Is "there exists at least $0$" the tautological quantifier?

Question

Is "there exists at least $0$" the tautological quantifier, that is, it always holds, no matter what? And also, is the negation of that quantifier the contradictory quantifier, that is, the quantifier that never holds? My real question is, are those quantifiers genuine quantifiers? We would need a definition of what a quantifier is, to answer that question.

Answer 1

See this Wikipedia page for a rather brief discussion of what are called counting quantifiers. I will follow its notation and write, e.g., $\exists^{\ge2}x\phi(x)$ as the formalisation of "there exist at least $2$ values of $x$ such that $\phi(x)$ holds" or "$\{x : \phi(x)\}$ has at least two elements". When the bound on the number of values is finite, you can just view quantifiers like this as short-hands. E.g.,

$$ (\exists^{=2}x\phi(x)) \Leftrightarrow (\exists x, y (x \neq y \land \phi(x) \land \phi(y) \land (\forall x(\phi(z) \Rightarrow z = x \lor z = y)))$$

and

$$ (\exists^{\ge2}x\phi(x)) \Leftrightarrow (\exists x, y (x \neq y \land \phi(x) \land \phi(y))$$

$\exists^{\ge0}x\phi(x)$ means that $\{x : \phi(x)\}$ has at least $0$ elements. But any set has at least $0$ elements, so you are quite right to say that $\exists^{\ge0}x\phi(x)$ is just tautologically true.

As regards how to define the counting quantifiers, you can just view them as short-hands for regular first-order logic statements, if the bound on the number of values is finite, (as discussed above and in the Wikipedia page). You can certainly extend the logical language to allow quantifiers like $\exists^{\infty}x\phi(x)$. How you go about this will depend on what you are trying to achieve. It will likely bring you into either the realm of infinitary logic or higher-order logic.

Answer 2

This is a genuine quantifier -- a counting quantifier. The superscript the existential counting quantifier has is "$\geq 0$" in your case.

And yes, you could say it is the tautological quantifier, and its negation is the contradictory quantifier, because very often we explicitly require the domain of discourse to be non-empty.

Answer 3

We can use the following general templates for such counting quantifiers:

"There are at least $n$ P's":

$$\exists^{\ge n} P(x) = \exists x_1 \exists x_2 ... \exists x_n (P(x_1) \land P(x_2) \land ... P(x_n) \land x_1 \neq x_2 \land x_1 \neq x_3 \land ... \land x_{n-1} \neq x_n )$$

and therefore:

"It is not true that there are at least $n$ P's ... i.e. "There are less than $n$ P's"

$$\exists^{< n} P(x) = $$

$$ \neg \exists^{\ge n} P(x) =$$

$$ \neg \exists x_1 \exists x_2 ... \exists x_n (P(x_1) \land P(x_2) \land ... P(x_n) \land x_1 \neq x_2 \land x_1 \neq x_3 \land ... \land x_{n-1} \neq x_n ) =$$

$$\forall x_1 \forall x_2 ... \forall x_n ((P(x_1) \land P(x_2) \land ... P(x_n)) \to \neg (x_1 \neq x_2 \land x_1 \neq x_3 \land ... \land x_{n-1} \neq x_n )) =$$

$$\forall x_1 \forall x_2 ... \forall x_n ((P(x_1) \land P(x_2) \land ... P(x_n)) \to (x_1 = x_2 \lor x_1 = x_3 \lor ... \lor x_{n-1} = x_n ))$$

Using generalized conjunctions and disjunctions, that respectively conjunct and disjunct all elements of a set of terms, we can write this as:

"There are at least $n$ P's":

$$\exists^{\ge n} P(x) = \exists x_1 \exists x_2 ... \exists x_n (\bigwedge \{ P_i | 1\leq i \leq n\} \land \bigwedge \{x_i \neq x_j | 1\leq i < n, i < j \leq n\} )$$

"There are less than $n$ P's"

$$\exists^{< n} P(x) = \forall x_1 \forall x_2 ... \forall x_n (\bigwedge \{ P_i | 1\leq i \leq n\} \to \bigvee \{x_i = x_j \ 1\leq i < n,i < j \leq n\} )$$

Now, a generalized conjunction of a set of terms is true iff all the terms are true, i.e. iff there is no term that is false. Hence, a generalized conjunction of zero terms is a tautology, since there is no term that is false. That is:

$$\bigwedge \emptyset = \top$$

On the other hand, a generalized disjunction of a set of terms is true iff there is at least one term that is true. Hence, a generalized disjunction of zero terms is a contradiction, since there is no term that are true. Thus:

$$\bigvee \emptyset = \bot$$

So with that, let's see what we get when we set $n=0$:

"There are at least $0$ P's":

$$\exists^{\ge 0} P(x) = \bigwedge \emptyset \land \bigwedge \emptyset = \top \land \top = \top $$

"There are less than $n$ P's"

$$\exists^{< 0} P(x) = \bigwedge \emptyset \to \bigvee \emptyset = \top \to \bot = \bot$$

So ... I am not sure I would call $\exists^{\ge 0}$ a 'tautological quantifier' and $\exists^{< 0}$ a 'contradictory quantifier', but they certainly are deeply connected to a tautology and contradiction respectively.