De Morgan's Laws show how negation changes logical connectives. The next major family of equivalences explains how conjunction and disjunction distribute through each other. Distributive laws for logical connectives are central in simplification, switching networks, normal forms, and proofs of equivalence. In this lesson, distributive laws for logical connectives will be used to rewrite compound statements without changing their truth values. :::theorem[Distributive Laws] Let $p$, $q$, and $r$ be primitive statements. Then $$\begin{equation*} p\land(q\lor r)\Leftrightarrow(p\land q)\lor(p\land r) \end{equation*}$$ and $$\begin{equation*} p\lor(q\land r)\Leftrightarrow(p\lor q)\land(p\lor r). \end{equation*}$$ These are called the Distributive Laws for logical connectives. ::: :::proof Given that $p$, $q$, and $r$ are primitive statements. To prove the distributive laws. The statement $p\land(q\lor r)$ is true exactly when $p$ is true and at least one of $q$ and $r$ is true. The statement $(p\land q)\lor(p\land r)$ is true exactly when $p$ and $q$ are both true, or $p$ and $r$ are both true. These conditions are the same. Therefore, $$\begin{equation*} p\land(q\lor r)\Leftrightarrow(p\land q)\lor(p\land r). \end{equation*}$$ The statement $p\lor(q\land r)$ is true when $p$ is true, or when both $q$ and $r$ are true. The statement $(p\lor q)\land(p\lor r)$ is true in the same cases. Therefore, $$\begin{equation*} p\lor(q\land r)\Leftrightarrow(p\lor q)\land(p\lor r). \end{equation*}$$ Hence, both distributive laws hold. $\square$ ::: :::definition[Conjunction Distributed Over Disjunction] For primitive statements $p$, $q$, and $r$, $$\begin{equation*} p\land(q\lor r)\Leftrightarrow(p\land q)\lor(p\land r). \end{equation*}$$ This law distributes $\land$ over $\lor$. ::: :::definition[Disjunction Distributed Over Conjunction] For primitive statements $p$, $q$, and $r$, $$\begin{equation*} p\lor(q\land r)\Leftrightarrow(p\lor q)\land(p\lor r). \end{equation*}$$ This law distributes $\lor$ over $\land$. ::: Use the comparator to test both logical distributive laws under the same truth-value assignment. Choose which law to study, then set values for $p$, $q$, and $r$. The two sides should always produce the same output when the law is valid. This is especially helpful for the second distributive law, because it has no direct ordinary arithmetic analogue. :::scientific-preview[Distributive Law Comparator] ::: :::example[Arithmetic Comparison] For real numbers $a$, $b$, and $c$, $$\begin{equation*} a(b+c)=ab+ac. \end{equation*}$$ This resembles $p\land(q\lor r)\Leftrightarrow(p\land q)\lor(p\land r)$. Logic also has $p\lor(q\land r)\Leftrightarrow(p\lor q)\land(p\lor r)$, which has no direct counterpart in ordinary arithmetic addition and multiplication. ::: :::solved-problem Rewrite $p\land(q\lor r)$ and $p\lor(q\land r)$ using distributive laws. ::: :::solution Given that $p$, $q$, and $r$ are primitive statements. Using the first distributive law, $$\begin{equation*} p\land(q\lor r)\Leftrightarrow(p\land q)\lor(p\land r). \end{equation*}$$ Using the second distributive law, $$\begin{equation*} p\lor(q\land r)\Leftrightarrow(p\lor q)\land(p\lor r). \end{equation*}$$ Hence, the required rewritten forms are $(p\land q)\lor(p\land r)$ and $(p\lor q)\land(p\lor r)$. ::: :::solved-problem Use laws of logic to verify $(p\lor q)\land(p\lor\neg q)\Leftrightarrow p$. ::: :::solution Given that $p$ and $q$ are primitive statements. To prove the equivalence, use the distributive law in reverse: $$\begin{align*} (p\lor q)\land(p\lor\neg q)&\Leftrightarrow p\lor(q\land\neg q)\\ &\Leftrightarrow p\lor F_0\\ &\Leftrightarrow p. \end{align*}$$ Hence, $(p\lor q)\land(p\lor\neg q)\Leftrightarrow p$. ::: :::solved-problem Simplify $\neg q\lor(\neg p\land q)$. ::: :::solution Given that $p$ and $q$ are primitive statements. Using distribution, $$\begin{align*} \neg q\lor(\neg p\land q)&\Leftrightarrow(\neg q\lor\neg p)\land(\neg q\lor q)\\ &\Leftrightarrow(\neg q\lor\neg p)\land T_0\\ &\Leftrightarrow\neg q\lor\neg p. \end{align*}$$ Hence, the simplified statement is $\neg p\lor\neg q$. ::: :::exercise Use distributive laws to rewrite: (1) $p\land(q\lor r)$. (2) $p\lor(q\land r)$. (3) $(p\lor q)\land(p\lor\neg q)$. ::: :::answer (1) $(p\land q)\lor(p\land r)$. (2) $(p\lor q)\land(p\lor r)$. (3) $p\lor(q\land\neg q)$, which simplifies to $p$. ::: :::faq[Frequently Asked Questions] Q: Can $\lor$ distribute over $\land$? A: Yes. In logic, both $\land$ distributes over $\lor$ and $\lor$ distributes over $\land$. Q: Can a distributive law be used in reverse? A: Yes. Logical equivalence works in both directions. Q: Why are distributive laws useful? A: They transform compound statements into forms that can often be simplified. ::: :::call-to-action[Use Distribution Carefully] subtitle: Practise both distributive laws before collecting the full list of standard laws of logic. button: Next Lesson | /discrete-mathematics/mathematical-logic/logical-equivalence-and-the-laws-of-logic/laws-of-logic button-ghost: Previous Lesson | /discrete-mathematics/mathematical-logic/logical-equivalence-and-the-laws-of-logic/de-morgans-laws :::