Is it possible to say that "there exist exactly $x$ elements in a domain of discourse"?

Question

I was wondering how it would be possible to say that there exist 'exactly' $x$ elements within our domain of discourse (without using the uniqueness quantifier?) $x$ can be any positive integer, but it must be a constant.

Answer 1

Here are some possible ways to make these kinds of numerical claims in general. In addition to claims involving 'exactly $n$' I'll also cover 'at least n' claim and 'at most n' claims, since thewy can all be realted to each other.

'At least $n$' (Method 1)

"There is at least $1$ object" : $\exists x \ x=x$ (most logics do not allow a simple $\exists x$ as a claim so we add the dummy $x=x$. Also note that most logics assume there is at least one object in the domain, so you don't have to use any symbolization at all!)

"There are at least $2$ objects" : $\exists x \exists y \ x \not = y$

"There are at least $3$ objects" : $\exists x \exists y \exists z ( x \not = y \land x \not = z \land y \not = z)$

Etc.

Note that with this method you need to use $n \choose 2$ non-identity claims so that increases rather quickly (e.g. to express there are at least $10$ objects, we would need $45$ non-identity statements! ... can we improve on that? Yes! But first let's discuss 'at most':

'At most $n$' (Method 1)

One method to do 'at most $n$' is to deny 'at least $n+1$'. So:

"There is at most $1$ object": $\neg \exists x \exists y \ x \not = y$

If you bring the negation inside, this is equivalent to: $\forall x \forall y \ x = y$

"There are at most $2$ objects": $\neg \exists x \exists y \exists z (x \not = y \land x \not = z \land y \not = z)$

Again, bringing the negation inside you get:

$\forall x \forall y \forall z (x = y \lor x = z \lor y = z)$

Etc.

OK, so here we get even more (non-)identity statements: $n+1 \choose 2$. Later we will see how we can do this more efficiently, but first:

'Exactly $n$' (Method 1)

One method is to recognize that 'Exactly $n$' is equivalent to 'at least $n$ and at most $n$'. Doing a straightforward conjunction, we thus get:

"There is exactly $1$ object" : $\exists x x=x \land \neg \exists x \exists y \ x \not = y$

or, equivalently: $\exists x \ x=x \land \forall x \forall y \ x = y)$

"There are exactly $2$ objects" : $\exists x \exists y \ x \not = y \land \neg \exists x \exists y \exists z (x \not = y \land x \not = z \land y \not = z)$

or, equivalently: $\exists x \exists y \ x \not = y \land \forall x \forall y \forall z (x = y \lor x = z \lor y = z)$

Etc.

'Exactly $n$' (Method 2)

OK, so these claims get really big really fast. Can we do better? Yes. Instead of just conjuncting the 'at least' and 'at most' claims, let's integrate these two ideas: To say there are exactly $n$ objects, we can say that there are $n$ different objects ... but no others:

"There is exactly $1$ object" : $\exists x (x=x \land \neg \exists y \ x \not = y)$

And now we can get rid of the 'dummy' $x=x$ as well, so we get $\exists x \neg \exists y \ x \not = y$

or, equivalently: $\exists x \forall y \ x = y$

"There are exactly $2$ objects" : $\exists x \exists y (x \not = y \land \neg \exists z (z \not = x \land z \not = y))$

or, equivalently: $\exists x \exists y (x \not = y \land \forall z (z = x \lor z = y))$ (This is Eric Wofsey's example for 'exactly $2$')

"There are exactly $3$ objects" : $\exists x \exists y \exists z (x \not = y \land x \not = z \land y \not = z \land \neg \exists w (w \not = x \land w \not = y \land w \not = z))$

or, equivalently: $\exists x \exists y \exists z (x \not = y \land x \not = z \land y \not = z \land \forall w (w = x \lor w = y \lor w = z))$

Etc.

Interestingly, we see that in the second half of the statement, we no longer get $n+1 \choose 2$ (non-)identity claims, but merely $n$ (non-)identity claims, because we end up saying: once you have your $n$ different objects, then any object is one of the $n$ objects, so you can't have any more than $n$. This is an idea that we can use to write the 'at least' and 'at most' claims more efficiently as well:

'At most $n$' (Method 2)

As just observed, we can say that there are exactly $n$ objects by saying that if you already have $n$ different objects, you can't get any other object. But to say that there are 'at most' $n$ objects, we don't have to require those $n$ objects to be different: we can simply say that we can pick $n$ objects, that may or may not be different, such that there is no object that is different from all those. So:

'There is at most $1$ object' : $\exists x \forall y \ y = x$

"There are at most $2$ objects" : $\exists x \exists y \forall z (z = x \lor z = y)$

Note that this is not claiming that $x$ and $y$ are different ... but they could be. Hence, you can have $1$ or $2$ different objects, but you definitely can't have $3$ or more!

"There are at most $3$ objects" : $\exists x \exists y \exists z \forall w (w = x \lor w = y \lor w = z)$

Etc.

So, in general, the claim is that there are $n$ objects such that every object has to be one of those objects, and that means that you can't have more than $n$ objects.

The nice thing about this expression is that it uses exactly $n$ identity statements, so it is quite a bit more efficient to write than the first method for expressing 'at most', especially for large $n$.

'At least $n$' (Method 2)

We pointed out earlier that 'at most $n$' is the negation of 'at least $n+1$', but that also means that 'at least $n+1$' is the negation of 'at most $n$'.

So:

"There are at least 2 P's": $\neg \exists x \forall y \ y = x$ which is equivalent to $\forall x \exists y \ y \not = x$

The latter statement says that for whatever object $x$ I pick, I can always find a different object. So, you need at least $2$ objects to make this true.

"There are at least $3$ objects": $\neg \exists x \exists y \forall z (z = x \lor z = y)$ which is equivalent to: $\forall x \forall y \exists z (z \not = x \land z \not= y)$

Etc.

So, to say that there are at least $n$ objects, we can say that no matter how you pick $n-1$ objects, you can always find an object different from all those. And, once again, this method saves us a lot of writing: only $n-1$ identity claims.

'Exactly $n$' (Method 3)

In the second method for 'exactly n', we saw that a claim like "There are exactly $3$ objects" translated into:

$\exists x \exists y \exists z (x \not = y \land x \not = z \land y \not = z \land \forall w (w = x \lor w = y \lor w = z))$

Thus, while in the second half we reduce the number of identity statements in comparison to the first method, we still get $3 \choose 2$ non-identity claims in the first half. Especially for large $n$ then, the most efficient way to express 'exactly $n$' seems to be the conjunction of the efficient ways to expressing 'at least $n$' and 'at most $n$'. That is:

"There are exactly $3$ objects" : $\forall x \forall y \exists z (z \not = x \land z \not= y) \land \exists x \exists y \exists z \forall w (w = x \lor w = y \lor w = z))$

"There are exactly $4$ objects" : $\forall x \forall y \forall z \exists w (w \not = x \land w \not = y \land w \not = z) \land \exists x \exists y \exists z \exists w \forall v (v = x \lor v = y \lor v = z \lor v = w))$

This method will have $2n-1$ (non-)identity claims.

Answer 2

Here's how you can do it for $x=2$: $$\exists a\exists b(\neg(a=b)\wedge \forall c(c=a\vee c=b)).$$ I'll leave it to you to figure out why this works and how to generalize to arbitrary values of $x$.