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The two place quantifier "All C are D" (where C and D represent classes) can be defined from the single place quantifier "For all X", like so: "For all X, if X is C, then X is D". Also, the two place quantifier "Some C are D" can be defined from the single place existential quantifier, like this: "There exists some X such that both X is C and X is D". But what about the two place quantifier "Most C are D"? Can it be defined using the single place quantifier "For most X", even augmented with the universal and existential quantifier and the identity relation $=$? If it cannot, I would like a proof.

user107952
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    What does "for most X" mean in the first place? if X ranged over 100 things, what number or higher would be "most"? I don't think "for most" can be defined precisely in logic. If a numeric threshold were defined it would only work for a single writer as others would not agree. Not to mention problems occurring for infinite X. – coffeemath Jul 21 '23 at 01:11

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It cannot be so defined, under any reasonable interpretation of "most."

Let's start with the simplest interpretation: for (possibly infinite) cardinals $\kappa,\lambda$, say $\kappa\triangleleft\lambda$ iff $\kappa+\kappa<\lambda$. For finite cardinals this corresponds to $\kappa$ being less than half of $\lambda$, and for infinite cardinals this corresponds to $\kappa$ simply being $<\lambda$. If "most $Y$ are $X$" is interpreted as "$\vert Y\setminus X\vert\triangleleft \vert Y\vert$" (= the non-$X$ things in $Y$ are small relative to $Y$), then we can argue as follows:

Let $\mathfrak{A}_1$ be a structure of size $\aleph_2$ with a unary predicate $U$ naming a set of size $\aleph_1$ and a unary predicate $V$ naming a subset of $U$ of size $\aleph_0$. Then in $\mathfrak{A}_1$, "Most $U$ are $\neg V$" is true. But now suppose we tweak our structure to get $\mathfrak{A}_2$, in which $V$ holds of all-but-$\aleph_0$-much of $U$. In $\mathfrak{A}_2$, "Most $U$ are $\neg V$" is false. However, for every formula $\varphi$ using just the first-order quantifers + UnaryMost, we have $$\vert\varphi^{\mathfrak{A}_1}\vert\triangleleft\vert\mathfrak{A}_1\vert \quad\iff\quad \vert\varphi^{\mathfrak{A}_2}\vert\triangleleft\vert\mathfrak{A}_2\vert.$$ So $\mathfrak{A}_1$ and $\mathfrak{A}_2$ are equivalent with respect tot he UnaryMost quantifier, but not with respect to the BinaryMost quantifier.

This general strategy of "distorting a small set" works for any version of the question I can think of. (It's most interesting over finite structures, where additional care is required since we don't literally get elementarily equivalent structures.)

Noah Schweber
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