Using the formalization system I wrote of earlier, let’s take a look at this famous question.

First we should note that this is a yes/no question which is different from the questions that I have earlier formalized. The earlier questions sought to identity a certain individual, but yes/no questions do not. Instead they ask whether something is the case or not. So this time I cannot use the (x=?) question phrase from earlier, since there is no individual to identify similarly to the earlier cases.

One idea is to simply add a question mark at the end. Like this:

F1. (∃x)(∃y)(Wxy∧Bxy∧Cxy∧x=a)?

Wxy ≡ x’s wife is identical with y

Bxy ≡ x beats y

Cxy ≡ x used to beat y^{1}

a ≡ you

But this fails to capture that it is not all of the things that are being asked whether they are the case or not. It is only the Bxy part that is being asked. The rest is stipulated as true. We can change the formalization to capture this, like this:

F1*. (∃x)(∃y)(Wxy∧Cxy∧x=a)∧(Bxy)?

The question mark is now understood as a predicate that works on whatever is *before/to the left of it*. (In parentheses for clarity.) Not to the right like with the other monadic predicates and propositional connectors (¬, ◊, □, etc.). In this case the question mark only functions on (Bxy) and not the rest of the formula.

Translated into LE:

There exists an x and there exists an y such that x’s wife is y and x used to beat y and x is identical with you and is it the case that x beats y?

### Answering yes/no questions

When answered in the positive, the answered version simply removes the question mark. Producing:

F2. (∃x)(∃y)(Wxy∧Cxy∧x=a)∧(Bxy)

Answering in the negative removes the question mark and adds a negation sign to the part of the formula that the question predicate is working on. Producing this:

F3. (∃x)(∃y)(Wxy∧Cxy∧x=a)∧¬(Bxy)

If the produced formula is true, then the question has been answered correctly. However since this question is loaded. Both of the produced formulas are false, that is, it is both false that:

F2. (∃x)(∃y)(Wxy∧Cxy∧x=a)∧(Bxy)

There exists an x and there exists an y such that x’s wife is y and x used to beat y and x is identical with you and x beats y.

and that:

F3. (∃x)(∃y)(Wxy∧Cxy∧x=a)∧¬(Bxy)

There exists an x and there exists an y such that x’s wife is y and x used to beat y and x is identical with you and it is not the case that x beats y.

Since they both imply the falsehood:

(∃x)(∃y)(Wxy∧Cxy∧x=a)

There exists an x and there exists an y such that x’s wife is y and x used to beat y and x is identical with you.

1Alternatively one could deepen to formalization to formalize the temporal aspect of this predicate. Though it doesn’t seem important here so I will leave it out.