After studying negation, we now join two statements to form larger statements. The connectives conjunction and disjunction correspond to the mathematical meanings of and and or. These meanings must be handled carefully because ordinary language sometimes uses or ambiguously. In this lesson, conjunction and disjunction will be treated through precise truth conditions. :::definition[Conjunction] Let $p$ and $q$ be statements. The conjunction of $p$ and $q$ is denoted by $p\land q$ and is read as $p$ and $q$. The statement $p\land q$ is true exactly when both $p$ and $q$ are true. ::: :::example[Conjunction] Let $$\begin{align*} p &: \text{The class starts at noon.}\\ q &: \text{The classroom has a projector.} \end{align*}$$ Then $p\land q$ means The class starts at noon and the classroom has a projector. This compound statement is true only when both component statements are true. ::: :::definition[Truth Table for Conjunction] Let $p$ and $q$ be statements. The truth table for $p\land q$ is $$\begin{equation*} \begin{array}{c|c|c} p & q & p\land q\\ \hline 0 & 0 & 0\\ 0 & 1 & 0\\ 1 & 0 & 0\\ 1 & 1 & 1 \end{array} \end{equation*}$$ ::: :::definition[Disjunction] Let $p$ and $q$ be statements. The disjunction of $p$ and $q$ is denoted by $p\lor q$ and is read as $p$ or $q$. In mathematical logic, unless otherwise stated, or is interpreted inclusively. ::: :::definition[Inclusive Or] The inclusive or statement $p\lor q$ is true when at least one of $p$ and $q$ is true. Thus, it is true when $p$ is true, when $q$ is true, or when both are true. ::: :::example[Inclusive Disjunction] Let $$\begin{align*} p &: \text{The student studies algebra.}\\ q &: \text{The student studies statistics.} \end{align*}$$ Then $p\lor q$ means The student studies algebra or the student studies statistics. This statement also allows the possibility that the student studies both subjects. ::: :::definition[Truth Table for Inclusive Disjunction] Let $p$ and $q$ be statements. The truth table for $p\lor q$ is $$\begin{equation*} \begin{array}{c|c|c} p & q & p\lor q\\ \hline 0 & 0 & 0\\ 0 & 1 & 1\\ 1 & 0 & 1\\ 1 & 1 & 1 \end{array} \end{equation*}$$ ::: :::definition[Exclusive Or] The exclusive or of $p$ and $q$ is true exactly when one of $p$ and $q$ is true, but not both. It is denoted by $p\underline{\lor}q$. ::: :::definition[Truth Table for Exclusive Disjunction] Let $p$ and $q$ be statements. The truth table for $p\underline{\lor}q$ is $$\begin{equation*} \begin{array}{c|c|c} p & q & p\underline{\lor}q\\ \hline 0 & 0 & 0\\ 0 & 1 & 1\\ 1 & 0 & 1\\ 1 & 1 & 0 \end{array} \end{equation*}$$ ::: Use the comparison table to study three connectives on the same truth-value row. Choose values for $p$ and $q$, then compare how $p\land q$, $p\lor q$, and exclusive or respond. Notice that inclusive disjunction accepts the case where both statements are true, while exclusive disjunction rejects that case. This comparison helps separate the ordinary word or from the precise mathematical meanings used in truth tables. :::scientific-preview[And Or Comparison Table] ::: :::example[Exclusive Disjunction] Let $$\begin{align*} p &: \text{A student selects calculus.}\\ q &: \text{A student selects discrete mathematics.} \end{align*}$$ Then $p\underline{\lor}q$ means The student selects calculus or discrete mathematics, but not both. This is different from inclusive disjunction because the case where both are selected is false. ::: :::solved-problem Let $p$ be true and $q$ be false. Determine the truth values of $p\land q$, $p\lor q$, and $p\underline{\lor}q$. ::: :::solution Let $p=1$ and $q=0$. For conjunction, both component statements must be true, so $$\begin{equation*} p\land q=0. \end{equation*}$$ For inclusive disjunction, at least one component statement must be true, so $$\begin{equation*} p\lor q=1. \end{equation*}$$ For exclusive disjunction, exactly one component statement must be true. Since $p=1$ and $q=0$, $$\begin{equation*} p\underline{\lor}q=1. \end{equation*}$$ Hence, the truth values are $0$, $1$, and $1$ respectively. ::: :::solved-problem Let $p:$ A number is divisible by $2$, and $q:$ A number is divisible by $3$. Interpret $p\land q$ and $p\lor q$ for the number $6$. ::: :::solution Let the number be $6$. Since $6$ is divisible by $2$, $p=1$. Since $6$ is divisible by $3$, $q=1$. Therefore, $$\begin{align*} p\land q &= 1\land 1\\ &=1,\\ p\lor q &= 1\lor 1\\ &=1. \end{align*}$$ Hence, both the conjunction and the inclusive disjunction are true for the number $6$. ::: :::exercise Let $p$ and $q$ be statements. Complete the truth values: (1) If $p=0$ and $q=0$, find $p\lor q$. (2) If $p=1$ and $q=1$, find $p\underline{\lor}q$. (3) If $p=0$ and $q=1$, find $p\land q$. (4) If $p=0$ and $q=1$, find $p\lor q$. ::: :::answer (1) $p\lor q=0$. (2) $p\underline{\lor}q=0$. (3) $p\land q=0$. (4) $p\lor q=1$. ::: :::faq[Frequently Asked Questions] Q: Does or always mean inclusive or in mathematics? A: Yes, unless exclusive or is explicitly stated. The statement $p\lor q$ allows both $p$ and $q$ to be true. Q: When is $p\land q$ false? A: It is false whenever at least one of $p$ and $q$ is false. Q: Why is exclusive or separated from ordinary disjunction? A: Exclusive or rejects the case where both statements are true, while inclusive disjunction accepts it. ::: :::call-to-action[Compare And and Or] subtitle: Review the truth tables for conjunction, inclusive disjunction, and exclusive disjunction before studying implication. button: Next Lesson | /discrete-mathematics/mathematical-logic/basic-logical-connectives-and-truth-tables/implication-and-biconditional button-ghost: Previous Lesson | /discrete-mathematics/mathematical-logic/basic-logical-connectives-and-truth-tables/negation-and-compound-statements :::