After reading other similar questions I got even more confused:

(1) “Every student in this class has visited Mexico”

$\forall x(S(x) \rightarrow M(x))$

For (example 1) we could either get:

  1. T-T (Is a student and has visited Mexico)

  2. F-T (Not a student and has visited Mexico)

  3. F-F (Not a student and hasn't visited Mexico)

And I was told this makes sense because example 1 can be re-written as "For every person x, if that person is a student then he has visited Mexico", and we can have the cases F-T, and F-F because they don't go against the evidence of our statement since we didn't say anything about when they're neither a student or went to Mexico.

(2) “Some student in this class has visited Mexico”

$\exists x(S(x) \land M(x))$ is right.

$\exists x(S(x) \rightarrow M(x))$ is wrong

  1. If person is a student, then he went to Mexico
  2. If a person is not a student, then he went to Mexico
  3. If a person is not a student, then he didn't go to Mexico.

In example 1 we came to the conclusion that we can have cases 2 and 3 because we didn't say anything about when they're neither a student or went to Mexico, but now for the existential quantifier with implication, we're not allowed to do that because we're not precisely capturing the desired values, since "there exists a person who is not a student" makes the statement true.

I really don't understand why implication works for Universal but doesn't work for Existential


It's because, as they say, if there is an object in the domain that is not a student, then $S(x)$ would be false, and given the way that we have defined the truth-functional implication, the $S(x) \rightarrow M(x)$ part would then be automatically true. Hence, the statement $\exists x (S(x) \rightarrow M(x))$ can be made true by simply some object in the domain that is not a student ... and so if there are in fact no students in the domain at all, or if there are students, but none of them have visited Mexico, the statement would still be true... which is of course not what we want. We do want to say that there is a student, and that they visited Mexico, i.e. $\exists x (S(x) \land M(x))$

For the universal, things are different. If we were to use $\forall x (S(x) \land M(x))$, then we are saying that everything in the domain is a student and went to Mexico ... so if we would have objects in the doamin other than student, e.g. bananas, then we end up pointing to those bananas and say: "that is a student, and it visited Mexico" ... which is clearly not what we want ... We are merely saying that all students visited Mexico, not that everything visited Mexico. That is, we should say that out of all objects: if it is a student, then 'it' went to Mexico ... i.e. $\forall x (S(x) \rightarrow M(x))$

  • $\begingroup$ That's my exact confusion, can't the same argument be used for the universal example? The statement is still true even though we're capturing people that are not students and haven't visited Mexico? And that still makes the statement true and not what we want? $\endgroup$ – user3252 Sep 28 '17 at 15:10
  • $\begingroup$ @user32523850925 Just added that to my post! $\endgroup$ – Bram28 Sep 28 '17 at 15:12
  • $\begingroup$ Thank you for adding to your post. The AND makes sense not to use with universal but I can't wrap my head around implication! But if we had a universe (Apple, Bob, John) then $\forall x(S(x) \rightarrow M(x))$ is true even though Apple is not a student and has not went to Mexico? Why are we allowed to capture such non sensical values? $\endgroup$ – user3252 Sep 28 '17 at 15:25
  • $\begingroup$ @user32523850925 But please recognize that it is a good thing that the sentence is true for Apple, because we are using a universal, and so the statement should be true for all objects. That is, the conditional is true for all objects, but only for the students it will be true that they went to Mexico. To be precise, Apple may have visited Mexico as well, but the statement does not force that to be true ... indeed it ends up saying nothing of interest for those objects that are not students, and again that's exactly what we want. $\endgroup$ – Bram28 Sep 28 '17 at 15:53

In the second case, if there were no students in the class the implication would be true. The normal reading of the sentence is clearly false if the class is empty. As long as there is at least one student in the class the implication works fine.

  • $\begingroup$ maybe I'm confused I thought $\forall \to S\subseteq M$ versus $\exists\to S\cap M\neq\emptyset$ $\endgroup$ – user451844 Sep 28 '17 at 15:10

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