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

Indi sen Hiç bir hasapçy buny çözmez, emma onuň bölekleri hasaplanmak üçindir. Aşakda birini synla ýa-da özüňi ýaz.

⌨ _Täze ✎ Ony çyz

Siziň rime problemaňyz

π Çöl

Tälim (4)

Her birini kagyzda synlap gör. Çözgüni barlap görmek üçin aç; barlap görlenler her bir ädim üçin çözgüççide açylyp bilner.

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. \(\sim p\to q\)
   _Işle Eşik

   Jawaby görkez

   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. \(p\to \sim p∨q\)
   _Işle Eşik

   Jawaby görkez

   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. \(\sim p↔\ \sim q\)
   _Işle Eşik

   Jawaby görkez

   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. \(p∧q\to \sim r↔p∧q∧r\)
   _Işle Eşik

   Jawaby görkez

   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

Özüň işle

Mugt hasap her bir ders üçin notlar goşýar, tamamlanan zadyňyzyň ýazgysyny, çözen problemaňyzy bir ýerde, we bu sahypa barada sorap biljek mugallymy. Matematika özüçe her kim üçin açyk, girseň ýa girmeseň.

_Gün _Giriş

Şu ýerde ulanylan simwollar

Eň täze düşündirme, surat, we her harpyň manysy üçin her bir simwoly basyň.

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_ksummation
Add a_k for k = 1 up to n.x \in Aelement of, subset
x belongs to A; every element of A is in B.A \cup Bunion, intersection, difference
In either; in both; in A but not B.\emptysetempty set, cardinality
The set with no elements; the number of elements of A.\forallfor all, there exists
Quantifiers: every x; at least one x.\negnot, and, or, implies, iff
Logical connectives.\blacksquare\ \text{or}\ \squareend of proof (halmos)
Marks the point where the statement has been established.a \equiv b \pmod ncongruent modulo n
n divides a − b; a and b have the same remainder.O(n^2)big-O notation
Grows no faster than n² (up to a constant), for large n.a \bmod nremainder
What is left after dividing a by n.

Nädip: 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

Adamlar soraýan soraglar

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.

What order should I take these lessons in?

Logic and proof first, because every later lesson proves things. Then induction, sets, relations and functions, counting, recurrences, graphs, Boolean algebra and automata, and finally algorithms and complexity, which use nearly everything before them.

Is discrete mathematics the same as the maths of computer science?

Largely. Data structures are graphs and trees, correctness arguments are induction, running times are recurrences and sums, circuits are Boolean algebra, and what a computer can do efficiently is the subject of the last lessons. It is also ordinary mathematics in its own right.

Do I need calculus for this course?

No. School algebra is enough. A little familiarity with limits helps when comparing growth rates, and the lesson on asymptotic analysis explains the one limit fact it uses.

Bu sahypanyň käbir bölegi şundan alyndy OpenStax Contemporary Mathematics (CC BY-NC-SA 4.0). Bu ýerde gysgaldylan we tekrar beýan edilen; hatalar biziňkidir. Bu dersiň çeşmesi kommersiýa däl ulanmak üçin lisenziyalanandygy üçin bu ders mugt, hiç zat satmaýar we şol bir lisenziya astynda paýlaşylýar.

_Ýaşa Discrete Math & Logic

Truth tablesSums and inductionProof by inductionAlgorithms and growth of functionsPredicate logic and quantifiersDirect proof, contrapositive and contradictionStrong induction and the well-ordering principleRelations: equivalence relations and partial ordersFunctions and cardinalityCounting: rules, choices and bijectionsThe pigeonhole principleSolving linear recurrence relationsGraphs: degrees, paths and connectivityTrees and spanning trees