Conservativity

In research on quantificational determiners across the world’s languages, and in cross-linguistic semantic investigations more broadly, one of the generalizations that seems to have held up the best is the generalization that determiner meanings are conservative (Barwise and Cooper 1981; Keenan and Stavi 1986). For a determiner meaning to be conservative is for it to satisfy the definition in (1).

A function $R ∈ D_{((e → t) → ((e → t) → t))}$ is conservative if, for any two characteristic functions $f$ and $g$ (i.e., any two elements of $D_{(e → t)}$), $R(f)(g) = \true$ if and only if $R(f)(f ∩ g) = \true$.

(We are abusing notation slightly by denoting the intersection of the sets characterized by $f$ and $g$ as ‘$f ∩ g$’, even though $f$ and $g$ are technically not sets, but characteristic functions of sets. That’s okay—what is meant by ‘$f ∩ g$’ is $\{x ∈ D_{e} ∣ f(x) = \true \text{ and } g(x) = \true\}_{CF}$—that is, the characteristic function of the set of entities of which both $f$ and $g$ are true (their “intersection”).)

What does it mean, intuitively, for a determiner meaning to be conservative? Well, in practice, $f$ will be the meaning of our noun, and $g$ will be the meaning of our verb phrase. So, for a determiner meaning $R$ to be conservative is for $R(f)$ to be true of a verb phrase meaning $g$ whenever it is true of the noun meaning and verb phrase meaning intersected ($f ∩ g$).

For example, consider the quantificational noun phrase some dog. When we combine this noun phrase with a verb phrase—slept—to get some dog slept, the resulting sentence is true if and only if the sentence some dog is a dog that slept is also true. And it doesn’t matter what noun or verb phrase we use: some cat purred, some glass broke, and some ship sank all have this property, which you can test for yourself.

Thus one way of saying intuitively what it means for a determiner meaning to be conservative is to say that you will never have to look at any entities outside of the set denoted by the noun that the determiner combines with when you determine if the noun and verb phrase meanings are in the right relation with each other. When you check if some dog slept is true, for example, you don’t care about the things that aren’t dogs; specifically, you don’t care about whether or not any non-dogs slept—you only care about whether or not any of the dogs slept. The same is true for every dog slept—it doesn’t matter here, either, what the non-dogs are doing. They could be sleeping, barking, you name it; all that matters is that all of the dogs slept.

Here is a hypothetical determiner that appears not to be conservative: all-non. Thus, say that all-non dogs slept just means that all of the non-dogs slept. Intuitively, this (hypothetical) determiner is non-conservative because checking if it gives a true sentence when it combines with a noun and a verb phrase requires you to check what the non-dogs are doing. Indeed, you can check if it is conservative by performing the test from above. Does all-non glasses broke mean the same thing as all-non glasses are glasses that broke? Intuitively, no! The first sentences requires all of the things that aren’t glasses to have broken, while the second sentence requires all of the non-glasses to additionally be glasses. (This second requirement seems to be a contradiction!¹)

Another hypothetical determiner is the determiner less-than. Thus, say that less-than dogs slept means that the number of dogs is less than the number of things that slept. For example, it is true if (a) there are three dogs, and (b) four things—not necessarily dogs—slept. This hypothetical determiner appears to be non-conservative, since less-than dogs slept doesn’t mean the same thing as less-than dogs are dogs that slept. The second sentence means that the number of dogs is less than the number of things which are dogs that slept—an impossibility!

A working hypothesis in linguistic semantics is that all natural language determiner meanings are conservative (Keenan and Stavi 1986).

Conservativity hypothesis: all natural-language determiner meanings are conservative.

Let’s go through our determiner meanings from earlier and show whether or not they satisfy the property defined in (1).

Every

$(λf.(λg.f_{SET} ⊆ g_{SET}))$

For every to be conservative, it has to be the case that, for any characteristic functions $f$ and $g$,

\[f_{SET} ⊆ g_{SET}\]

is true if and only if

\[f_{SET} ⊆ f_{SET} ∩ g_{SET}\]

is true.

Does this hold up? Well, $f_{SET} ⊆ g_{SET}$ means that everyting in $f_{SET}$ is also in $g_{SET}$. Meanwhile, $f_{SET} ⊆ f_{SET} ∩ g_{SET}$ means that everything in $f_{SET}$ is both in $f_{SET}$ and in $g_{SET}$. Since everything in $f_{SET}$ is always in $f_{SET}$, no matter what $f_{SET}$ is, both statements boil down to the same thing—that everything in $f_{SET}$ is also in $g_{SET}$. So, yes, $f_{SET} ⊆ g_{SET}$ is true whenever $f_{SET} ⊆ f_{SET} ∩ g_{SET}$ is true, and vice versa. And as a result, the meaning of every—(3)—is conservative.

Some

$(λf.(λg.f_{SET} ∩ g_{SET} ≠ \varnothing))$

For some to be conservative, it has to be the case that, for any characteristic functions $f$ and $g$,

\[f_{SET} ∩ g_{SET} ≠ \varnothing\]

is true if and only if

\[f_{SET} ∩ f_{SET} ∩ g_{SET} ≠ \varnothing\]

is true.

Note that in the second statement, we have intersected $g_{SET}$ with $f_{SET}$ twice: $f_{SET} ∩ f_{SET} ∩ g_{SET}$. This is the same as intersecting it only once, since once you’ve intersected a set with $f_{SET}$, you’ve already gotten rid of all the things that aren’t in $f_{SET}$. So, $f_{SET} ∩ f_{SET} ∩ g_{SET}$ is the same set as $f_{SET} ∩ g_{SET}$, no matter what. As a result, saying that $f_{SET} ∩ g_{SET}$ is not the empty set is the same thing as saying that $f_{SET} ∩ f_{SET} ∩ g_{SET}$ is not the empty set, which is it what it takes for some to be conservative. So, yeah, some is conservative.

No

$(λf.(λg.f_{SET} ∩ g_{SET} = \varnothing))$

For no to be conservative, it has to be the case that, for any characteristic functions $f$ and $g$,

\[f_{SET} ∩ g_{SET} = \varnothing\]

is true if and only if

\[f_{SET} ∩ f_{SET} ∩ g_{SET} = \varnothing\]

is true.

The argument that no is conservative is therefore exactly the same as the argument that some is conservative, modulo the fact that, here, we are saying that intersecting the relevant two sets does give you the empty set.

Three

$(λf.(λg.|f_{SET} ∩ g_{SET}| ≥ 3))$

For three to be conservative, it has to be the case that, for any characteristic functions $f$ and $g$,

\[|f_{SET} ∩ g_{SET}| ≥ 3\]

is true if and only if

\[|f_{SET} ∩ f_{SET} ∩ g_{SET}| ≥ 3\]

is true.

We can say it’s conservative using essentially the same argument as we did for some and no.

All-non

What about a hypothetical determiner, like all-non, which we claimed not to be conservative. Can we assign it a meaning that we can then reason about formally? Let’s assign it the meaning in (7).

$(λf.(λg.(D - f_{SET}) ⊆ g_{SET}))$

This meaning guarantees that when all-non combines with a noun and then a verb phrase, the sentence that results will be true just in case all of the entities in the model’s domain ($D$) of which the noun is false are also in the set characterized by the verb phrase. E.g., all-non dogs slept will be true if all of the things in the domain which aren’t in $\ct{dog}_{SET}$ are in $\ct{sleep}_{SET}$.

Now, for the demonstration. For all-non to be conservative, it has to be the case that, for any characteristic functions $f$ and $g$,

\[(D - f_{SET}) ⊆ g_{SET}\]

is true if and only if

\[(D - f_{SET}) ⊆ (f_{SET} ∩ g_{SET})\]

is true.

Well, the first statement requires that everything in the domain which is not in $f_{SET}$ is in $g_{SET}$. Meanwhile, the second statement requires that everything in the domain which is not in $f_{SET}$ is in $f_{SET}$, as well as in $g_{SET}$. The second statement therefore makes a stronger requirement: it additionally requires that everything in the domain which is not in $f_{SET}$ is in $f_{SET}$. This can only be true if $D - f_{SET} = \varnothing$, which forces $D$ to be a subset of $f_{SET}$. In other words, it can only be true if every entity in the domain is in $f_{SET}$.

Since these two statements impose different requirements on $f_{SET}$ and $g_{SET}$, we can conclude that the hypothetical meaning in (7) is not conservative.

Less-than

What about the other hypothetical, apparently non-conservative determiner, less-than? Let’s assign it the meaning in (8).

$(λf.(λg.|f_{SET}| < |g_{SET}|))$

This meaning guarantees that when less-than combines with a noun and then a verb phrase, the sentence that results will be true just in case the number of entities of which the noun is true is less than the number of entities of which the verb phrase is true. E.g., less-than ships sank will be true if the number of ships is less than the number of things that sank.

For less to be conservative, it has to be the case that, for any characteristic functions $f$ and $g$,

\[|f_{SET}| < |g_{SET}|\]

is true if and only if

\[|f_{SET}| < |f_{SET} ∩ g_{SET}|\]

is true.

Well, because $f_{SET} ∩ g_{SET}$ has to be a subset of $f_{SET}$—this is true anytime you intersect some set with another set—it must be the case that $|f_{SET} ∩ g_{SET}| ≤ |f_{SET}|$. But then it can’t be true that $|f_{SET}| < |f_{SET} ∩ g_{SET}|$. So, while the first statement above

\[|f_{SET}| < |g_{SET}|\]

might or might not hold true, depending on the model, the second statement

\[|f_{SET}| < |f_{SET} ∩ g_{SET}|\]

is a contradiction! As a result, the two statements don’t mean the same thing. Therefore, we can conclude that the hypothetical determiner meaning in (8) is not conservative.

References

Barwise, Jon, and Robin Cooper. 1981. “Generalized Quantifiers and Natural Language.” Linguistics and Philosophy 4 (2): 159–219. https://doi.org/10.1007/BF00350139.

Keenan, Edward L., and Jonathan Stavi. 1986. “A Semantic Characterization of Natural Language Determiners.” Linguistics and Philosophy 9 (3): 253–326. https://doi.org/10.1007/BF00630273.

Footnotes

Or it at least appears to require that everything be a glass, since that way, there aren’t any non-glasses to speak of.↩︎

--- title: "Conservativity" bibliography: ../../semantics.bib format: html: css: ../styles.css html-math-method: mathjax mathjax-config: loader: {load: ['[tex]/bussproofs','[tex]/bbox','[tex]/colorbox']} tex: packages: {'[+]': ['bussproofs','bbox','colorbox']} --- ::: {.hidden} $$ \newcommand{\expr}[3]{\begin{array}{c} #1 \\ \bbox[lightblue,5px]{#2} \end{array} ⊢ #3} \newcommand{\ct}[1]{\bbox[font-size: 0.8em]{\mathsf{#1}}} \newcommand{\abbr}[1]{\bbox[transform: scale(0.95)]{\mathtt{#1}}} \def\true{\ct{T}} \def\false{\ct{F}} $$ ::: In research on quantificational determiners across the world's languages, and in cross-linguistic semantic investigations more broadly, one of the generalizations that seems to have held up the best is the generalization that determiner meanings are *conservative* [@barwise_generalized_1981;@keenan_semantic_1986]. For a determiner meaning to be conservative is for it to satisfy the definition in (@ex-conservativity-def). (@ex-conservativity-def) A function $R ∈ D_{((e → t) → ((e → t) → t))}$ is [*conservative*]{.underline} if, for any two characteristic functions $f$ and $g$ (i.e., any two elements of $D_{(e → t)}$), $R(f)(g) = \true$ if and only if $R(f)(f ∩ g) = \true$. (We are abusing notation slightly by denoting the intersection of the sets characterized by $f$ and $g$ as '$f ∩ g$', even though $f$ and $g$ are technically not sets, but characteristic functions of sets. That's okay---what is meant by '$f ∩ g$' is $\{x ∈ D_{e} ∣ f(x) = \true \text{ and } g(x) = \true\}_{CF}$---that is, the characteristic function of the set of entities of which both $f$ and $g$ are true (their "intersection").) What does it mean, intuitively, for a determiner meaning to be conservative? Well, in practice, $f$ will be the meaning of our noun, and $g$ will be the meaning of our verb phrase. So, for a determiner meaning $R$ to be conservative is for $R(f)$ to be true of a verb phrase meaning $g$ whenever it is true of the noun meaning and verb phrase meaning intersected ($f ∩ g$). For example, consider the quantificational noun phrase *some dog*. When we combine this noun phrase with a verb phrase---*slept*---to get *some dog slept*, the resulting sentence is true if and only if the sentence *some dog is a dog that slept* is also true. And it doesn't matter what noun or verb phrase we use: *some cat purred*, *some glass broke*, and *some ship sank* all have this property, which you can test for yourself. Thus one way of saying intuitively what it means for a determiner meaning to be conservative is to say that you will never have to look at any entities outside of the set denoted by the noun that the determiner combines with when you determine if the noun and verb phrase meanings are in the right relation with each other. When you check if *some dog slept* is true, for example, you don't care about the things that aren't dogs; specifically, you don't care about whether or not any non-dogs slept---you only care about whether or not any of the *dogs* slept. The same is true for *every dog slept*---it doesn't matter here, either, what the non-dogs are doing. They could be sleeping, barking, you name it; all that matters is that all of the *dogs* slept. Here is a hypothetical determiner that appears *not* to be conservative: *all-non*. Thus, say that *all-non dogs slept* just means that all of the non-dogs slept. Intuitively, this (hypothetical) determiner is non-conservative because checking if it gives a true sentence when it combines with a noun and a verb phrase requires you to check what the non-dogs are doing. Indeed, you can check if it is conservative by performing the test from above. Does *all-non glasses broke* mean the same thing as *all-non glasses are glasses that broke*? Intuitively, no! The first sentences requires all of the things that aren't glasses to have broken, while the second sentence requires all of the non-glasses to additionally be glasses. (This second requirement seems to be a contradiction!^[ Or it at least appears to require that everything be a glass, since that way, there aren't any non-glasses to speak of. ]) Another hypothetical determiner is the determiner *less-than*. Thus, say that *less-than dogs slept* means that the number of dogs is less than the number of things that slept. For example, it is true if (a) there are three dogs, and (b) four things---not necessarily dogs---slept. This hypothetical determiner appears to be non-conservative, since *less-than dogs slept* doesn't mean the same thing as *less-than dogs are dogs that slept*. The second sentence means that the number of dogs is less than the number of things which are dogs that slept---an impossibility! A working hypothesis in linguistic semantics is that all natural language determiner meanings are conservative [@keenan_semantic_1986]. (@ex-conservativity-hyp) Conservativity hypothesis: all natural-language determiner meanings are conservative. Let's go through our determiner meanings from earlier and show whether or not they satisfy the property defined in (@ex-conservativity-def). ### Every (@ex-every-sem) $(λf.(λg.f_{SET} ⊆ g_{SET}))$ For *every* to be conservative, it has to be the case that, for any characteristic functions $f$ and $g$, $$f_{SET} ⊆ g_{SET}$$ is true [if and only if]{.underline} $$f_{SET} ⊆ f_{SET} ∩ g_{SET}$$ is true. Does this hold up? Well, $f_{SET} ⊆ g_{SET}$ means that everyting in $f_{SET}$ is also in $g_{SET}$. Meanwhile, $f_{SET} ⊆ f_{SET} ∩ g_{SET}$ means that everything in $f_{SET}$ is both in $f_{SET}$ and in $g_{SET}$. Since everything in $f_{SET}$ is always in $f_{SET}$, no matter what $f_{SET}$ is, both statements boil down to the same thing---that everything in $f_{SET}$ is also in $g_{SET}$. So, yes, $f_{SET} ⊆ g_{SET}$ is true whenever $f_{SET} ⊆ f_{SET} ∩ g_{SET}$ is true, and vice versa. And as a result, the meaning of *every*---(@ex-every-sem)---is conservative. ### Some (@ex-some-sem) $(λf.(λg.f_{SET} ∩ g_{SET} ≠ \varnothing))$ For *some* to be conservative, it has to be the case that, for any characteristic functions $f$ and $g$, $$f_{SET} ∩ g_{SET} ≠ \varnothing$$ is true [if and only if]{.underline} $$f_{SET} ∩ f_{SET} ∩ g_{SET} ≠ \varnothing$$ is true. Note that in the second statement, we have intersected $g_{SET}$ with $f_{SET}$ twice: $f_{SET} ∩ f_{SET} ∩ g_{SET}$. This is the same as intersecting it only once, since once you've intersected a set with $f_{SET}$, you've already gotten rid of all the things that aren't in $f_{SET}$. So, $f_{SET} ∩ f_{SET} ∩ g_{SET}$ is the same set as $f_{SET} ∩ g_{SET}$, no matter what. As a result, saying that $f_{SET} ∩ g_{SET}$ is not the empty set is the same thing as saying that $f_{SET} ∩ f_{SET} ∩ g_{SET}$ is not the empty set, which is it what it takes for *some* to be conservative. So, yeah, *some* is conservative. ### No (@ex-no-sem) $(λf.(λg.f_{SET} ∩ g_{SET} = \varnothing))$ For *no* to be conservative, it has to be the case that, for any characteristic functions $f$ and $g$, $$f_{SET} ∩ g_{SET} = \varnothing$$ is true [if and only if]{.underline} $$f_{SET} ∩ f_{SET} ∩ g_{SET} = \varnothing$$ is true. The argument that *no* is conservative is therefore exactly the same as the argument that *some* is conservative, modulo the fact that, here, we are saying that intersecting the relevant two sets *does* give you the empty set. ### Three (@ex-three-sem) $(λf.(λg.|f_{SET} ∩ g_{SET}| ≥ 3))$ For *three* to be conservative, it has to be the case that, for any characteristic functions $f$ and $g$, $$|f_{SET} ∩ g_{SET}| ≥ 3$$ is true [if and only if]{.underline} $$|f_{SET} ∩ f_{SET} ∩ g_{SET}| ≥ 3$$ is true. We can say it's conservative using essentially the same argument as we did for *some* and *no*. ### All-non What about a hypothetical determiner, like *all-non*, which we claimed *not* to be conservative. Can we assign it a meaning that we can then reason about formally? Let's assign it the meaning in (@ex-all-non-sem). (@ex-all-non-sem) $(λf.(λg.(D - f_{SET}) ⊆ g_{SET}))$ This meaning guarantees that when *all-non* combines with a noun and then a verb phrase, the sentence that results will be true just in case all of the entities in the model's domain ($D$) of which the noun is false are also in the set characterized by the verb phrase. E.g., *all-non dogs slept* will be true if all of the things in the domain which aren't in $\ct{dog}_{SET}$ *are* in $\ct{sleep}_{SET}$. Now, for the demonstration. For *all-non* to be conservative, it has to be the case that, for any characteristic functions $f$ and $g$, $$(D - f_{SET}) ⊆ g_{SET}$$ is true [if and only if]{.underline} $$(D - f_{SET}) ⊆ (f_{SET} ∩ g_{SET})$$ is true. Well, the first statement requires that everything in the domain which is not in $f_{SET}$ *is* in $g_{SET}$. Meanwhile, the second statement requires that everything in the domain which is not in $f_{SET}$ is in $f_{SET}$, as well as in $g_{SET}$. The second statement therefore makes a stronger requirement: it additionally requires that everything in the domain which is *not* in $f_{SET}$ *is* in $f_{SET}$. This can only be true if $D - f_{SET} = \varnothing$, which forces $D$ to be a subset of $f_{SET}$. In other words, it can only be true if every entity in the domain is in $f_{SET}$. Since these two statements impose different requirements on $f_{SET}$ and $g_{SET}$, we can conclude that the hypothetical meaning in (@ex-all-non-sem) is *not* conservative. ### Less-than What about the other hypothetical, apparently non-conservative determiner, *less-than*? Let's assign it the meaning in (@ex-less-than-sem). (@ex-less-than-sem) $(λf.(λg.|f_{SET}| < |g_{SET}|))$ This meaning guarantees that when *less-than* combines with a noun and then a verb phrase, the sentence that results will be true just in case the number of entities of which the noun is true is less than the number of entities of which the verb phrase is true. E.g., *less-than ships sank* will be true if the number of ships is less than the number of things that sank. For *less* to be conservative, it has to be the case that, for any characteristic functions $f$ and $g$, $$|f_{SET}| < |g_{SET}|$$ is true [if and only if]{.underline} $$|f_{SET}| < |f_{SET} ∩ g_{SET}|$$ is true. Well, because $f_{SET} ∩ g_{SET}$ has to be a subset of $f_{SET}$---this is true anytime you intersect some set with another set---it must be the case that $|f_{SET} ∩ g_{SET}| ≤ |f_{SET}|$. But then it can't be true that $|f_{SET}| < |f_{SET} ∩ g_{SET}|$. So, while the first statement above $$|f_{SET}| < |g_{SET}|$$ might or might not hold true, depending on the model, the second statement $$|f_{SET}| < |f_{SET} ∩ g_{SET}|$$ is a contradiction! As a result, the two statements don't mean the same thing. Therefore, we can conclude that the hypothetical determiner meaning in (@ex-less-than-sem) is not conservative.