neg q$, conclude $\neg p$. It is established by the tautology $((p \rightarrow q) \wedge \neg q) \rightarrow \neg p$.

Level 2 Spot the Invalid Step

A student writes: “From $p \rightarrow q$ and $q$, I conclude $p$.” Is this step valid?

(A) Yes, by modus ponens.

(B) Yes, by modus tollens.

(C) No, this is affirming the consequent, which is not a rule of inference.

(D) No, but it becomes valid if $p$ and $q$ are both true.

Thought Process

Check whether $((p \rightarrow q) \wedge q) \rightarrow p$ is a tautology. Test the row $p$ false, $q$ true.

Show Answer

Answer: (C). The conditional $((p \rightarrow q) \wedge q) \rightarrow p$ is false in the row $p = \text{F}$, $q = \text{T}$, so it is not a tautology and not a valid rule. The error is named affirming the consequent.

  • (A) is wrong: modus ponens needs $p$ as a premise, not $q$.
  • (B) is wrong: modus tollens needs $\neg q$.
  • (D) is wrong: validity is a property of the argument form, not of one truth assignment.
Level 3 Two-Step Chain

Show that $r$ follows from the premises $p$, $p \rightarrow q$, and $q \rightarrow r$.

Thought Process

Use $p$ with the first conditional to reach $q$, then $q$ with the second conditional to reach $r$. Each arrow is one modus ponens.

Show Answer
Step Statement Justification
1 $p$ Premise
2 $p \rightarrow q$ Premise
3 $q$ Modus ponens, from steps 2 and 1
4 $q \rightarrow r$ Premise
5 $r$ Modus ponens, from steps 4 and 3

The two conditionals could also be combined first by hypothetical syllogism into $p \rightarrow r$, then closed with one modus ponens against $p$. Both paths reach $r$.

Level 4 Disjunction and Conditionals

Show that $s$ follows from the premises $p \vee q$, $\neg p$, $q \rightarrow r$, and $r \rightarrow s$.

Thought Process

The first two premises strip the disjunction down to $q$. Then chain the two conditionals to climb from $q$ to $s$.

Show Answer
Step Statement Justification
1 $p \vee q$ Premise
2 $\neg p$ Premise
3 $q$ Disjunctive syllogism, from steps 1 and 2
4 $q \rightarrow r$ Premise
5 $r$ Modus ponens, from steps 4 and 3
6 $r \rightarrow s$ Premise
7 $s$ Modus ponens, from steps 6 and 5
Level 5 Quantified Argument

Domain: all integers. Premises: “Every integer that is a multiple of $4$ is even” and “$12$ is a multiple of $4$.” Conclude that $12$ is even, naming each rule used.

Thought Process

Translate the general premise as a universal conditional. Bring it down to the element $12$ with universal instantiation, then close with modus ponens.

Show Answer

Let $M(x)$ be “$x$ is a multiple of $4$” and $V(x)$ be “$x$ is even.”

Step Statement Justification
1 $\forall x\,(M(x) \rightarrow V(x))$ Premise
2 $M(12) \rightarrow V(12)$ Universal instantiation, from step 1
3 $M(12)$ Premise
4 $V(12)$ Modus ponens, from steps 2 and 3

Therefore $12$ is even. Steps 2 through 4 together are an instance of universal modus ponens.


Mastery Checklist

Novice (Level 1-2):

Competent (Level 3-4):

Proficient (Level 5):


Connections

Looking back:

Looking ahead:

Real-world connections:


Resources

Resource Reference
Primary text Rosen, Discrete Mathematics and Its Applications, 8th ed., Section 1.6 Rules of Inference (p. 73-83), including Table 1 Rules of Inference and Table 2 for quantified statements.
Open companion Levin, Discrete Mathematics: An Open Introduction (3rd ed.), Rules of Inference and Proofs: https://discrete.openmathbooks.org/dmoi3/sec_seq-arithmetic.html
Next tree node Introduction to Proofs (the section this unlocks).


Last updated: 2026-06-16