Truth Tables for the Conditional and Biconditional

Learning Objectives

After completing this section, you should be able to:

1. Use and apply the conditional to construct a truth table.
2. Use and apply the biconditional to construct a truth table.
3. Use truth tables to determine the validity of conditional and biconditional statements.

Use and Apply the Conditional to Construct a Truth Table

A conditional is a logical statement of the form if \(p\), then \(q\). The conditional statement in logic is a promise or contract. The only time the conditional, \(p\to q,\) is false is when the contract or promise is broken.

For example, consider the following scenario. A child’s parent says, “If you do your homework, then you can play your video games.” The child really wants to play their video games, so they get started right away, finish within an hour, and then show their parent the completed homework. The parent thanks the child for doing a great job on their homework and allows them to play video games. Both the parent and child are happy. The contract was satisfied; true implies true is true.

Now, suppose the child does not start their homework right away, and then struggles to complete it. They eventually finish and show it to their parent. The parent again thanks the child for completing their homework, but then informs the child that it is too late in the evening to play video games, and that they must begin to get ready for bed. Now, the child is really upset. They held up their part of the contract, but they did not receive the promised reward. The contract was broken; true implies false is false.

So, what happens if the child does not do their homework? In this case, the hypothesis is false. No contract has been entered, therefore, no contract can be broken. If the conclusion is false, the child does not get to play video games and might not be happy, but this outcome is expected because the child did not complete their end of the bargain. They did not complete their homework. False implies false is true. The last option is not as intuitive. If the parent lets the child play video games, even if they did not do their homework, neither parent nor child are going to be upset. False implies true is true.

The truth table for the conditional statement below summarizes these results.

\(p\)	\(q\)	\(p\to q\)
T	T	T
T	F	F
F	T	T
F	F	T

Condensed — the full section is in OpenStax Contemporary Mathematics.

Use and Apply the Biconditional to Construct a Truth Table

The biconditional, \(p↔q\), is a two way contract; it is equivalent to the statement \((p\to q)∧(q\to p).\) A biconditional statement, \(p↔q,\) is true whenever the truth value of the hypothesis matches the truth value of the conclusion, otherwise it is false.

The truth table for the biconditional is summarized below.

\(p\)	\(q\)	\(p↔q\)
T	T	T
T	F	F
F	T	F
F	F	T

Constructing Truth Tables for Biconditional Statements

Try it.

Assume both of the following statements are true: \(p\): The plumber fixed the leak, and \(q\): The homeowner paid the plumber $150.00. Create a truth table to determine the truth value of each of the following biconditional statements.

1. \(p↔q\)
2. \(p↔\ \sim q\)
3. \(\sim p↔\ \sim q\)

Solution

1. Because \(p\) is true and \(q\) is true, the statement \(p↔q\) is “The plumber fixed the leak if and only if the homeowner paid them $150.00.” Because both \(p\) and \(q\) are true, the leak was fixed and the plumber was paid, meaning both parties satisfied their end of the bargain. The biconditional statement is true, as indicated by the truth table representing this case: T ↔ T = T.

   \(p\)	\(q\)	\(p↔q\)
   T	T	T

2. \(p\to \sim q\) translates to the statement, “The plumber fixed the leak if and only if the homeowner did not pay them $150.” If the plumber fixed the leak and the homeowner did not pay them, the homeowner will have broken their end of the contract. The biconditional statement is false, as indicated by the truth table representing this case:
   T ↔ F = F.

   \(p\)	\(q\)	\(\sim q\)	\(p↔\ \sim q\)
   T	T	F	F

3. \(\sim p↔\ \sim q\) translates to the statement, “The plumber did not fix the leak if and only if the homeowner did not pay them $150.” In this case, neither party—the plumber nor the homeowner—entered into the contract. The leak was not repaired, and the plumber was not paid. No agreement was broken. The biconditional statement is true, as indicated by the truth table representing this case: F ↔ F = T.

   \(p\)	\(q\)	\(\sim p\)	\(\sim q\)	\(\sim p↔\ \sim q\)
   T	T	F	F	T

Condensed — the full section is in OpenStax Contemporary Mathematics.

Key Concepts

• The conditional statement, if \(p\) then \(q\), is like a contract. The only time it is false is when the contract has been broken. That is, when \(p\) is true, and \(q\) is false.

  Conditional
  \(p\)	\(q\)	\(p\to q\)
  T	T	T
  T	F	F
  F	T	T
  F	F	T

• The biconditional statement, \(p\) if and only if \(q\), it true whenever \(p\) and \(q\) have matching true values, otherwise it is false.

  Biconditional
  \(p\)	\(q\)	\(p↔q\)
  T	T	T
  T	F	F
  F	T	F
  F	F	T

• Know how to construct truth tables involving conditional and biconditional statements.
• Use truth tables to analyze conditional and biconditional statements and determine their validity.

Video

• Logic Part 8: The Conditional and Tautologies
• Logic Part 11B Biconditional and Summary of Truth Value Rules in Logic
• Logic Part 13: Truth Tables to Determine if Argument is Valid or Invalid

Practice (4)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. Assume both of the following statements are true: \(p\): My sibling washed the dishes, and \(q\): My parents paid them $5.00. Create a truth table to determine the truth value of each of the following conditional statements.

   1. \(p\to q\)
   2. \(p\to \ \sim q\)
   3. \(\sim p\to q\)
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   1. Because \(p\) is true and \(q\) is true, the statement \(p\to q\) is, “If my sibling washed the dishes, then my parents paid them $5.00.” My sibling did wash the dishes, since \(p\) is true, and the parents did pay the sibling $5.00, so the contract was entered and completed. The conditional statement is true, as indicated by the truth table representing this case:
      T → T = T.

      \(p\)	\(q\)	\(p\to q\)
      T	T	T

   2. \(p\to \sim q\) translates to the statement, “If my sibling washed the dishes, then my parents did not pay them $5.00.” \(p\) is true, but \(\sim q\) is false. The sibling completed their end of the contract, but they did not get paid. The contract was broken by the parents. The conditional statement is false, as indicated by the truth table representing this case:
      T → F = F.

      \(p\)	\(q\)	\(\sim q\)	\(p\to \ \sim q\)
      T	T	F	F

   3. \(\sim p\to q\) translates to the statement, “If my sibling did not wash the dishes, then my parents paid them $5.00.” \(\sim p\) is false, but \(q\) is true. The sibling did not do the dishes. No contract was entered, so it could not be broken. The parents decided to pay them $5.00 anyway. The conditional statement is true, as indicated by the truth table representing this case: F → T = T.

      \(p\)	\(q\)	\(\sim p\)	\(\sim p\to q\)
      T	T	F	T

2. Construct a truth table to analyze all possible outcomes for each of the following statements then determine whether they are valid.

   1. \(p∧q\to \sim q\)
   2. \(p\to \sim p∨q\)
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   1. Applying the dominance of connectives, the statement \(p∧q\to \ \sim q\) is equivalent to \((p∧q)\to (\sim q).\) So, the columns of the truth table will include \(p\), \(q\), \(p∧q,\) \(\sim q\), and \(p∧q\to \ \sim q.\) Because there are only two basic propositions, \(p\) and \(q\), the table will have \(2(2)=4\) rows of truth values to account for all the possible outcomes. The statement is not valid because the last column is not all true.

      \(p\)	\(q\)	\(p∧q\)	\(\sim q\)	\(p∧q\to \sim q\)
      T	T	T	F	F
      T	F	F	T	T
      F	T	F	F	T
      F	F	F	T	T

   2. Applying the dominance of connectives, the statement \(p\to \ \sim p∨q\) is equivalent to \((p)\to \ ((\sim p)∨q).\) So, the columns of the truth table will include \(p\), \(q\), \(\sim p\), \(\sim p∨q,\) and \(p\to \ (\sim p∨q).\) Because there are only two basic propositions, \(p\) and \(q\), the table will have \(2(2)=4\) rows of truth values to account for all the possible outcomes. The statement is not valid because the last column is not all true.

      \(p\)	\(q\)	\(\sim p\)	\(\sim p∨q\)	\(p\to (\sim p∨q)\)
      T	T	F	T	T
      T	F	F	F	F
      F	T	T	T	T
      F	F	T	T	T

3. Assume both of the following statements are true: \(p\): The plumber fixed the leak, and \(q\): The homeowner paid the plumber $150.00. Create a truth table to determine the truth value of each of the following biconditional statements.

   1. \(p↔q\)
   2. \(p↔\ \sim q\)
   3. \(\sim p↔\ \sim q\)
   ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

   1. Because \(p\) is true and \(q\) is true, the statement \(p↔q\) is “The plumber fixed the leak if and only if the homeowner paid them $150.00.” Because both \(p\) and \(q\) are true, the leak was fixed and the plumber was paid, meaning both parties satisfied their end of the bargain. The biconditional statement is true, as indicated by the truth table representing this case: T ↔ T = T.

      \(p\)	\(q\)	\(p↔q\)
      T	T	T

   2. \(p\to \sim q\) translates to the statement, “The plumber fixed the leak if and only if the homeowner did not pay them $150.” If the plumber fixed the leak and the homeowner did not pay them, the homeowner will have broken their end of the contract. The biconditional statement is false, as indicated by the truth table representing this case:
      T ↔ F = F.

      \(p\)	\(q\)	\(\sim q\)	\(p↔\ \sim q\)
      T	T	F	F

   3. \(\sim p↔\ \sim q\) translates to the statement, “The plumber did not fix the leak if and only if the homeowner did not pay them $150.” In this case, neither party—the plumber nor the homeowner—entered into the contract. The leak was not repaired, and the plumber was not paid. No agreement was broken. The biconditional statement is true, as indicated by the truth table representing this case: F ↔ F = T.

      \(p\)	\(q\)	\(\sim p\)	\(\sim q\)	\(\sim p↔\ \sim q\)
      T	T	F	F	T

4. Construct a truth table to analyze all possible outcomes for each of the following statements, then determine whether they are valid.

   1. \(p∧q↔p∧\sim q\)
   2. \(p∨q↔\sim p∨q\)
   3. \(p\to q↔\sim q\to \sim p\)
   4. \(p∧q\to \sim r↔p∧q∧r\)
   ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

   1. Applying the dominance of connectives, the statement \(p∧q↔\ p∧\sim q\) is equivalent to \((p∧q)↔\ (p∧(\sim q)).\) So, the columns of the truth table will include \(p\), \(q\), \(p∧q,\) \(\sim q\), \(p∧\sim q\) and \((p∧q)↔(p∧\sim q).\) Because there are only two basic propositions, \(p\) and \(q\), the table will have \(2(2)=4\) rows of truth values to account for all the possible outcomes. The statement is not valid because the last column is not all true.

      \(p\)	\(q\)	\(p∧q\)	\(\sim q\)	\(p∧\sim q\)	\((p∧q)↔(p∧\sim q)\)
      T	T	T	F	F	F
      T	F	F	T	T	F
      F	T	F	F	F	T
      F	F	F	T	F	T

   2. Applying the dominance of connectives, the statement \(p∨q↔\sim p∨q\) is equivalent to \((p∨q)↔\ ((\sim p)∨q).\) So, the columns of the truth table will include \(p\), \(q\), \(p∨q,\) \(\sim p\), \(\sim p∨q,\) and \((p∨q)↔(\sim p∨q).\) Because there are only two basic propositions, \(p\) and \(q\), the table will have \(2(2)=4\) rows of truth values to account for all the possible outcomes. The statement is not valid because the last column is not all true.

      \(p\)	\(q\)	\(p∨q\)	\(\sim p\)	\(\sim p∨q\)	\((p∨q)↔(\sim p∨q)\)
      T	T	T	F	T	T
      T	F	T	F	F	F
      F	T	T	T	T	T
      F	F	F	T	T	F

   3. Applying the dominance of connectives, the statement \(p\to q↔\ \sim q\to \sim p\) is equivalent to \((p\to q)↔\ ((\sim q)\to (\sim p)).\) So, the columns of the truth table will include \(p\), \(q\), \(p\to q,\) \(\sim q\), \(\sim p\), \(\sim q\to \sim p,\) and \((p\to q)↔(\sim q\to \ \sim p).\) Because there are only two basic propositions, \(p\) and \(q\) the table will have \(2(2)=4\) rows of truth values to account for all the possible outcomes. The statement is valid because the last column is all true.

      \(p\)	\(q\)	\(p\to q\)	\(\sim q\)	\(\sim p\)	\(\sim q\to \sim p\)	\((p\to q)↔(\sim q\to \sim p)\)
      T	T	T	F	F	T	T
      T	F	F	T	F	F	T
      F	T	T	F	T	T	T
      F	F	T	T	T	T	T

   4. Applying the dominance of connectives, the statement \(p∧q\to \ \sim r\ ↔\ p∧q∧r\) is equivalent to \(((p∧q)\to (\sim r))\ ↔\ ((p∧q)∧r).\) So, the columns of the truth table will include \(p\), \(q\), \(r\), \(\sim r\), \(p∧q,\) \((p∧q)∧\ \sim r,\) \((p∧q)∧r,\) and \(((p∧q)\to (\sim r))\ ↔\ ((p∧q)∧r).\) Because there are three basic propositions, \(p\), \(q\), and \(r\), the table will have \(2(2)(2)=8\) rows of truth values to account for all the possible outcomes. The statement is not valid because the last column is not all true.

      \(p\)	\(q\)	\(r\)	\(\sim r\)	\(p∧q\)	\((p∧q)\to \sim r\)	\((p∧q)∧r\)	\((p∧q\to \sim r)↔(p∧q∧r)\)
      T	T	T	F	T	F	T	T
      T	T	F	T	T	T	F	T
      T	F	T	F	F	T	F	F
      T	F	F	T	F	T	F	F
      F	T	T	F	F	T	F	F
      F	T	F	T	F	T	F	F
      F	F	T	F	F	T	F	F
      F	F	F	T	F	T	F	F

Symbols used here

n!

factorial
n × (n−1) × … × 1; the number of orderings of n things. 0! = 1.

\binom{n}{k}

binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

x \in A,\ A \subseteq B

element of, subset
x belongs to A; every element of A is in B.

A \cup B,\ A \cap B,\ A \setminus B

union, intersection, difference
In either; in both; in A but not B.

\emptyset,\ |A|

empty set, cardinality
The set with no elements; the number of elements of A.

\forall,\ \exists

for all, there exists
Quantifiers: every x; at least one x.

\neg,\ \wedge,\ \vee,\ \Rightarrow,\ \Leftrightarrow

not, and, or, implies, iff
Logical connectives.

\blacksquare\ \text{or}\ \square

end of proof (halmos)
Marks the point where the statement has been established.

a \equiv b \pmod n

congruent modulo n
n divides a − b; a and b have the same remainder.

O(n^2),\ \Theta,\ \Omega

big-O notation
Grows no faster than n² (up to a constant), for large n.

a \bmod n

remainder
What is left after dividing a by n.

How to: Truth Tables for the Conditional and Biconditional

1. Use and apply the conditional to construct a truth table.
2. Use and apply the biconditional to construct a truth table.
3. Use truth tables to determine the validity of conditional and biconditional statements.
4. Or,
5. Because
6. Applying the dominance of connectives, the statement
7. Applying the dominance of connectives, the statement
8. Because

Questions people ask

What makes mathematics "discrete"?

It deals with separate, countable objects — integers, graphs, statements — rather than continuous quantities. No limits, no infinitesimals; instead induction, counting and logic.

How does a proof by induction work?

Show the statement for the first case, then show that whenever it holds for n it holds for n + 1. Like dominoes: the first falls, and each knocks over the next.

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Parts of this page are adapted from OpenStax Contemporary Mathematics (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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