Universal Specification
Throughout our discussion of arguments so far, we haven’t made use of any quantified statements — though we devoted several sections to quantifiers in the previous chapter, so we certainly got some mileage out of them there. Here, we start to discuss how quantified statements can be used in arguments.
The reason we want to do this is that many of the results we’re going to come across are stated in the language of quantifiers. For example, consider the Pythagorean Theorem:
$$\text{If any triangle has leg lengths } a \text{ and } b \text{, and hypotenuse length } c \text{, then } a^2 + b^2 = c^2.$$Notice that implicit in this statement is the universal quantifier “if any.” We could write the Pythagorean Theorem using our current mathematical symbology as follows:
\[ \begin{array}{rl} \mathcal{U}\text{: } &\text{All planar triangles.} \\ p(t)\text{: } &t \text{ is a right triangle with leg lengths } a \text{ and } b, \text{ hypotenuse } c. \\ q(t)\text{: } &a^2 + b^2 = c^2. \end{array} \]$$\forall t\ [p(t) \to q(t)]$$The Pythagorean Theorem works for every single conceivable right triangle in the plane — not just some specific kind, but every single one. That’s the power of the theorem: it lets us compute a side length of any right triangle when the other two are known. Since scientists, engineers, architects, and mathematicians all need to calculate lengths of triangles constantly, this theorem has come in handy very often. It would be near-useless if it only applied to one triangle, like a $3$-$4$-$5$ triangle — it wouldn’t be nearly as widely known or applicable as it is today.
This is why we care about quantifiers: they let us extend results beyond a single example to potentially infinitely many of them. So much of what’s calculable in engineering, science, and mathematics is possible only because the underlying theorems are so extensive — and that’s precisely because they’re quantified.
Here, we discuss Universal Specification, one way to use quantifiers in arguments that lets us go from broadly true statements to specifically true statements.
Motivating Examples
At a particular school, one of the most loved teachers by the students is Ms. Lippy, a very creative and sometimes eccentric teacher who loves the color green. Suppose we knew:
\[ \begin{array}{l} \text{Everything Ms. Lippy owns is green.} \\[0.75em] \text{Ms. Lippy owns a car.} \end{array} \]What, if anything, can we figure out? Based on the first piece of information, we can sort every object into two categories: objects owned by Ms. Lippy, and objects not owned by Ms. Lippy. If an object isn’t owned by Ms. Lippy, we don’t know anything about its color — it could be green, or some other color entirely. The only objects we’re told anything about are those Ms. Lippy owns: they’re all green!
We also know Ms. Lippy’s car is an object she owns. As such, we’re guaranteed to know Ms. Lippy’s car is green!
One of Ms. Lippy’s favorite students is named Billy, who has a very wild and active imagination. As part of his education, he’s required to take Ms. Lippy’s art class. Suppose we know:
\[ \begin{array}{l} \text{Every duck that Billy draws is blue.} \\[0.75em] \text{Billy drew a duck for his art class assignment.} \end{array} \]What, if anything, can we conclude? Just as before, we have two categories: ducks drawn by Billy, and ducks not drawn by Billy. Since Billy drew a duck for his assignment, that duck must have been drawn blue — there are no exceptions to the first piece of information, so that duck is guaranteed to be blue, because it was drawn by Billy, and every duck Billy draws is blue.
Billy isn’t the only student in Ms. Lippy’s art class — she manages a lot of students across all of her classes. Consider:
\[ \begin{array}{l} \text{Every duck that Billy draws is blue.} \\[0.75em] \text{One duck submitted for Ms. Lippy's art class was not blue.} \end{array} \]What can we conclude here? Examining this a little more closely than before, we infer the implication “If Billy draws a duck, then that duck is blue,” which we can represent as $\text{billy} \to \text{blue}$ — using words instead of single letters for clarity: “billy” for “Billy drew a duck,” and “blue” for “The duck is blue.”
What we know is that a duck was submitted that was not blue — that is, $\neg \text{blue}$. Since we know $\text{billy} \to \text{blue}$ and $\neg \text{blue}$, Modus Tollens tells us we must have $\neg \text{billy}$. So, Billy did not draw that particular duck — because if he had, it would definitely have been blue.
The Rule of Universal Specification
Thinking back to the previous examples, the general strategy was to figure out what classifications were in use, then figure out which classification an object belonged to. Once we knew an object’s category, we knew it had a certain property, since that property was shared by every object in the category.
Consider an open statement $p(x)$ defined on some universe $\mathcal{U}$.
If $p(x) = 1$ for every replacement of $x$ by every element within $\mathcal{U}$, then $p(x)$ takes on truth value $1$ when $x$ is replaced by a specifically chosen element within $\mathcal{U}$, which we’ll refer to as $c$.
In other words, if $\forall x \in \mathcal{U}\ [p(x)] = 1$ and $c \in \mathcal{U}$, then $p(c) = 1$ as well.
This is the formal statement of what we were trying to say in the previous three examples. When we spoke of “categories” or “kinds,” we were dealing with inclusion within the universe $\mathcal{U}$ — an element $c$ belonging to the category is the same thing as saying $c \in \mathcal{U}$; not belonging is $c \notin \mathcal{U}$ (read “not a member of,” or “not an element of” — analogous to the inequality $\neq$ symbol).
The next part of this rule is to notice there’s an implicit implication: $c \in \mathcal{U} \to p(c) = 1$. So, if $c$ is an element of the universe, then $p(c)$ is true. But if $c$ is not an element of the universe, we don’t know whether $p(c) = 0$ or $p(c) = 1$, because either way, the implication is trivially true. The argument
\[ \begin{array}{l} \forall x \in \mathcal{U}\ [p(x)] \\ c \in \mathcal{U} \\ \hline \therefore p(c) \end{array} \]is valid. Just like with the other rules of inference, this is a valid rule usable in the analysis of a mathematical argument. The intuition is that if every member of a group satisfies some property, then picking any element from that group means the chosen element satisfies that property too.
Universal Specification and Modus Ponens
With the quantified statement $\forall x \in \mathcal{U}\ [p(x)] = 1$, $p(x)$ may represent a primitive statement, or a compound one. Very often, $p(x)$ represents an implication — for example, if $p(x)$ represents $a(x) \to b(x)$, we could rewrite the quantified expression as $\forall x \in \mathcal{U}\ [a(x) \to b(x)] = 1$.
Let’s re-examine the “everything Ms. Lippy owns is green” example. First, we figure out the applicable universe of discourse: we’ll use $\mathcal{U}$ for every possible object in existence, since this problem is fundamentally about objects, whether or not they’re owned by Ms. Lippy, and whether or not they’re green. We pick out the propositions
\[ \begin{array}{rl} \ell(x)\text{: } &x \text{ is an object owned by Ms. Lippy.} \\ g(x)\text{: } &x \text{ is green.} \end{array} \]The phrase “everything” in “everything Ms. Lippy owns is green” suggests the universal quantifier, applied to every possible object. That statement, and the fact that being owned by Ms. Lippy implies being green, rewrites as
$$\forall x \in \mathcal{U}\ [\ell(x) \to g(x)].$$Next, “Ms. Lippy owns a car” tells us the object in question — her car, which we’ll call $c$ — exists, so $c \in \mathcal{U}$. It also tells us that $c$ is owned by Ms. Lippy, so $\ell(c)$. We now have three true propositions to serve as premises:
$$\forall x \in \mathcal{U}\ [\ell(x) \to g(x)] \qquad c \in \mathcal{U} \qquad \ell(c).$$Let’s analyze this argument:
\[ \begin{array}{lll} \textbf{Step} & \textbf{Proposition} & \textbf{Reason} \\ (1) & \forall x \in \mathcal{U}\ [\ell(x) \to g(x)] & \text{Premise} \\ (2) & c \in \mathcal{U} & \text{Premise} \\ (3) & \ell(c) \to g(c) & \text{Universal Specification on (1) and (2)} \\ (4) & \ell(c) & \text{Premise} \\ (5) & \therefore g(c) & \text{Modus Ponens on (3) and (4)} \end{array} \]Our final conclusion is $g(c)$, corresponding to “Ms. Lippy’s car is green” — the same conclusion we reached before, now derived formally.
The argument
\[ \begin{array}{l} \forall x \in \mathcal{U}\ [a(x) \to b(x)] \\ a(c) \\ \hline \therefore b(c) \end{array} \]is valid — a combination of Modus Ponens and the Rule of Universal Specification, for any open statements $a(x)$ and $b(x)$ defined on some universe $\mathcal{U}$.
Universal Specification and Modus Tollens
If we can combine the Rule of Universal Specification with Modus Ponens, surely we can combine it with Modus Tollens too.
Let’s re-examine the “one duck was not blue” example, where the universe of discourse is all ducks, denoted $D$ instead of $\mathcal{U}$. We pick out the propositions
\[ \begin{array}{rl} s(x)\text{: } &x \text{ is a duck drawn by Billy.} \\ t(x)\text{: } &x \text{ is blue.} \end{array} \]giving us the premises $\forall x \in D\ [s(x) \to t(x)]$ and $\neg t(d)$, where $d$ is the non-blue duck submitted for the assignment. Now we use the rules of inference to make a valid deduction:
\[ \begin{array}{lll} \textbf{Step} & \textbf{Proposition} & \textbf{Reason} \\ (1) & \forall x \in D\ [s(x) \to t(x)] & \text{Premise} \\ (2) & d \in D & \text{Premise} \\ (3) & s(d) \to t(d) & \text{Rule of Universal Specification on (1) and (2)} \\ (4) & \neg t(d) & \text{Premise} \\ (5) & \therefore \neg s(d) & \text{Modus Tollens on (3) and (4)} \end{array} \]Our conclusion, $\neg s(d)$, represents “The duck was not drawn by Billy” — again matching the conclusion we reached before.
Notice that we listed $d \in D$ as a premise, even though it wasn’t explicitly one of the argument’s premises — inclusion in the universe is often an implicit assumption, since analyzing an object not in the universe wouldn’t tell us anything useful.
It’s also worth pointing out that even though one of the premises is a quantified statement, we’re able to extract a non-open statement from it using the Rule of Universal Specification: $s(d) \to t(d)$ is not an open statement, because both $s(d)$ and $t(d)$ have definite truth values, since $d$ is a specified member of $D$, not a placeholder like $x$.
The argument
\[ \begin{array}{l} \forall x \in \mathcal{U}\ [a(x) \to b(x)] \\ \neg b(c) \\ \hline \therefore \neg a(c) \end{array} \]is valid — combining the Rule of Universal Specification with Modus Tollens. In both cases, once we have a non-open statement, we can use any of the other rules of inference we’ve learned — we’re not limited to Modus Ponens and Modus Tollens.
For a more mathematical example, consider the propositions
\[ \begin{array}{rl} s(x)\text{: } &\text{The opposite angles of quadrilateral } x \text{ are supplementary.} \\ p(x)\text{: } &\text{The perpendicular bisectors of the sides of } x \text{ are all concurrent.} \\ c(x)\text{: } &\text{Quadrilateral } x \text{ is a cyclic quadrilateral.} \end{array} \]Here, the universe $\mathcal{U}$ is all planar quadrilaterals, and let $q$ represent quadrilateral $ABCD$. Consider the argument
\[ \begin{array}{l} \forall x\ [(s(x) \lor p(x)) \to c(x)] \\ \neg c(q) \\ \hline \therefore \neg s(q) \end{array} \]We use the rules of inference to determine validity:
\[ \begin{array}{lll} \textbf{Step} & \textbf{Proposition} & \textbf{Reason} \\ (1) & \forall x\ [(s(x) \lor p(x)) \to c(x)] & \text{Premise} \\ (2) & \neg c(q) & \text{Premise} \\ (3) & \neg (s(q) \lor p(q)) & \text{Modus Tollens + Universal Specification on (1) and (2)} \\ (4) & \neg s(q) \land \neg p(q) & \text{DeMorgan's Law on (3)} \\ (5) & \therefore \neg s(q) & \text{Conjunctive Simplification on (4)} \end{array} \]So the argument is valid — if planar quadrilateral $ABCD$ isn’t cyclic, its opposite angles must not be supplementary.
Arguing by the Converse and Inverse
The argument
\[ \begin{array}{l} \forall x \in \mathcal{U}\ [a(x) \to b(x)] \\ b(c) \\ \hline \therefore a(c) \end{array} \]is invalid, since it combines argument by the converse with the Rule of Universal Specification. Likewise, the argument
\[ \begin{array}{l} \forall x \in \mathcal{U}\ [a(x) \to b(x)] \\ \neg a(c) \\ \hline \therefore \neg b(c) \end{array} \]is invalid, combining argument by the inverse with the same rule.
One should be careful when analyzing arguments — just as arguing by the converse or inverse is a fallacy without quantified statements, it’s equally fallacious with them.
Consider the universe $Q$ of all planar quadrilaterals, along with
\[ \begin{array}{rl} s(x)\text{: } &x \text{ is a square.} \\ r(x)\text{: } &\text{Every angle of } x \text{ is a right angle.} \end{array} \]Let $q$ represent quadrilateral $ABCD$, and consider the argument
\[ \begin{array}{l} \forall x \in Q\ [s(x) \to r(x)] \\ \neg s(x) \\ \hline \therefore \neg r(q) \end{array} \]We can find many counterexamples showing this argument is invalid, since it’s essentially arguing by the inverse — just because a quadrilateral isn’t a square doesn’t mean it doesn’t have all right angles. One such example is a rectangle whose sides measure $4$ units and $2$ units; another is a rectangle whose sides measure $2.718$ units and $3.142$ units. Every angle of every rectangle is a right angle.