$p \to q$ that means (among others)

$p$ is a sufficient condition for $q$.

To show the sufficiency, I teach my study by determining the set for $p$, the set for $q$ first and comparing their cardinal numbers. If the former has lower cardinal number then $p \to q$ is a correct proportion rather than $q \to p$.

For example,

  1. p: I am in Tokyo, q: I am in Japan. The set for $p$ just contains a single city Tokyo but the set for $q$ contains many cities such as Tokyo, Osaka, Sapporo, etc. As the former set has small cardinal number then "I am in Tokyo." is a sufficient condition for "I am in Japan." or $p\to q$.

  2. p: $x=2$, q: $x^2=4$. The set for $p$ just contains a single element 2 and the set for $q$ contains 2 elements (2 and -2). Therefore, "$x=2$" is a sufficient condition for "$x^2=4$" or $p\to q$.


Now consider the following

p: I am a vegetarian.

q: I don't eat pork.

The students are asked to determine the correct implication whether "$p \to q$" or "$q \to p$".

My attempt

  • the set for p is {vegetarian}

  • the set for q is the set of people not eating pork = {vegetarian, Moslem, people who are allergic to pork, etc}

As the cardinal number of p is lower than q then $p\to q$ is the correct implication.

My student attempt

  • the set of p is the set meats the vegetarian don't eat = {pork, beef, fish, etc}

  • the set of q is {pork}

As the cardinal number of q is lower than p then "$q \to p$" is the correct implication.

I realize that my attempt is correct and the student's attempt is wrong.

As a teacher, how should I explain their fallacy in determining the set?

  • 1
    $\begingroup$ I don't see how this can have an answer, because I don't see how you can determine the 'right' sets in this case. The student has come up with two suitable sets and applied the rule you told them. I think the problem is that you've tried to teach them a method that can't be taught explicitly. It seems to me the only way to determine the sets is to already understand the implication. $\endgroup$
    – Jessica B
    Jul 18, 2016 at 5:45
  • 4
    $\begingroup$ I must confess that I find what you are trying to do very confusing, and possibly not even correct (but the ambiguities of your explanation prevent me from judging correctness). Also, I think this is making a simple idea more complex than it needs to be. $p \rightarrow q$ means that you can logically go (or "drive" if students like cars) from $p$ to $q.$ It also means that $p$ is stronger (has more information, etc.) than $q,$ and that $q$ is weaker (has less information, etc.) than $p.$ Of course, "stronger" and "weaker" here are used in their non-strict sense. $\endgroup$ Jul 18, 2016 at 20:45
  • $\begingroup$ One problem is that this approach seems to assume that either p => q or q => p, but of course that is typically not true. The most that can be said is that if p,q are predicates which define finite sets and if it is known that either p => q or q => p is true, then looking at the cardinalities of the corresponding finite sets can determine which. But -- this is clearly not a robust approach to teaching implication. At best, it can help explain the difference between a statement and its converse. $\endgroup$ Jul 21, 2016 at 11:33

3 Answers 3


First, although you talk a bunch about cardinality, I don't see how that makes sense, so I'm going to assume you mean that you have them determine if the set corresponding to p is a subset of the set corresponding to q. (Otherwise, in your second example, you'd also have $x=2$ implies $x^2=9$, for instance.)

In formal terms, your method requires translating the sentences into predicates with a free variable and then comparing the sets defined by those predicates. With the math example, this is easy because there is a free variable.

With the English examples, though, it's less obvious. In the implication "I am in Tokyo"/"I am in Japan" there are really two potential variables: the underlying form is "X is in Y". The actual comparison you want is the one with a single free variable and two predicates: you want to compare "X am in Tokyo" to "X am in Japan", that is, your sets are "the set of people in Tokyo" versus "the set of people in Japan".

Instead you've chosen to work with the single predicate "I am in Y" for two different values of Y. There's no way to make that work in pure logic: the reason it works in your example is that "X am in Y" is monotone in the predicate Y. Which means that it happens to work in that example, but for a completely different reason than the method you're trying to teach.

Your students are doing exactly the same thing with "I am a vegetarian" and "I don't eat pork". They follow you in parsing this as "I don't eat {pork, chicken, ...}" and "I don't eat pork". Again, this is a two place predicate "X don't eat Y". If you think your first example is right, you have to think their parsing, of "I don't eat Y" is correct. The problem is that "X don't eat Y" is antimonotone in Y.

If you aren't going to insist that you compare on the common free variable, there's no way to make this work without getting into questions of monotonicity of predicates, which is actually (while quite interesting, and potentially accessible to children - I recently saw a Dr. Seuss themed talk on the subject by Larry Moss) rather complicated.


I would not recommend to teach this method since there are some downsides. Take

  • $A(x) \iff x \text{ is divisible by } 2$
  • $B(x) \iff x \text{ is divisible by } 42$

Is $A(x) \implies B(x)$ or $B(x) \implies A(x)$? Since there are infinitely many $x\in\mathbb Z$ fullfiling $A(x)$ and $B(x)$ this cannot be answered unless your students already know about the cardinality of infinite sets. Actually the cardinality of $\{x\in \mathbb Z: A(x)\}$ and $\{x\in \mathbb Z: B(x)\}$ is the same. Does this mean, that $A(x) \iff B(x)$?

Here is another example:

  • $C(x) \iff x = 23$
  • $D(x) \iff x = 42 \text{ or } x = 102$

There are two objects fulfilling $D(x)$ and one object for which $C(x)$ is true. So we have $C(x) \implies D(x)$, right?!

The set-theoretic counterpart of the implication is the inclusion. You have $A(x)\implies B(x)$ whenever $\{x:A(x)\} \subseteq \{x:B(x)\}$. However, the fact that the cardinality of $\{x:A(x)\}$ is less or equal to the cardinality of $\{x:B(x)\}$ does not imply $\{x:A(x)\} \subseteq \{x:B(x)\}$. That's the reason why your method does not work in the above examples.

  • 6
    $\begingroup$ "Some downsides" is a rather mild way of putting it — the method is simply wrong, and very often leads to false conclusions. $\endgroup$ Jul 18, 2016 at 20:07

The elements of the set describing a statement are things that "make the statement true." The statement "I am in Japan" is described by the set {Tokyo, Osaka, Sapporo, ...} because "I am in Osaka" makes "I am in Japan" true. The statement "$x^2=4$" is described by the set {-2,2} because "$x=-2$" makes "$x^2=4$" true.

The statement "I don't eat pork" is described by the set {vegetarian, Moslem, a person who is allergic to pork,..} because "I am a Moslem" makes "I don't eat pork" true.

The student claims that the statement "I am a vegetarian" is described by the set {pork, beef, fish, ...}. This is incorrect: "I don't eat beef" does not (necessarily) make "I am a vegetarian" true, because, for example, I could be allergic to beef but still not be a vegetarian.

  • 1
    $\begingroup$ Unlike other answers, this answers the questions instead of dismissing the method. $\endgroup$
    – Amy B
    Jul 21, 2016 at 6:45

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