Distributive laws are part of a larger collection called the laws of logic. These laws allow compound statements to be rewritten step by step while preserving logical equivalence. They play the same role in logic that algebraic identities play in algebra. In this lesson, laws of logic will be organized and applied to simplify compound statements and justify each transformation. :::definition[Logical Equivalence by Tautology] Let $s_1$ and $s_2$ be two statements. Then $s_1$ and $s_2$ are logically equivalent if $s_1\leftrightarrow s_2$ is a tautology. Thus, $s_1\Leftrightarrow s_2$ means that $s_1$ and $s_2$ have the same truth value under every assignment of truth values to their primitive statements. ::: :::definition[Logical Non-Equivalence] Let $s_1$ and $s_2$ be two statements. Then $s_1$ and $s_2$ are not logically equivalent if there exists at least one truth assignment for which $s_1$ and $s_2$ have different truth values. We write $s_1\not\Leftrightarrow s_2$. ::: :::theorem[Negation Preserves Logical Equivalence] Let $s_1$ and $s_2$ be two statements. If $s_1\Leftrightarrow s_2$, then $\neg s_1\Leftrightarrow\neg s_2$. ::: :::proof Given that $s_1\Leftrightarrow s_2$. To prove that $\neg s_1\Leftrightarrow\neg s_2$. Since $s_1$ and $s_2$ have the same truth value for every truth assignment, their negations also have the same truth value for every truth assignment. Hence, $\neg s_1\Leftrightarrow\neg s_2$. $\square$ ::: :::theorem[Transitivity of Logical Equivalence] Let $s_1$, $s_2$, and $s_3$ be statements. If $s_1\Leftrightarrow s_2$ and $s_2\Leftrightarrow s_3$, then $s_1\Leftrightarrow s_3$. ::: :::proof Given that $s_1\Leftrightarrow s_2$ and $s_2\Leftrightarrow s_3$. To prove that $s_1\Leftrightarrow s_3$. For every truth assignment, $s_1$ and $s_2$ have the same truth value, and $s_2$ and $s_3$ have the same truth value. Therefore, $s_1$ and $s_3$ have the same truth value for every truth assignment. Hence, $s_1\Leftrightarrow s_3$. $\square$ ::: :::definition[Standard Laws of Logic] For statements $p$, $q$, and $r$, the standard laws include $$\begin{align*} \neg\neg p&\Leftrightarrow p,\\ p\lor q&\Leftrightarrow q\lor p,\\ p\land q&\Leftrightarrow q\land p,\\ p\lor(q\lor r)&\Leftrightarrow(p\lor q)\lor r,\\ p\land(q\land r)&\Leftrightarrow(p\land q)\land r,\\ p\lor p&\Leftrightarrow p,\\ p\land p&\Leftrightarrow p. \end{align*}$$ These are the double negation, commutative, associative, and idempotent laws. ::: :::definition[Laws Involving Tautology and Contradiction] Let $T_0$ denote any tautology and let $F_0$ denote any contradiction. Then $$\begin{align*} p\lor F_0&\Leftrightarrow p,\\ p\land T_0&\Leftrightarrow p,\\ p\lor\neg p&\Leftrightarrow T_0,\\ p\land\neg p&\Leftrightarrow F_0,\\ p\lor T_0&\Leftrightarrow T_0,\\ p\land F_0&\Leftrightarrow F_0. \end{align*}$$ These are the identity, inverse, and domination laws. ::: :::definition[Absorption Laws] For statements $p$ and $q$, $$\begin{equation*} p\lor(p\land q)\Leftrightarrow p \end{equation*}$$ and $$\begin{equation*} p\land(p\lor q)\Leftrightarrow p. \end{equation*}$$ These are called the Absorption Laws. ::: :::solved-problem Verify $p\lor(p\land q)\Leftrightarrow p$. ::: :::solution Given that $p$ and $q$ are primitive statements. To verify the absorption law, use a truth table: $$\begin{equation*} \begin{array}{c|c|c|c} p&q&p\land q&p\lor(p\land q)\\ \hline 1&1&1&1\\ 1&0&0&1\\ 0&1&0&0\\ 0&0&0&0 \end{array} \end{equation*}$$ The final column equals the column for $p$. Hence, $p\lor(p\land q)\Leftrightarrow p$. ::: :::solved-problem Simplify $[(p\lor q)\land(p\lor\neg q)]\lor q$. ::: :::solution Given that $p$ and $q$ are primitive statements. Using distribution and the inverse law, $$\begin{align*} (p\lor q)\land(p\lor\neg q)&\Leftrightarrow p\lor(q\land\neg q)\\ &\Leftrightarrow p\lor F_0\\ &\Leftrightarrow p. \end{align*}$$ Thus, $$\begin{equation*} [(p\lor q)\land(p\lor\neg q)]\lor q\Leftrightarrow p\lor q. \end{equation*}$$ Hence, the simplified statement is $p\lor q$. ::: :::exercise Use laws of logic to simplify: (1) $p\lor[p\land(p\lor q)]$. (2) $p\lor q\lor(\neg p\land\neg q\land r)$. (3) $[(p\lor q)\land(p\lor\neg q)]$. ::: :::answer (1) $p$. (2) $p\lor q\lor r$. (3) $p$. ::: :::faq[Frequently Asked Questions] Q: Why are laws of logic useful? A: They allow simplification without constructing a full truth table each time. Q: What do $T_0$ and $F_0$ mean? A: $T_0$ denotes a tautology, and $F_0$ denotes a contradiction. Q: Should every simplification step have a reason? A: Yes. Each replacement should be justified by a named law or proved equivalence. ::: :::call-to-action[Use Laws with Reasons] subtitle: Practise naming each law used in a simplification before studying duality in logical equivalence. button: Next Lesson | /discrete-mathematics/mathematical-logic/logical-equivalence-and-the-laws-of-logic/duality-in-logical-equivalence button-ghost: Previous Lesson | /discrete-mathematics/mathematical-logic/logical-equivalence-and-the-laws-of-logic/distributive-laws-for-logical-connectives :::