A universal statement applies to every element of its universe. Universal specification is the rule that allows us to use such a statement for a particular object. This rule is the bridge between quantified premises and ordinary propositional inference rules such as Modus Ponens, Modus Tollens, Hypothetical Syllogism, and Disjunctive Syllogism. In this lesson, universal specification in quantified arguments is used to test validity and identify common conditional errors. :::definition[Universal Specification] The Rule of Universal Specification states that if an open statement is true for every element of a universe, then it is true for each particular element of that universe. If $a$ is a specific element of the universe and $\forall x\,p(x)$ is true, then $p(a)$ is true. ::: :::definition[Rule of Universal Specification] The rule is written as $$\begin{equation*} \frac{\forall x\,p(x)}{\therefore p(a)}. \end{equation*}$$ Here $a$ must be an element of the universe under consideration. ::: :::theorem[Universal Specification] Let $p(x)$ be an open statement over a universe $U$. If $\forall x\in U,\ p(x)$ is true and $a\in U$, then $p(a)$ is true. ::: :::proof Given that $\forall x\in U,\ p(x)$ is true. To prove that $p(a)$ is true for a particular element $a\in U$. Since the statement is true for every element of $U$, it is true for the element $a$. Therefore $p(a)$ is true. Hence the rule of universal specification is valid. $\square$ ::: :::definition[Universal Specification with Modus Ponens] The combined pattern $$\begin{equation*} \frac{\forall x[p(x)\to q(x)]\qquad p(a)}{\therefore q(a)} \end{equation*}$$ uses universal specification first to obtain $p(a)\to q(a)$, and then uses Modus Ponens. ::: :::definition[Universal Specification with Modus Tollens] The combined pattern $$\begin{equation*} \frac{\forall x[p(x)\to q(x)]\qquad \neg q(a)}{\therefore \neg p(a)} \end{equation*}$$ uses universal specification first to obtain $p(a)\to q(a)$, and then uses Modus Tollens. ::: Universal specification lets a universal rule act on a named object. We observe the rule $\forall x[p(x)\to q(x)]$ after it is applied to a particular object $a$. Choose a reasoning pattern and compare the premise about $a$ with the conclusion that is claimed. The checker marks Modus Ponens and Modus Tollens as valid, while the converse and inverse patterns are rejected. This matters because a universal statement gives only the conditional direction that was proved. :::scientific-preview[Universal Specification Rule Checker] ::: :::example[Mathematics Professors and Calculus] Let the universe be all people. Define $m(x):x$ is a mathematics professor and $c(x):x$ has studied calculus. The argument “All mathematics professors have studied calculus. Leona is a mathematics professor. Therefore Leona has studied calculus” has symbolic form $$\begin{equation*} \frac{\forall x[m(x)\to c(x)]\qquad m(l)}{\therefore c(l)}. \end{equation*}$$ ::: :::solved-problem Verify the validity of the mathematics-professor argument. ::: :::solution Given that $\forall x[m(x)\to c(x)]$ and $m(l)$. To prove that $c(l)$. By universal specification, $$\begin{equation*} m(l)\to c(l). \end{equation*}$$ Since $m(l)$ is true, Modus Ponens gives $c(l)$. Therefore Leona has studied calculus. Hence the argument is valid. ::: :::solved-problem Let the universe be all triangles in the plane. Define $p(t):t$ has two sides of equal length, $q(t):t$ is an isosceles triangle, and $r(t):t$ has two angles of equal measure. Let triangle $XYZ$ be denoted by $c$. Verify the argument: $\neg r(c)$, $\forall t[p(t)\to q(t)]$, $\forall t[q(t)\to r(t)]$, therefore $\neg p(c)$. ::: :::solution Given that $\neg r(c)$, $\forall t[p(t)\to q(t)]$, and $\forall t[q(t)\to r(t)]$. To prove that $\neg p(c)$. By universal specification, $p(c)\to q(c)$. Again, by universal specification, $q(c)\to r(c)$. Since $p(c)\to q(c)$ and $q(c)\to r(c)$, Hypothetical Syllogism gives $p(c)\to r(c)$. Since $\neg r(c)$ is true, Modus Tollens gives $\neg p(c)$. Therefore triangle $XYZ$ does not have two sides of equal length. Hence the argument is valid. ::: :::solved-problem Let the universe be the student body at a particular college. Define $j(x):x$ is a junior, $s(x):x$ is a senior, and $p(x):x$ is enrolled in a physical education class. Let Mary Gustin be denoted by $m$. Verify the argument: no junior or senior is enrolled in a physical education class; Mary is enrolled in a physical education class; therefore Mary is not a senior. ::: :::solution Given that $\forall x[(j(x)\lor s(x))\to\neg p(x)]$ and $p(m)$. To prove that $\neg s(m)$. By universal specification, $(j(m)\lor s(m))\to\neg p(m)$. Since $p(m)$ is true, $\neg\neg p(m)$ is true. By Modus Tollens, $\neg(j(m)\lor s(m))$ follows. By De Morgan’s law, $\neg j(m)\land\neg s(m)$. By simplification, $\neg s(m)$. Therefore Mary Gustin is not a senior. Hence the argument is valid. ::: :::example[Invalid Converse from a Universal Statement] Let $p(x):x$ is a square and $q(x):x$ has four sides. The argument $$\begin{equation*} \frac{\forall x[p(x)\to q(x)]\qquad q(c)}{\therefore p(c)} \end{equation*}$$ is invalid. It has the form of affirming the consequent. A quadrilateral may have four sides without being a square. ::: :::example[Invalid Inverse from a Universal Statement] With $p(x):x$ is a square and $q(x):x$ has four sides, the argument $$\begin{equation*} \frac{\forall x[p(x)\to q(x)]\qquad \neg p(c)}{\therefore \neg q(c)} \end{equation*}$$ is invalid. It has the form of denying the antecedent. A figure may fail to be a square and still have four sides. ::: :::solved-problem Determine whether each argument is valid or invalid. (1) All mail carriers carry a can of mace. Mrs. Bacon is a mail carrier. Therefore Mrs. Bacon carries a can of mace. (2) All law-abiding citizens pay their taxes. Mr. Pelosi pays his taxes. Therefore Mr. Pelosi is a law-abiding citizen. (3) All people concerned about the environment recycle plastic containers. Margarita is not concerned about the environment. Therefore Margarita does not recycle plastic containers. ::: :::solution Given the first argument has the form $\frac{\forall x[p(x)\to q(x)]\qquad p(a)}{\therefore q(a)}$, it is valid by universal specification and Modus Ponens. The second argument has the form $\frac{\forall x[p(x)\to q(x)]\qquad q(a)}{\therefore p(a)}$, so it is the converse error and invalid. The third argument has the form $\frac{\forall x[p(x)\to q(x)]\qquad \neg p(a)}{\therefore \neg q(a)}$, so it is the inverse error and invalid. ::: :::exercise Fill in the missing conclusion and identify the rule. (1) All integers are rational numbers. The real number $\pi$ is not rational. Therefore what follows? (2) All librarians know the Library of Congress Classification System. Margaret is a librarian. Therefore what follows? (3) All rectangles are equiangular. Quadrilateral $MNPQ$ is not equiangular. Therefore what follows? ::: :::answer (1) $\pi$ is not an integer, by universal specification and Modus Tollens. (2) Margaret knows the Library of Congress Classification System, by universal specification and Modus Ponens. (3) Quadrilateral $MNPQ$ is not a rectangle, by universal specification and Modus Tollens. ::: :::faq[Frequently Asked Questions] Q: What does universal specification allow? A: It allows a universal statement to be applied to a particular element of the universe. Q: Does $\forall x[p(x)\to q(x)]$ and $q(a)$ imply $p(a)$? A: No. That is the converse error. Q: Does $\forall x[p(x)\to q(x)]$ and $\neg p(a)$ imply $\neg q(a)$? A: No. That is the inverse error. ::: :::call-to-action[Apply Universal Premises] subtitle: Practise applying universal statements to particular objects before proving universal conclusions. button: Next Lesson | /discrete-mathematics/mathematical-logic/quantifiers-definitions-and-proofs-of-theorems/universal-generalization-in-theorem-proofs button-ghost: Previous Lesson | /discrete-mathematics/mathematical-logic/quantifiers-definitions-and-proofs-of-theorems/proof-by-exhaustion-and-theorem-terminology :::