Truth Value of a Statement

In propositional logic, the truth value of a statement refers to whether that statement is true or false based on its correspondence with reality or formal mathematical definitions. In essence, determining the truth value of a statement allows us to evaluate whether its assertion actually holds.

Classical logic relies on the principle of bivalence, which establishes that there are only two possible truth values: true (typically denoted by T or 1) and false (denoted by F or 0). Because of this fundamental property, we refer to it as a bivalent logic system.

Truth Value of a Simple Statement

A simple statement, also known as an atomic statement or atomic proposition, is a declarative sentence that cannot be broken down into smaller constituent statements. Finding the truth value of a simple statement depends directly on its relationship to verifiable facts or, in formal mathematical systems, the underlying axiomatic framework.

Consider the following examples:

• "Paris is the capital of France" is a true statement, as it aligns with verifiable geographic and political facts.
• "The Sun revolves around the Earth" is a false statement, since it contradicts established astronomical science.
• "15 is a prime number" is false, because 15 is divisible by 3 and 5.
• "Every even number is divisible by 2" is true, by definition of an even integer.

In each case, we determine whether the statement is true or false by evaluating it against accepted definitions or factual reality. This fundamental process forms the foundation for analyzing more complex, multi-part logical assertions.

The truth value of a statement p is often denoted as v(p) or T(p). Thus, if p is true, then v(p) = T, and if p is false, v(p) = F.

Truth Value of a Compound Statement

In mathematical reasoning, we frequently combine simple statements using logical connectives to create compound statements (also referred to as molecular statements). These logical operators allow us to construct complex formulas that establish relationships between different conditions.

The standard logical operators include: the negation (¬ or ~), which reverses the truth value; the conjunction (∧), representing the logical "and"; the disjunction (∨), representing the inclusive "or"; the conditional statement (→), representing "if... then"; and the biconditional statement (↔), representing "if and only if."

Learn more about logical connectives

When finding the truth value of a compound statement, the outcome depends entirely on two factors: the truth values assigned to each component statement and the specific logical operators connecting them.

To systematically determine all possible truth values of a compound proposition, we construct a truth table. A truth table organizes every possible combination of inputs from the atomic statements and computes the overall truth value of the compound expression for each case.

Below is the reference truth table summarizing the standard rules for each primary logical connective:

p	q	¬p	p ∧ q	p ∨ q	p → q	p ↔ q
T	T	F	T	T	T	T
T	F	F	F	T	F	F
F	T	T	F	T	T	F
F	F	T	F	F	T	T

This reference chart serves as the fundamental baseline for evaluating multi-step compound expressions involving multiple connectives.

How to Determine the Truth Value

In practice, we can easily find the truth value of a compound statement without building a full table whenever the individual truth values of its components are already known.

For instance, the statement "the number 4 is even and positive" is true because both component statements are true and joined by a conjunction. Similarly, "If 8 is greater than 10, then 10 is less than 8" is true because the hypothesis (antecedent) is false, which automatically makes any conditional statement true regardless of the conclusion's truth value.

However, when evaluating statement variables or expressions with unknown truth assignments, we must construct a complete truth table to explore every logical possibility.

Example 1

Logical expression: (p ∨ q) → (p ∧ ¬q)

To determine the truth value of this compound statement across all scenarios, we build its truth table:

p	q	p ∨ q	¬q	p ∧ ¬q	(p ∨ q) → (p ∧ ¬q)
T	T	T	F	F	F
T	F	T	T	T	T
F	T	T	F	F	F
F	F	F	T	F	T

If you need a refresher on setting up the rows and columns, check out this guide:

How to construct truth tables step by step

Let's analyze what each row of the table reveals:

• In the first row, where both variables are true, the hypothesis (p ∨ q) is true, but the conclusion (p ∧ ¬q) is false. A true hypothesis leading to a false conclusion results in a false conditional statement.
• In the second row, where p is true and q is false, both the hypothesis and conclusion are true, making the conditional statement true.
• In the third row, where p is false and q is true, the hypothesis is true but the conclusion is false, yielding a final value of false.
• Finally, in the fourth row where both p and q are false, the hypothesis is false. By definition, a conditional statement with a false hypothesis is always true, regardless of the conclusion.

Example 2

Logical expression: (p ∧ q) → (p ∨ q)

The truth table for this statement is:

p	q	p ∧ q	p ∨ q	(p ∧ q) → (p ∨ q)
T	T	T	T	T
T	F	F	T	T
F	T	F	T	T
F	F	F	F	T

Notice that regardless of the initial truth values assigned to p and q, the final result of the conditional is always true. This is intuitively sound: if both statements are true (conjunction), then it is necessarily true that at least one of them is true (disjunction). In every case where the conjunction is false, the implication remains true due to a false hypothesis.

Example 3

Logical expression: p ↔ ¬p

The truth table for this statement is:

p	¬p	p ↔ ¬p
T	F	F
F	T	F

A biconditional statement requires both components to share the exact same truth value to be true. However, a statement p and its negation ¬p always hold opposite values. Therefore, this biconditional statement is false under all conditions.

Classifying Statements by Truth Value

Examining the final column of a compound statement's truth table allows us to classify it into one of three core categories:

• Tautology: a statement that is true for every possible combination of truth values of its individual variables. Tautologies represent foundational logical laws. Example 2, (p ∧ q) → (p ∨ q), is a tautology because it evaluates to true across all rows.
• Contradiction: a statement that is false for every possible combination of truth values of its components. It represents a complete logical impossibility. Example 3, p ↔ ¬p, is a contradiction because a statement can never be logically equivalent to its own negation.
• Contingency: a statement that evaluates to true under some truth assignments and false under others. Most natural-language assertions fall into this category, as their validity depends on specific conditions. Example 1, (p ∨ q) → (p ∧ ¬q), is a contingency because its truth value varies depending on p and q.

Practice Problems

Determine the truth values for the following problems.

1. If p is false, q is true, and r is false, determine the truth value of: (p ∨ q) ∧ r.
2. Given the truth values v(p) = T, v(q) = F, and v(r) = T, evaluate: ¬p → (q ∧ r).
3. Is there any truth value assignment for p and q that makes the statement (p → q) ∨ p false?
4. Under what truth values of p and q is the biconditional statement (p ∨ q) ↔ ¬(¬p ∧ ¬q) true?
5. For what truth value(s) of p, if any, is the conditional statement p → ¬p true?
6. Without using a truth table, explain whether it is possible for ¬(p ∨ q) to be true while ¬p ∧ ¬q is false. Justify your reasoning.

Solution 1

We are given that p is false (F), q is true (T), and r is false (F). First, evaluate the disjunction p ∨ q. Since p is F and q is T, the disjunction evaluates to true, as an inclusive "or" only requires at least one component to be true. Next, evaluate the conjunction with r: (p ∨ q) ∧ r. Since r is false, the conjunction of T and F yields F. Therefore, the overall statement is false.

v((p ∨ q) ∧ r) = F

Solution 2

We are given that v(p) = T, v(q) = F, and v(r) = T. First, find the negation ¬p: because p is true, ¬p is false. Next, evaluate the conjunction q ∧ r: with q as F and r as T, the conjunction evaluates to F. Finally, evaluate the conditional statement ¬p → (q ∧ r). A conditional statement with a false hypothesis (¬p = F) is vacuously true, regardless of the truth value of the conclusion. Therefore, the compound statement is true.

Solution 3

We analyze the disjunction (p → q) ∨ p. If p is true, the second component of the disjunction is T, which immediately makes the entire disjunction true. If p is false, the conditional statement p → q has a false hypothesis and is therefore true, making the disjunction of T and F evaluate to T. Testing all truth value combinations confirms this: (p = T, q = T) yields T; (p = T, q = F) yields T; (p = F, q = T) yields T; and (p = F, q = F) yields T. Because every case evaluates to true, no combination makes the statement false. This can also be verified by constructing its truth table, proving it is a tautology.

Solution 4

Consider the biconditional statement (p ∨ q) ↔ ¬(¬p ∧ ¬q). First, we can simplify the right side using De Morgan's Laws: ¬(¬p ∧ ¬q) is logically equivalent to ¬(¬p) ∨ ¬(¬q), which simplifies to p ∨ q by the double negation law. Substituting this back into the original expression gives (p ∨ q) ↔ (p ∨ q). Since any statement is logically equivalent to itself, this is a tautology. Therefore, the biconditional statement is true under all possible truth value assignments of p and q.

Solution 5

Consider p → ¬p. If p is true, then ¬p is false, and the conditional statement T → F evaluates to false. If p is false, then ¬p is true, and the conditional statement F → T evaluates to true. Therefore, the statement p → ¬p is true only when p is false.

Solution 6

We want to determine whether ¬(p ∨ q) can be true while ¬p ∧ ¬q is false. By De Morgan's Laws, the expressions ¬(p ∨ q) and ¬p ∧ ¬q are logically equivalent. By definition, logically equivalent statements must share the identical truth value across every possible assignment. Consequently, it is impossible for one expression to be true while the other is false; whenever ¬(p ∨ q) is true, ¬p ∧ ¬q must also be true. Therefore, the scenario described cannot occur.