Nested conditionals show how equivalent statements may be written in different forms. Simplification of compound statements uses the laws of logic to find a shorter or cleaner equivalent form. This is a central skill in discrete mathematics because long formulas occur in proofs, circuits, program conditions, and switching networks. In this lesson, simplification of compound statements will be performed step by step with named equivalence laws. :::definition[Simplification of a Compound Statement] Let $s$ be a compound statement. To simplify $s$ means to find a logically equivalent statement that uses fewer connectives, fewer negations, or fewer repeated components. Thus, if $s\Leftrightarrow t$ and $t$ is shorter or structurally simpler than $s$, then $t$ is a simplification of $s$. ::: A simplification is not a numerical calculation. It is a chain of logical equivalences. Each step must preserve truth values under every assignment. Students often try to cancel symbols by appearance, but logical simplification requires named laws such as De Morgan's Laws, distributive laws, identity laws, inverse laws, and absorption laws. Use the checker to practise selecting a valid next step in a simplification chain. Choose a preset expression and then select the transformation you think is justified. The feedback names the law that makes the step valid. This reinforces the habit of simplifying by equivalence laws rather than by visual cancellation. :::calculator[Simplification Step Checker] ::: :::solved-problem Simplify $(p\lor q)\land\neg(\neg p\land q)$. ::: :::solution Given that $p$ and $q$ are primitive statements. Using De Morgan's Law and the Double Negation Law, $$\begin{align*} (p\lor q)\land\neg(\neg p\land q)&\Leftrightarrow(p\lor q)\land(p\lor\neg q)\\ &\Leftrightarrow p\lor(q\land\neg q)\\ &\Leftrightarrow p\lor F_0\\ &\Leftrightarrow p. \end{align*}$$ Hence, the simplified statement is $p$. ::: :::solved-problem Simplify $\neg\{[(p\lor q)\land r]\lor\neg q\}$. ::: :::solution Given that $p$, $q$, and $r$ are primitive statements. Using De Morgan's Law, $$\begin{align*} \neg\{[(p\lor q)\land r]\lor\neg q\}&\Leftrightarrow\neg[(p\lor q)\land r]\land q\\ &\Leftrightarrow[\neg(p\lor q)\lor\neg r]\land q\\ &\Leftrightarrow[(\neg p\land\neg q)\lor\neg r]\land q\\ &\Leftrightarrow[q\land(\neg p\land\neg q)]\lor(q\land\neg r)\\ &\Leftrightarrow F_0\lor(q\land\neg r)\\ &\Leftrightarrow q\land\neg r. \end{align*}$$ Hence, the simplified statement is $q\land\neg r$. ::: :::solved-problem Simplify $\neg\{[(p\lor q)\land r]\to\neg q\}$. ::: :::solution Given that $p$, $q$, and $r$ are primitive statements. Using the negation of an implication, $$\begin{align*} \neg\{[(p\lor q)\land r]\to\neg q\}&\Leftrightarrow[(p\lor q)\land r]\land q\\ &\Leftrightarrow q\land(p\lor q)\land r\\ &\Leftrightarrow q\land r. \end{align*}$$ Hence, the simplified statement is $q\land r$. ::: :::solved-problem Simplify $[[[(p\land q)\land r]\lor[(p\land r)\land\neg r]]\lor\neg q]\to s$ as far as the antecedent can be simplified directly. ::: :::solution Given that $p$, $q$, $r$, and $s$ are primitive statements. Simplify the antecedent: $$\begin{align*} [[(p\land q)\land r]\lor[(p\land r)\land\neg r]]\lor\neg q&\Leftrightarrow(p\land q\land r)\lor[p\land(r\land\neg r)]\lor\neg q\\ &\Leftrightarrow(p\land q\land r)\lor(p\land F_0)\lor\neg q\\ &\Leftrightarrow(p\land q\land r)\lor F_0\lor\neg q\\ &\Leftrightarrow(p\land q\land r)\lor\neg q. \end{align*}$$ Therefore, a simplified form of the original statement is $[(p\land q\land r)\lor\neg q]\to s$. ::: :::exercise Negate each statement and simplify the result: (1) $p\land(q\lor r)\land(\neg p\lor\neg q\lor r)$. (2) $(p\land q)\to r$. (3) $p\to(\neg q\land r)$. (4) $p\lor q\lor(\neg p\land\neg q\land r)$. ::: :::answer (1) $\neg p\lor(\neg q\land\neg r)\lor(p\land q\land\neg r)$. (2) $p\land q\land\neg r$. (3) $p\land(q\lor\neg r)$. (4) $\neg p\land\neg q\land\neg r$. ::: :::faq[Frequently Asked Questions] Q: What makes a statement simplified? A: It is logically equivalent but uses fewer connectives, fewer negations, or fewer repeated parts. Q: Can there be more than one simplified form? A: Yes. Different equivalent forms may be useful for different purposes. Q: Why should each step have a reason? A: Reasons prevent invalid symbolic cancellation and show that truth values are preserved. ::: :::call-to-action[Simplify Step by Step] subtitle: Practise named-law simplification before applying logical equivalence to switching networks. button: Next Lesson | /discrete-mathematics/mathematical-logic/logical-equivalence-and-the-laws-of-logic/switching-networks-and-logical-equivalence button-ghost: Previous Lesson | /discrete-mathematics/mathematical-logic/logical-equivalence-and-the-laws-of-logic/nested-conditionals-and-program-logic :::