The laws of logic often appear in pairs. When conjunction and disjunction are interchanged, and tautology and contradiction are interchanged, one valid law often produces another valid law. This symmetry is called duality. In this lesson, duality in logical equivalence will be used to form dual statements and understand why many laws of logic occur in matching pairs. :::definition[Dual of a Statement] Let $s$ be a statement that contains no logical connectives other than $\land$ and $\lor$. The dual of $s$, denoted by $s^d$, is obtained by replacing each occurrence of $\land$ by $\lor$, each occurrence of $\lor$ by $\land$, each occurrence of $T_0$ by $F_0$, and each occurrence of $F_0$ by $T_0$. ::: Use the builder to practise the rule for forming a dual statement. Choose a statement form and compare it with the result after switching and with or, and switching tautology with contradiction. Primitive statements and their negations stay fixed. This procedure explains why many laws of logic occur in matching pairs. :::calculator[Dual Statement Builder] ::: :::definition[Dual of a Primitive Statement] If $p$ is a primitive statement, then $$\begin{equation*} p^d=p. \end{equation*}$$ Thus, a primitive statement remains unchanged when the dual is formed. ::: :::definition[Dual of a Negated Primitive Statement] If $p$ is a primitive statement, then $$\begin{equation*} (\neg p)^d=\neg p. \end{equation*}$$ Thus, primitive statements and negated primitive statements remain unchanged when the dual is formed. ::: :::example[Dual Pair] The statements $$\begin{equation*} p\lor\neg p \end{equation*}$$ and $$\begin{equation*} p\land\neg p \end{equation*}$$ are duals of each other. The first statement is a tautology, and the second statement is a contradiction. ::: :::example[Dual of a Compound Statement] Let $$\begin{equation*} s=(p\land\neg q)\lor(r\land T_0). \end{equation*}$$ Replacing $\land$ by $\lor$, replacing $\lor$ by $\land$, and replacing $T_0$ by $F_0$ gives $$\begin{equation*} s^d=(p\lor\neg q)\land(r\lor F_0). \end{equation*}$$ ::: :::theorem[Principle of Duality] Let $s$ and $t$ be statements involving only $\land$, $\lor$, $T_0$, and $F_0$, together with primitive statements and their negations. If $s\Leftrightarrow t$, then $s^d\Leftrightarrow t^d$. ::: :::proof Given that $s\Leftrightarrow t$. To prove that $s^d\Leftrightarrow t^d$. The operation of taking a dual interchanges $\land$ and $\lor$, and also interchanges $T_0$ and $F_0$. The standard laws of logic occur in dual pairs. For example, the identity law $p\lor F_0\Leftrightarrow p$ has the dual $p\land T_0\Leftrightarrow p$. Therefore, if an equivalence is valid, then the corresponding dual equivalence is also valid. Hence, $s^d\Leftrightarrow t^d$. $\square$ ::: :::corollary Every law of logic involving only $\land$, $\lor$, $T_0$, and $F_0$ has a corresponding dual law. ::: :::proof Given that a law of logic has the form $s\Leftrightarrow t$. To prove that its dual is also a law. By the Principle of Duality, $s^d\Leftrightarrow t^d$. Therefore, the dual of a valid law is also valid. Hence, laws of logic occur in dual pairs. $\square$ ::: :::solved-problem Write the dual of $(p\lor q)\land(p\lor\neg q)\Leftrightarrow p$. ::: :::solution Given that $(p\lor q)\land(p\lor\neg q)\Leftrightarrow p$. To write the dual, replace $\lor$ by $\land$ and $\land$ by $\lor$. The dual of the left side is $$\begin{equation*} (p\land q)\lor(p\land\neg q). \end{equation*}$$ The dual of the right side is $p$. Therefore, the dual equivalence is $$\begin{equation*} (p\land q)\lor(p\land\neg q)\Leftrightarrow p. \end{equation*}$$ ::: :::solved-problem Find the dual of $p\to q$. ::: :::solution To find the dual of a statement containing $\to$, first rewrite the implication using only $\neg$, $\lor$, and $\land$: $$\begin{equation*} p\to q\Leftrightarrow\neg p\lor q. \end{equation*}$$ Thus, $$\begin{equation*} (p\to q)^d\Leftrightarrow(\neg p\lor q)^d\Leftrightarrow\neg p\land q. \end{equation*}$$ Hence, the dual of $p\to q$ is $\neg p\land q$. ::: :::exercise Write the dual of each statement: (1) $q\to p$. (2) $p\to(q\land r)$. (3) $p\lor(q\land F_0)$. (4) $(p\land q)\lor T_0$. ::: :::answer (1) $\neg q\land p$. (2) $\neg p\land(q\lor r)$. (3) $p\land(q\lor T_0)$. (4) $(p\lor q)\land F_0$. ::: :::faq[Frequently Asked Questions] Q: Does negation change when taking a dual? A: No. Primitive statements and their negations remain unchanged. Q: Can I take the dual of an implication directly? A: First rewrite the implication using $\neg$, $\lor$, and $\land$, then take the dual. Q: Why is duality useful? A: It lets one valid law produce another valid law by a systematic interchange of symbols. ::: :::call-to-action[Recognise Dual Laws] subtitle: Practise forming duals before using substitution rules to replace equivalent substatements. button: Next Lesson | /discrete-mathematics/mathematical-logic/logical-equivalence-and-the-laws-of-logic/substitution-rules-for-logical-equivalence button-ghost: Previous Lesson | /discrete-mathematics/mathematical-logic/logical-equivalence-and-the-laws-of-logic/laws-of-logic :::