Express the multiple choice question symbolically

Symbolize the following argument using first order logic symbols. It's not necessary to solve the puzzle.

Multiple Choice Question

A: There are no correct answers B: All answers are correct C: There is exactly one correct answer D: None of the above

Let the three propositions be represented by variables A, B, C, and D. These propositions are characterized by references to each of the other propositions, and in the case of A B and C, to themselves. We will not fuss about the self-referentiality of these propositions (that's the point of such humorous brain-teasers), and hope simply that they do not give rise to an inconsistent system. Then we have:

• A ≡ ¬A & ¬B & ¬C & ¬D
• B ≡ A & B & C & D
• C ≡ (A & ¬B & ¬C & ¬D) V (¬A & B & ¬C & ¬D) V (¬A & ¬B & C & ¬D) V (¬A & ¬B & ¬C & D)
• D ≡ ¬A & ¬B & ¬C

Exercise. Prove A V B V C V D (i.e. show that the question does have a valid answer) from the premisses above.

• Looks good to me! That you can do it in the propositional fragment of 1OL doesn't mean it's not done in 1OL symbols. – Paul Ross Feb 12 '14 at 22:31
• Not to mention that this is necessary, if the first-order propositions you wish to define are meant to be logically to logical formulae involving those same propositions. (There is obviously no quantification over propositions or predicates of propositions in first order logic.) – Niel de Beaudrap Feb 13 '14 at 1:32

Let α be the set of choices/options. Let C be a unary function from α to {true, false}. Then A-D say:

A. ¬∃x ∈ α C(x).

B. ∀x ∈ α C(x).

C. ∃x ∈ α C(x) ∧ ∀y ∈ α C(y) → y = x.

D. [¬∃x ∈ α C(x)] ∧ [∀x ∈ α C(x)] ∧ [∃x ∈ α C(x) ∧ ∀y ∈ α C(y) → y = x].

Consider a concrete example for illustration:

Question. Animals X eat which of the following?

1. fruits
2. vegetables
3. meat
4. candy

A. There are no correct answers
C. There is exactly one correct answer
D. None of the above

My interpretation of the question is that "answers" is referring to (in this example) 1-4, not A-D. If that's the case, then A-D say:

A says: animals X eat neither fruits nor vegetables nor meat nor candy.
B says: animals X eat fruits, vegetables, meat, and candy.
C says: animals X eat one and only one of: fruits, vegetables, meat, and candy.
D says: [the denial of the conjunction of A-C] (consistent?, I think so).

• Maybe there might be more to this answer! It's not clear that α is a well-formed set, so we should probably try to construct it. Suppose Correct is something like a Tarskian Truth predicate, and that it's okay to help ourselves to it. Let's relax the condition that α is "the" set of choices or options and instead just use Q as a generic parameter for a set of sentences. Let 'A' be the sentence scheme ' ¬∃x ∈ Q Correct(x)', let 'B' be the sentence scheme '∀x ∈ Q Correct(x)', let 'C' be '∃x ∈ Q . Correct(x) ∧ ∀y ∈ Q (Correct(y) → y = x)' and 'D' be the same substitution. – Paul Ross Feb 13 '14 at 21:56
• Now the problem we have is that while there is absolutely no problem in using these schematic definitions, they're just a collection of sentence schemes, not in themselves well-defined sentences. In order to get well-formed sentences, we need to substitute in values for Q. What values can Q possibly take? Well, since the Qs need to be well-defined sets, one thing Q certainly can't be is the very same set with the instances of 'A', 'B', 'C' and 'D' featuring itself as the parameter, because then you could define an indirect "Liar Set" with just 'A'! – Paul Ross Feb 13 '14 at 22:52
• You can get some fun technical stuff by playing with well-defined sets of sentences Q; I think there might be something interesting in running a revision-theory (plato.stanford.edu/entries/truth-revision) strategy towards getting the right answer! But I guess this is way beyond what the original questioner was interested in. – Paul Ross Feb 13 '14 at 23:41
• Paul, you are crazy! But thanks, I learned something. Now, since my answer has been superseded by Niel's, I think it won't hurt to try to modify my answer; I think D in particular is wrong and I think we (me and (you and Niel)) have interpreted the question differently. – Hunan Rostomyan Feb 14 '14 at 4:02

The problem with doing this in a first-order theory is that the choices are what Saul Kripke, in his 1975 paper "Outline of a Theory of Truth" called Ungrounded sentences. Neither sentence says anything substantial except in reference to each other.

• A is simply false, since if A then ¬A, and thus A & ¬A.
• Since A is false, then so is B.
• D says that none of the above are true, so the informative content of its assertion in the context of both A and B being false is simply that C is false.
• C says that exactly one of A, B, C or D are true. That one can't be A or B. So, since C can't also say of itself that it is false (since that would be inconsistent for the same reasons as A), it must therefore say that it is true and D is false.
• So, C if and only if not D, and vice versa.

(Edit: as Niel has quite rightly pointed out, since D being true and C false would make C also true, D must be false as well, so in fact C simply has to be true, despite saying nothing more than that C is the only one of the sentences that's true!)

There is nothing Paradoxical about this state of affairs. The case where exactly one of A, B, C or D are true is an absolutely consistent response; it's just that there's nothing in the world involved in the assertion. What this suggests is that there's nothing in fact involved in the content of A, B C and D in formulating this puzzle, but simply that Niel's answer above is exactly right in restricting its interest to Propositional interpretations, and that we will ultimately be interested in those propositional interpretations satisfying ¬A, ¬B, C and ¬D.

• Actually, there is a way to decide between them. (Hint: what happens if C is false?) – Niel de Beaudrap Feb 13 '14 at 1:48
• Of course! #*Forehead slap*#. So I wonder then if, since the sentences need not be ungrounded in the sense I was thinking there might be a syntactic theory along Hunan's lines borrowing from a Tarskian truth construction. – Paul Ross Feb 13 '14 at 21:46