The quantifier "most"

Question

It is well known that the quantifier "most", understood as "more than half the ps" is not first-order definable. This is one of the results Barwise and Cooper (1981: p122-3), in "Formal Semantics: the essential readings" by Portner and Partee) discuss. Furthermore, many so-called proportional quanitifers have been proven not to be first order-definable (Keenan 1996: "The semantics of determiners"). My question is whether "most" can be defined in plural (first-order) logic. Is there a proof that "most" and other proportional quantifiers can be defined in plural (first-order) logic? Does a definition of these quantifiers require full 2nd order logic?

Answer 1

That depends on whether you have the predicate is larger than $>$ defined for plural individuals. The two-place most quantifier (most ps are qs) is definable if and only if $>$ is definable.

First, a word on notation. Let an unadorned letter $a, b, c$ refer to a plural individual in either a free or bound variable. Let a letter with a subscript zero $x_0$ refer to a singular individual. $a_0$ and $a$ are not related in any way.

We write two-place most in the following way:

$$ \mathop{\text{M}} x_0 : x_0 \in p \mathop. x_0 \in q $$

Or in prose For most values of the variable x such that x is in p, x is in q.

Let $a > b$ mean that a plural individual $a$ is strictly larger than a plural individual $b$.

There are more things in $a$ than there are in $b$ if most of the things in $a$ or $b$ but not both are in $a$.

Intuitively, the common elements of $a$ and $b$ don't count for or against our claim of mostness since they count for the $a$ side and the $b$ side equally. We need to examine the elements in the symmetric difference.

$$ a > b \;\;\Longleftrightarrow\;\; \mathop{\text{M}} x_0 : x_0 \in a \oplus x_0 \in b \mathop. x_0 \in a $$

We can define mostness in terms of $>$ if we allow ourselves to use the relative complement $\setminus$.

$$ \mathop{\text{M}} x_0 : x_0 \in a \mathop. x_0 \in b \;\;\Longleftrightarrow\;\; b > a \setminus b $$

The relative complement is itself definable.

$$ x_0 \in a \setminus b \;\;\Longleftrightarrow\;\; x_0 \in a \land x_0 \notin b $$

We can rewrite the RHS by introducing two new variables

$$ \forall c \mathop. b > c \lor (\exists x_0 \mathop. x_0 \in c \land (x_0 \notin a \setminus b)) $$

$$ \forall c \mathop. b > c \lor (\exists x_0 \mathop. x_0 \in c \land (x_0 \notin a \lor x_0 \in b)) $$

So, putting this all together in prose:

Most $a$s are $b$s if and only if for all groups $c$ that are subsets of the non-$b$ $a$s, $c$ is strictly smaller than $b$.

If we have either $>$ or the mostness quantifier, we can rewrite our expression and get the other.

The most natural way to get $>$ to is to define in terms of equality and the existence of injections (which is second order).

If you have the ability to talk about two plural individuals having the same size $a \simeq b$, then you can construct $>$ by claiming that $b$ is the same size as a proper subset of $a$ and $a$ is not the same size as a proper subset of $b$.

But ... we also need to quantify over functions to talk about two plural individuals being the same size.