# Does this english-language sentence translate to the drinker paradox?

There are (one or) two students such that if they pass the exam, then every student passes the exam.

I'm tasked to formalize the above sentence and give either a proof or counterexample. If we take the task "formalize the sentence" as "construct a formula that best captures the real world meaning", then I think that $$\exists s1 \exists s2 (student(s1) \land student(s2) \land ((passes(s1) \land passes(s2)) \rightarrow (\forall x (student(x) \rightarrow passes(x)))))$$ is an intuitive formalization that seems to perfectly model the statement.

However, our professor insists that the correct formalization is $$\exists s1 \exists s2 ((student(s1) \land student(s2) \land passes(s1) \land passes(s2)) \rightarrow \forall x (student(x) \rightarrow passes(x))),$$ which is supposed to refer to the so-called drinker paradox. This formula is valid, i.e. true in all interpretations, since it's just the drinker paradox in disguise. However, I can't see how this would be the "correct" formalization of the sentence above. In terms of natural language, the sentence is obviously incorrect. As I understand this article, the whole paradox seems to stem from an "error" in formalizing statements, namely relating this type of sentences to the theorem $$\exists x (\neg D(x) \lor \forall y ( D(y)))$$, which states that either there is some $$x$$ that does not have property $$D$$ (i.e. a counterexample), or $$D(y)$$ holds for all $$y$$, i.e. all objects in the domain have the property.

Going back to my own answer: that formula is not universally true: we can easily construct a counterexample with a domain of 3 students where 2 pass, but one doesn't. This seems way better aligned with the real world, given that there clearly are no two students that "make everyone else pass" in any given exam. It seems to me that my answer better captures the original sentence. In the "drinker paradox version", "being a student" and "passing the exam" are treated as one property (in the sense that both together form the antecedent of the implication). Here we have two separate properties where one ("student") has to hold and if the other ("passes") also holds, we assert that the universal statement holds. Or, in other words, to me it seems reasonable to understand There are (one or) two students such that ... to mean that, for the whole sentence to be true, there must first be two students (which may be identical).

Is there some error in my argument, or is there some reason why formalizing this sentence is bound to end up in an instance of the drinker paradox?

• If you're just asking whether the second chunk formalises the given natural-language sentence, the answer is Yes. Mar 24 at 16:09
• Thanks, that's exactly what I'm asking - it seems totally reasonable to me, but our professor insists that the correct formalization is the first chunk (i.e. drinker paradox). I'm trying to get to grips with why they'd think so when there is an intuitive formalization that (to me) seems to perfectly model the statement. Mar 24 at 16:14
• I don't think the first formulation is exactly the drinker's paradox. It says there are two things in the universe such that if they are both students and both passed the exam, then all students passed. This is certainly true if the universe has at least two things in it; for example, let $s_1$ be the Rock of Gibraltar and let $s_2$ be France. But in the drinker's paradox you can't instantiate the existential with the Rock of Gibraltar, it has to be someone in the pub. Mar 24 at 16:35
• The second formulation is analogous to the drinker's paradox, and it is true as long as there are at least two students. For if everyone passed, instantiate with any two students; but if someone failed, instantiate with one failed student and any other student. (Your supposed counterexample isn't a counterexample.) Mar 24 at 16:37
• Thanks a lot @DavidK, that clears it up! Mar 24 at 17:19

Suppose we have an interpretation over domain $$D = \{a, b, c\}$$ where $$I(student) = \{a, b, c\}$$ and $$I(passes) = \{a, b\}$$. Then we can instantiate $$s1$$ and $$s2$$ with a pair of students where one did not pass (e.g. $$b, c$$). In this case, the subformula $$(passes(b) \land passes(c)) \rightarrow \forall x(student(x) \rightarrow passes(x))$$ is actually true since $$\neg(passes(a) \land passes(c))$$ is true.