After proving implication and equivalence between predicates, we now examine related conditional forms. Converse, inverse, and contrapositive of quantified implications are obtained by switching and negating the predicate parts of a universal implication. Only the contrapositive is always logically equivalent to the original implication. In this lesson, quantified implications will be compared through definitions, proof, and counterexamples. :::definition[Quantified Implication] A statement of the form $$\begin{equation*} \forall x[p(x)\to q(x)] \end{equation*}$$ is called a quantified implication. It says that for every element $x$ in the universe, if $p(x)$ is true, then $q(x)$ is true. ::: :::definition[Converse] The converse of $\forall x[p(x)\to q(x)]$ is $\forall x[q(x)\to p(x)]$. The converse is obtained by interchanging the hypothesis and the conclusion. ::: :::definition[Inverse] The inverse of $\forall x[p(x)\to q(x)]$ is $\forall x[\neg p(x)\to\neg q(x)]$. The inverse is obtained by negating both the hypothesis and the conclusion. ::: :::definition[Contrapositive] The contrapositive of $\forall x[p(x)\to q(x)]$ is $\forall x[\neg q(x)\to\neg p(x)]$. The contrapositive is obtained by interchanging the hypothesis and conclusion and negating both. ::: :::theorem[Equivalence of an Implication and Its Contrapositive] Let $p(x)$ and $q(x)$ be predicates over the same universe. Then $\forall x[p(x)\to q(x)]$ is logically equivalent to $\forall x[\neg q(x)\to\neg p(x)]$. ::: :::proof Given that $p(x)$ and $q(x)$ are predicates over the same universe. To prove that $\forall x[p(x)\to q(x)]$ is logically equivalent to $\forall x[\neg q(x)\to\neg p(x)]$. For each element $a$ in the universe, the implication $p(a)\to q(a)$ is logically equivalent to its contrapositive $\neg q(a)\to\neg p(a)$. Therefore, the two statements have the same truth value for every element $a$. Hence $\forall x[p(x)\to q(x)]$ is logically equivalent to $\forall x[\neg q(x)\to\neg p(x)]$. $\square$ ::: :::example[Squares and Equilateral Quadrilaterals] Let the universe be the set of all quadrilaterals in the plane. Let $s(x):x$ is a square and $e(x):x$ is equilateral. The statement $\forall x[s(x)\to e(x)]$ means every square is equilateral. Its converse is $\forall x[e(x)\to s(x)]$. Its inverse is $\forall x[\neg s(x)\to\neg e(x)]$. Its contrapositive is $\forall x[\neg e(x)\to\neg s(x)]$. ::: :::solved-problem Let the universe be the set of all real numbers. Let $p(x):|x|>3$ and $q(x):x>3$. Study $\forall x[p(x)\to q(x)]$, its converse, inverse, and contrapositive. ::: :::solution Given that $p(x):|x|>3$ and $q(x):x>3$. The statement $\forall x[p(x)\to q(x)]$ means that for every real number $x$, if $|x|>3$, then $x>3$. This statement is false. For $x=-4$, $|-4|=4>3$, but $-4>3$ is false. Therefore $x=-4$ is a counterexample. The converse is $\forall x[q(x)\to p(x)]$. This says if $x>3$, then $|x|>3$, which is true. The inverse is $\forall x[\neg p(x)\to\neg q(x)]$. It is logically equivalent to the converse and is true. The contrapositive is $\forall x[\neg q(x)\to\neg p(x)]$. It is logically equivalent to the original statement and is false. ::: A quantified implication is false only when some element makes the hypothesis true and the conclusion false. Enter a finite sample of real numbers and compare the original implication with its converse, inverse, and contrapositive. The calculator reports whether each universal statement survives the sample and identifies the first counterexample when it fails. Notice that the original and contrapositive match, while the converse and inverse match. :::calculator[Conditional Form Comparator] ::: :::solved-problem Let the universe be the set of all real numbers. Let $p(x):|x|>3$, $q(x):x>3$, and $r(x):x<-3$. Show that $p(x)$ is equivalent to $r(x)\lor q(x)$. ::: :::solution Given that $p(x):|x|>3$, $q(x):x>3$, and $r(x):x<-3$. For a real number $x$, the inequality $|x|>3$ holds exactly when $x>3$ or $x<-3$. Therefore, $$\begin{equation*} |x|>3\quad\text{if and only if}\quad (x<-3)\lor(x>3). \end{equation*}$$ Hence $\forall x[p(x)\leftrightarrow(r(x)\lor q(x))]$ is true. ::: The converse and inverse of a quantified implication are logically equivalent to each other, but they are not necessarily equivalent to the original statement. The original implication and its contrapositive are equivalent. Therefore, when proving an implication, proving the contrapositive is valid, but proving the converse usually proves a different statement. :::exercise For each statement, write the converse, inverse, and contrapositive, and determine the truth value of the original statement: If $m>n$, then $m^2>n^2$ over positive integers; if $a>b$, then $a^2>b^2$ over integers; if $m$ divides $n$ and $n$ divides $p$, then $m$ divides $p$ over integers; $\forall x[(x>3)\to(x^2>9)]$ over real numbers. ::: :::answer The first original statement is true over positive integers. The second is false, for example $a=1,b=-2$. The third is true by transitivity of divisibility. The fourth is true, while its converse is false because $x=-4$ gives $x^2>9$ but not $x>3$. ::: :::faq[Frequently Asked Questions] Q: Which form is always equivalent to the original implication? A: The contrapositive is always equivalent to the original implication. Q: Is the converse always false? A: No. It may be true or false, but it is not guaranteed by the original implication. Q: Why does the inverse match the converse? A: The inverse is the contrapositive of the converse, so they have the same truth value. ::: :::call-to-action[Compare Conditional Forms] subtitle: Practise forming converse, inverse, and contrapositive statements before studying general laws for quantified statements. button: Next Lesson | /discrete-mathematics/mathematical-logic/predicates-and-quantifiers/logical-laws-for-quantified-statements button-ghost: Previous Lesson | /discrete-mathematics/mathematical-logic/predicates-and-quantifiers/logical-equivalence-and-implication-of-predicates :::