Examples

Some simple, intuitive examples using English as the language to model demonstrate the core principles behind pregroups and their use in linguistic domains.

Let L = {John, Mary, the, dog, cat, met, barked, at}, let T = {N, S, N₀}, and let the following typing relation hold:

${\displaystyle {\textit {John}}:N\qquad {\textit {Mary}}:N\qquad {\textit {the}}:N\cdot N_{0}^{l}\qquad {\textit {dog}}:N_{0}\qquad {\textit {cat}}:N_{0}}$

${\displaystyle {\textit {met}}:N^{r}\cdot S\cdot N^{l}\qquad {\textit {barked}}:N^{r}\cdot S\qquad {\textit {at}}:S^{r}\cdot N^{rr}\cdot N^{r}\cdot S\cdot N^{l}}$

A sentence S that has type T is said to be grammatical if ${\displaystyle T\leq S}$. We can prove this by use of a chain of ${\displaystyle \leq }$. For example, we can prove that ${\displaystyle {\textit {John}}\ {\textit {met}}\ {\textit {Mary}}:N\cdot N^{r}\cdot S\cdot N^{l}\cdot N}$ is grammatical by proving that ${\displaystyle N\cdot N^{r}\cdot S\cdot N^{l}\cdot N\leq S}$:

${\displaystyle N\cdot N^{r}\cdot S\cdot N^{l}\cdot N~\leq ~S\cdot N^{l}\cdot N~\leq ~S}$

by first using contraction on ${\displaystyle N\cdot N^{r}}$ and then again on ${\displaystyle N^{l}\cdot N}$. A more convenient notation exists, however, that indicates contractions by connecting them with a drawn link between the contracting types (provided that the links are nested, i.e. don't cross). Words are also typically placed above their types to make the proof more intuitive. The same proof in this notation is simply

A more complex example proves that the dog barked at the cat is grammatical:

Purely formal semantics

The purely formal semantics for PG consists of a logical language defined according to the following rules:

• Given a set of atomic terms T = {a, b, ...} and atomic function symbols F = {f_m, g_n, ...} (where subscripts are meta-notational indicating arity), and variables x, y, ..., all constants, variables, and well-formed function applications are basic terms (a function application is well-formed when the function symbol is applied to the appropriate number of arguments, which can be drawn from the atomic terms, variables, or can be other basic terms)
• Any basic term is a term
• Given any variable x, [x] is a term
• Given any terms m and n, ${\displaystyle m\cdot n}$ is a term

Some examples of terms are f(x), g(a,h(x,y)), ${\displaystyle g(x,b)\cdot [x]}$. A variable x is free in a term t if [x] does not appear in t, and a term with no free variables is a closed term. Terms can be typed with pregroup types in the obvious manner.

The usual conventions regarding α conversion apply.

For a given language, we give an assignment I that maps typed words to typed closed terms in a way that respects the pregroup structure of the types. For the English fragment given above we might therefore have the following assignment (with the obvious, implicit set of atomic terms and function symbols):

${\displaystyle {\begin{aligned}I({\textit {John}}:N)&=j:E\\I({\textit {Mary}}:N)&=m:E\\I(the:N\cdot N_{0}^{l})&=\iota (p)\cdot [p]:E\cdot E_{0}^{l}\\I(dog:N_{0})&=dog:E_{0}\\I(cat:N_{0})&=cat:E_{0}\\I(met:N^{r}\cdot S\cdot N^{l})&=[x]\cdot met(x,y)\cdot [y]:E^{r}\cdot T\cdot E^{l}\\I(barked:N^{r}\cdot S)&=[x]\cdot barked(x):E^{r}\cdot T\\I(at:S^{r}\cdot N^{rr}\cdot N^{r}\cdot S\cdot N^{l})&=[x]\cdot y\cdot [y]\cdot at(x,z)\cdot [z]:T^{r}\cdot E^{rr}\cdot E^{r}\cdot T\cdot E^{l}\end{aligned}}}$

where E is the type of entities in the domain, and T is the type of truth values.

Together with this core definition of the semantics of PG, we also have a reduction rules that are employed in parallel with the type reductions. Placing the syntactic types at the top and semantics below, we have

For example, applying this to the types and semantics for the sentence ${\displaystyle {\textit {John}}\ {\textit {met}}\ {\textit {Mary}}:N\cdot (N^{r}\cdot S\cdot N^{l})\cdot N}$ (emphasizing the link being reduced)

For the sentence ${\displaystyle {\textit {the}}\ {\textit {dog}}\ {\textit {barked}}\ {\textit {at}}\ {\textit {the}}\ {\textit {cat}}:(N\cdot N_{0}^{l})\cdot N_{0}\cdot (N^{r}\cdot S)\cdot (S^{r}\cdot N^{rr}\cdot N^{r}\cdot S\cdot N^{l})\cdot (N\cdot N_{0}^{l})\cdot N_{0}}$: