# Can a element of a set be also a subset? (Set theory in predicate caculus)

I'm not even sure if this is the question I have to make...

I'm trying to formalize (using first-order predicate calculus) a list of rules in a grammar of certain natural language.

The first one is the following (rule about usage of question particle in a sentence):

(1) If an interrogative sentence do not have a question pronoun and the last word of the sentence has the letter a, o or y, and the last letter of the last word of the sentence is a short vowel or a consonant, then the sentence needs the question particle yy.

Well, at first I tought it would be easy to write this using the first-order predicate calculus, but right at the beginning i already encountered some problems.

My first try began like this ("If an interrogative sentence do not have a question pronoun"):

$$\forall x.Sentence(x)\ \land\ Sentence(x) \in InterrogativeSentences \to \lnot\exists y.QuestionPronoun(y) \in Sentence(x).$$

My first problem is that I'm not sure if I can even use set theory symbols in predicate logic (i.e. $$\in$$). The second problem is that first I define x as a sentence and a element of the set InterrogativeSentences, but riht after I said that y is a QuestionPronoun and it is a element of the sentence x. Well, I can't have an element being an element of another element, right?

I tought that maybe an element could be a subset too, but probably not, since I read a lot about set theory and didn't see anything like this.

But even solving this problem I would have another right after, since I have to be able to see not only the sentence as a set of words (to be able to have a Pronoun as an element of the sentence), but as a set of letters, since i would have to be able to get the letters inside the word to define what question particle to use (as described in rule (1)).

How can those problems be solved? How would you write the (1) using predicate logic? Is it even possible?

Thank you.

The standard syntax of first order logic has no simbol $$\in$$, because the language is built with terms (names of objects) and predicates (properties and relations of objects).

Thus, to say that $$x$$ is a sentence, we have to write (as you said) $$\text {Sentence}(x)$$ and not $$x \in \text {Sentence}$$.

In any case, we cannot write $$\text {QuestionPronoun}(y) \in \text {Sentence}(x)$$, because $$\text {Sentence}(x)$$ is a formula (a phrase) and not the name of a set.

What we can do to formalize "an interrogative sentence do not have a question pronoun" is to use suitable predicates :

$$∀x \ [\text {Sentence}(x) \land \text {Interrogative}(x) → ¬∃y \ (\text {QuestionPronoun}(y) \land \text {Occur}(y,x))]$$

where I have added the binary relation $$\text {Occur}(y,x)$$ to express the fact that particle $$y$$ occurr into sentence $$x$$.

• Thank you! This is exactly what I was trying to do! I have only one question now. At the begging you wrote Exists an X, such that Sentence(x) and Interrogative(x). With this it seems possible to have a Interrogative that it's not a sentence. How could I change that? One solution is simply put that "Exists x, such that InterrogativeSentence(x), but I would like to separate them. Is there a way of saying basically the same as you said, but in a way that makes impossible to imagine an Interogative(x) that is not a Sentence(x)? Mar 23, 2019 at 20:39
• @RicardoPotozky - correct; use the predicate $\text {InterrogativeSentence}(x)$. Mar 24, 2019 at 8:16

Set theory does not prohibit an element $$a$$ of a set $$A$$ from also appearing as a subset of $$A$$. The simplest example is $$A = \lbrace \emptyset\rbrace$$. Then $$\emptyset\in A$$ and trivially $$\emptyset\subset A$$.