# Is my expression for this quantifier correct?

Quantifier: “for exactly $$n$$ distinct $$x$$ $$P(x)$$ holds.” (Where $$n \in \Bbb{N}$$.)

My expression: $$\exists x_1, \ldots ,x_n ((P(x) \rightarrow (\exists i \in \{1,\ldots ,n\} )(x=x_i)) \wedge (\forall i,j \in \{ 1, \ldots ,n\} )(i \neq j \rightarrow x_i \neq x_j))$$

Is it correct? Is it okay to use quantifiers for indexing as I’ve used above?

Is it now correct? $$\exists x_1, \ldots ,x_n (P(x_1)\wedge \cdots\wedge P(x_n)\wedge (P(x) \rightarrow \bigvee_{i =1}^n x=x_i) \wedge \bigwedge_{i,j=1}^n (i\neq j \rightarrow x_i\neq x_j))$$

If you're trying to work within a formal system of predicate logic (I presume you are, since otherwise you might as well just use the sentence in words which gets the idea across just as precisely and much more readably) then you can't do this since the natural numbers $$\mathbb{N}$$ might not be things that exist in the structures you're working with. The natural numbers are things that when we meta-reason about the logic, not elements in the structure that $$\forall$$ can draw from.
If we're treating $$n$$ as a fixed known number, we can work this with a chain of disjunctions: something like $$\forall x: P(x) \rightarrow \bigvee_{i=1}^n x=x_i$$ where that big V is just shorthand for the chain of disjunctions $$x=x_1 \vee x=x_2 \vee ...$$
P.s: Since you're trying to translate 'Exactly $$n$$ distinct $$x$$,' you'll also need to include a clause saying that each $$x_i$$ has property $$P$$.
• Can’t we use natural numbers from set theory and logic (from $\emptyset$) for indexing with $\forall$ and $\exists$? – Atom Jun 4 '19 at 5:25