So, i aktualy roet sum filosofy for a chanj and the analytik kiend. Heer is the paeper i handed in for mie eksam in Filosofikal Lojik. I plan on ekspanding the paeper to a book-ish length projekt. I had to eksklood multipl things in moe paeper bekus i ran out of spaes (limited to 15 pajes, = 15·2400 symbols).

## Archive for the ‘Erotetic’ Category

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.

Formalization of questions and answers is not a much discussed nor studied topic. Though there is a branch of logic dealing with it, erotetic logics. I admit not to have read much on the issue, in fact, almost nothing. This is because there is almost nothing on the internet about it, and the few books that deal with it are not to find in the danish library system.

The idea is to formalize questions and the answering of such. With formalization we have a framework for understanding when a question has been answered. It seems to me that we only have our intuitions to rely on until a relevant logic has been created, and intuitions are too often not trust worthy.

### Formalization of questions

Consider the question:

Q1. Who is credited with discovering the element Uranium?

How should we formalize this? How about:

Q1F. (∃x)(Ux∧x=?)

Ux ≡ is credited with discovering the element Uranium

Though there is a hidden second condition for a correct answer to this question. Note the word “who”. That that word is used implies that the answer refers to a person. So we may add:

Q1F*. (∃x)(Ux∧Px∧x=?)

Ux ≡ is credited with discovering the element Uranium

Px ≡ is a person

Consider a more complex question such as:

Q2. Which scientist is credited with discovering the element Uranium and whose name is Martin and who is german?

Formalization:

Q2F. (∃x)(Ux∧Sx∧Mx∧Gx∧x=?)

Ux ≡ is credited with discovering the element Uranium

Sx ≡ is a scientist

Mx ≡ is named Martin

Gx ≡ is german

Note how we can now formally make sense of/explain complexity of questions. Complexity correlates with the number of predicates in the formalization. But since there are multiple possible formalizations one need to be careful with this correlation. It may be the case that it corresponds perfectly or nearly so with the deepest possible formalization of a question.

### Answers to questions

When should we say that an answer has been answered truthfully? Iff substituting the question mark “?” with the answer gives a true formula.^{1}

Consider the correct answer to the first question above:

A1. Martin Heinrich Klaproth

And its formalization:

A1F. m

m ≡ Martin Heinrich Klaproth

Now substituting “?” with “m”:

Q1W. (∃x)(Ux∧x=m)

(Q1W) is true, therefore, the answer is correct.

Similarly with the second question:

Q2W. (∃x)(Ux∧Sx∧Mx∧Gx∧x=m)

Again this is true.

### Translating to Logic English

Logic English (LE) being the language where to formalizations are translated, as with propositional logic and predicate logic. This is a precise language that corresponds perfectly with logic. (Another thing that I am working on.)

Consider again the first question unanswered and answered:

Q1F. (∃x)(Ux∧x=?)

Q1W. (∃x)(Ux∧x=m)

I propose this translation to LE:

There exists one x such that x is credited with discovering the element Uranium and x is identical with who?

There exists one x such that x is credited with discovering the element Uranium and x is identical with Martin Heinrich Klaproth.

### Questions as expressing propositions

It is clear from my chosen formalization and translation that I think questions express propositions. So they can express something that is true or false like other sentences. This should not be too surprising. Consider the loaded question fallacy. That happens *exactly* when one asks a question that expresses a false proposition. A proposition expressed by the question is false iff it is not the case that there exists an x such that [predicates]. The clause with the “x=?” is ignored when evaluating truth values.

So, by accepting that some questions express propositions we can make sense of the loaded question fallacy.

### Multiple correct answers

It is possible formalize the multiple correct answers aspect of questions. Consider the question:

Q3. What is x identical with in the equation x^{2}=4?

Formally:

Q3F. (∃x)(x^{2}=4∧x=?)

There are two correct answers to this question:

A3a. 2

A3b. -2

Substituting “?” with the answers produces:

Q3Wa. (∃x)(x^{2}=4∧x=2)

Q3Wb. (∃x)(x^{2}=4∧x=-2)

Which are both true formula. So, there are more than one true answer. We could formalize this as (with help from set theory):

Q3F*. (∃x)(x^{2}=4)∧((∀y)(y^{2}=4⇒y∈A∧M_{A}>1))∧(x=?)

M_{S} ≡ the number of members of S

A ≡ (the set of) answers

We can also translate this to LE:

There exists an x such that x^{2}=4, and for all y, that y^{2}=4 logically implies that y is a member of A and the number of members of A is more than 1, and x is identical with what?

It seems to me to be best to always have the question clause “x=?” at the end of a formula. Otherwise a translation of the formula into LE would produce a sentence that has a question mark in the middle of it. That would be grammatically incorrect. I strive to make LE grammatically correct or close to. The goal being that LE is readable by people that can read english.

An idea is to get rid of the first part of (Q3F*) and only keep the part with the universal quantifier (∀x), but that would produce problems with the question clause at the end. Like this:

Q3F**. (∀y)(y^{2}=4⇒y∈A∧M_{A}>1)∧(x=?)

“x” is not mentioned at all before the question clause at the end.

### Notes

1More precisely, a formula that when interpreted with the supplied interpretation keys expresses a true proposition.