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Optional division Lambek calculus grammars

Tikhon Pshenitsyn

Department of Mathematical Logic and Theory of Algorithms

Faculty of Mathematics and Mechanics

Lomonosov Moscow State University

Abstract. In this paper we investigate formal properties of the Lam-

bek calculus enriched with two new connectives named optional divisions

(ODLC). Linguistic motivation to add them is describing optional argu-

ments of verbs, e.g. the book in Tim reads the book. We present two

theorems regarding recognizing power of the ODLC; in particular, we

show that ﬁnite intersections of context-free languages can be generated

by Lambek grammars with optional divisions.

1 Introduction

1.1 Lambek calculus

In 1958 Joachim Lambek presented in [4] a new formalism named Lambek cal-

culus. The purpose of this calculus is to describe syntax of natural languages in

terms of categories. Categories in the Lambek calculus are types which are built

of primitive types P r =p1, p2, . . . using three binary connectives •,\, /. E.g.

(p1\p2)/p3is the type of the Lambek calculus. Formally speaking, every element

in P r is a type, and if Aand Bare types then A\B, A/B , A •Bare also types.

We shall denote the set of all types by T p.

In this paper small letters p, q, . . . and strings composed of them (e.g. np, cp)

range over primitive types. Capital letters A, B, . . . usually range over types.

Capital Greek letters range over ﬁnite (possibly empty) sequences of types. Se-

quents are of the form Γ→A, where Γis a nonempty.

The Lambek calculus deals with sequents. There is only one axiom A→A.

Rules are the following:

Π→A Γ, B, ∆ →C

Γ, Π, A\B , ∆ →C(\ →)A, Π →B

Π→A\B(→ \)Γ, A, B, ∆ →C

Γ, A •B, ∆ →C(• →)

Π→A Γ, B, ∆ →C

Γ, B/A, Π, ∆ →C(/→)Π, A →B

Π→B/A (→/)Γ→A ∆ →B

Γ, ∆ →A•B(→ •)

It can be proven that such rules are admissible:

Γ, B, ∆ →C

Γ, A, A\B, ∆ →C

Γ, B, ∆ →C

Γ, B/A, A, ∆ →C

Π→A Γ, A, ∆ →C

Γ, Π, ∆ →C(cut)

2 Tikhon Pshenitsyn

The sequent T1, . . . , Tn→Tis derivable in the Lambek calculus iﬀ it can be

obtained from axioms by rules. We denote the fact of derivability of a sequent

in the Lambek calculus as follows: L`T1, . . . , Tn→T.

In order to describe some natural language (for instance, English) one assigns

several types to each English word; besides, a distinguished type Thas to be

chosen. Then a sequence of English words w1. . . wnis said to be a grammatically

correct sentence iﬀ one can choose a corresponding type Tifor each of wisuch

that the sequent T1, . . . , Tn→Tis derivable in the Lambek calculus. Here are

some deﬁnitions formalizing this idea:

Deﬁnition 1. A Lambek grammar is a triple hΣ, T , i, where Σis a ﬁnite set

(alphabet), T∈T p is some distinguished type, and ⊆Σ×T p is a ﬁnite binary

relation (i.e., one assignes a ﬁnite set of types to any symbol in Σ).

Given some grammar G, we shall denote the set of types Tfor which ∃a∈Σ:

aT by T p(G). It follows from the deﬁnition that T p(G) is ﬁnite.

Deﬁnition 2. The language L(G)generated by the Lambek grammar G=hΣ, T , i

is the following set of strings over Σ:

{a1. . . an∈Σ+| ∃ T1, . . . , Tn∈T p :∀i= 1, . . . , n ai Ti, L `T1, . . . , Tn→T}.

Lambek gives following examples how types may be assigned to words (note

that the alphabet Σin linguistic examples consists of words):

John works

np np\s→s

John likes Jane

np (np\s)/np np →s

Here sis a distinguished type which can be understood as the marker of

declarative sentences. Sentences John works and John likes Jane are correct

in this simple grammar, because corresponding sequents are derivable in the

Lambek calculus.

1.2 Deﬁnition of optional divisions

One can notice that connectives \and /(named left division and right divi-

sion resp.) in the latter examples correspond to obligatory arguments of verbs.

Current rules imply that in order to obtain a correct sentence a verb must take

all its arguments. However, there are so called optional arguments considered in

linguistics, for instance:

Tim helped. & Tim helped Helen.

Tim reads. & Tim reads the book.

Optional division Lambek calculus grammars 3

In order to describe them we can simply assign two types to verbs, e.g.

helped np\s, (np\s)/np. However, let us consider another example from [3]:

Tim bet (Helen) (ﬁve dollars) (that they hired Alex). (1)

Each of three arguments can be omitted in this sentence, so there can be 8

variants. Thus we have to add 6 types for bet:np\s, (np\s)/np, (np\s)/cp,

(np\s)/cp/np, (np\s)/np/np, (np\s)/cp/np/np (cp denotes a subordinate clause).

It seems more reasonable to add new connectives which would represent

such optionality. Therefore we present new operations named optional divisions.

There are left and right ones, which are denoted by

∠

and ∠resp. Four rules for

them are added:

Γ, B, ∆ →C

Γ, A

∠

B, ∆ →C(

∠

0→)Π→A Γ, B, ∆ →C

Γ, Π, A

∠

B, ∆ →C(

∠

1→)

Γ, B, ∆ →C

Γ, B∠A, ∆ →C(∠0→)Π→A Γ, B, ∆ →C

Γ, B∠A, Π, ∆ →C(∠1→)

These rules represent the idea that one can freely choose whether to apply a

division or not. For instance, one assigns types to above English words as follows:

Tim reads the book

np (np\s)∠np np

It is easily seen that we can derive both sequents:

np, (np\s)∠np →s np, (np\s)∠np, np →s

These sequents correspond to sentences: Tim reads and Tim reads the book.

Optional divisions work as markers of an optional argument (the book) of a verb

(reads). By analogy, the type (np\s)∠cp∠np∠np is assigned to the word bet

which allows to generate 8 possible variants of (1).

1.3 Optional divisions and conjunction

Kanazawa in [2] (1992) deﬁned another binary operation ∧named conjunction,

rules of which are

Γ, Ai, ∆ →C

Γ, A1∧A2, ∆ →C(∧ →), i = 1,2Π→A1Π→A2

Π→A1∧A2

(→ ∧)

In this work, if we extend the Lambek calculus by ∧, we shall denote its

types by T p(∧) and the calculus itself by L(∧). We deﬁne L(∧)-grammars and

languages generated by them the same way as in Deﬁnition 1. L(∧) also admits

cut rule which shall be used in proofs.

There is an obvious connection between conjunction and optional divisions.

Let us add two new rules for latter:

A, Π →B Π →B

Π→A

∠

B(→

∠

)Π, A →B Π →B

Π→B∠A(→∠)

Then optional divisions can be equivalently deﬁned in terms of conjunction:

4 Tikhon Pshenitsyn

Deﬁnition 3.

B

∠

A:= A∧(B\A)

A∠B:= A∧(A/B)

If we consider the Lambek calculus extended by optional divisions, we call it

optional division Lambek calculus (ODLC), and we denote its types by T p(

∠

,∠).

We deﬁne ODLC-grammars and languages generated by them the same way as it

is done in Deﬁnition 1 and Deﬁnition 2. Note that the cut rule is also admissible

in the ODLC.

2 Main results

It is seen from Deﬁnition 3 that the ODLC is just a fragment of the Lambek

calculus with conjunction. It is interesting to investigate the recognizing power

of the ODLC (i.e. a property which shows how many languages can be generated

by ODLC-grammars) and to compare it with the recognizing power of L(∧). In

this work two proven results are presented:

Theorem 1. Any context-free language can be generated by an ODLC-grammar

hΣ, T , isuch that ∀a∈Σ|{T|aT}| ≤ 1.

Speaking less formally, this theorem proposes that any context-free language

can be generated by an ODLC-grammar in which at most one type is assigned

to each symbol.

Theorem 2. Any ﬁnite intersection of context-free languages can be generated

by some ODLC-grammar.

These theorems show that optional divisions are quite powerful and that

they inherit some properties of conjunction. Section 3 is concerned with proofs

of these theorems. In Section 4 obtained results and some open questions are

discussed.

3 Proofs

We prove both theorems by analysing free group interpretations of types and

sequents.

3.1 Free group interpretation

Let F G stand for the free group generated by all primitive types {pi|i∈N}. The

identity element is denoted by ε.

Deﬁnition 4. In [5] the mapping J K :T p →F G is deﬁned as follows:

Optional division Lambek calculus grammars 5

JpiK:= pi

JA/BK:= JAKJBK−1JA•BK:= JAKJBK

JB\AK:= JBK−1JAK JA1. . . AnK:= JA1K. . . JAnK

It is called the (free) group interpretation.

Now we extend it to types with conjunction, i.e. T p(∧). Let now the inter-

pretation be a ﬁnite subset of F G; its deﬁnition is also inductive and it inherits

Deﬁnition 4. The only diﬀerence is that piis now understood to be the set con-

taining one element pi, and all operations are understood to be set operations

which are applied to each set element (or pair of set elements). We add the

following deﬁnition for the conjunction case:

JA∧BK:= JAK∪JBK.

Example 1. J(p∧q)•(r∧s)/tK={prt−1, pst−1, qrt−1, qst−1}.

The following lemma is proven for the Lambek calculus:

Lemma 1. If L`Γ→C, then JΓK=JCK.

We also extend it to L(∧). In order to do that we deﬁne a set LimT p ⊆T p(∧)

as follows:

Deﬁnition 5. LimT p is the least set satisfying such requirements: P r ⊆LimT p;

if A, B ∈LimT p, then A•B, A ∧B∈LimT p; if A∈LimT p, C ∈T p, then

A/C, C\A∈LimT p.

Lemma 2. If L(∧)`A1. . . An→B1∧... ∧Bkand Ai∈LimT p, Bj∈T p then

∀j= 1, . . . , k JA1. . . AnK⊇JBjK.

Proof. It is easily seen that ∀i= 1, . . . , k L(∧)`A1. . . An→Bi. Considering

only sequents of the last form one notices that no type or fragment of a type in

an antecedent can “split” and get in diferent branches of a proof tree (it would be

possible if there appeared a conjunction in the succedent, but the last does not

happen, because no conjunctions appear in denominators of antecedent types).

So for each occurrence of such fragment as A∧Bin the sequent A1. . . An→Bi

antecedent one can replace it by one of its conjuncts Aor Bat the very begin-

ning of derivation and then derive a sequent in the ordinary Lambek caclulus.

Applying Lemma 2 we obtain the required. ut

Therefore, if for some jJA1. . . AnK6⊇ JBjKthen L(∧)6`A1. . . An→B1∧... ∧

Bk.

3.2 Context-free grammars

Deﬁnition 6. A context-free grammar Gis a 4-tuple hN, Σ , P, Si, where N

and Σare disjoint ﬁnite sets, Pis a ﬁnite set of productions, and S∈N. See

deﬁnition of a production and of a language generated by Gin [1].

We say that two grammars (possibly of diﬀerent types, e.g. a Lambek one

and a context-free one) are equivalent iﬀ they generate the same language.

It is proven that any context-free language can be generated by a Lambek

grammar and vice versa (see [5]), so we shall use any of these formalisms when

speaking about context-free languages.

6 Tikhon Pshenitsyn

3.3 Interpretable Lambek grammars

Here we present a new class of Lambek grammars which it is convenient to work

with.

Deﬁnition 7. Lambek grammar G=hΣ, t, iis said to be interpretable iﬀ t∈Pr ,

and ∀k≥1∀Ti∈T p(G),1≤i≤ksuch requirements are satisﬁed:

(a) JT1. . . TkK6=ε;

(b) JT1. . . TkK=t⇒L`T1, . . . , Tk→t;

(c) tdoes not appear in denominators of types in T p(G).

Theorem 3. Any context-free grammar without empty word is equivalent to an

interpretable Lambek grammar.

In order to prove this theorem we start from a context-free language Land

present an interpretable grammar that generates it.

Deﬁnition 8. A context-free grammar G=hΣ, N , P, Siis said to be in the

Rosenkrantz normal form (RNF) if each rule in Pis of one of the following

forms:

1. A→aαd (a, d ∈Σ, α ∈(Σ∪N)∗);

2. A→a(a∈Σ).

Theorem 4 (Rozenkrantz). Any context-free grammar without empty word

is equivalent to a context-free grammar in the RNF.

See the proof in [6].

Proof (of Theorem 3). Let us consider a context-free grammar G1=hΣ, N, P, Si

in the RNF generating L. We introduce |Σ|new nonterminal symbols, which

shall be denoted by A={Ta}a∈Σ;N0:= N∪A. We also deﬁne the function

b: (N∪Σ)+→(N0)+by the following equalities:

b

A=A(A∈N); ba=Ta(a∈Σ); [w1w2=cw1cw2.

The next step is to construct G2={Σ, N 0, P 0, S}such that P0={A→

ab

δ, A →aδ ∈P} ∪ {Ta→a|a∈Σ}. Obviously, L(G1) = L(G2). One notices

that G2does not have right recursions (see deﬁnition of this notion in [1]).

Let’s further assume without loss of generality that N0is a subset of P r.

We construct a Lambek grammar G3=hΣ, S, ias follows: a T iﬀ for some

A, B1, . . . , Bm∈N0we have T=A/Bm/ . . . /B1and A→aB1. . . Bm∈P0, m ≥

0 (note that every rule in G2is of this form).

It is argued that G3satisﬁes the property (a) from Deﬁnition 7, i.e. ∀n

∀T1, . . . , Tn∈T p(G3)JT1K. . . JTnK6=ε. The main observation for proving this

fact is that a reduction between interpretations of T p(G3) types corresponds to

composition of rules in the context-free grammar G2.

We deﬁne a new set T pExt as the least set satisfying following criteria:

T pExt ⊇T p(G3); if A/Bm/ . . . /B1, B1/Ck/ . . . /C1∈T pExt then

A/Bm/ . . . /B2/Ck/ . . . /C1∈T pExt. Speaking informally, types in T pExt are

composed of types in T p(G3) using the transitivity rule: A/p, p/C →A/C.

Optional division Lambek calculus grammars 7

Lemma 3. For each T=A/Bn/ . . . /B1∈T pExt there is a rule of the form

A→σB1. . . Bn, σ ∈Σ+which is got as a composition of several rules of G2.

Proof. We shall use induction on the number of applications of the transitivity

rule when constructing T. If T∈T p(G3) then it follows from the deﬁnition of G3.

Let T=A/Bm/ . . . /B2/Ck/ . . . /C1, and let it be composed of A/Bm/ . . . /B1,

B1/Ck/ . . . /C1∈T pExt. Applying the induction hypothesis one can show that

there are rules A→σB1. . . Bm,B1→τ C1. . . Ckfor some σ, τ ∈Σ+which are

got by a composition of several rules of G2. Then we can simply apply the second

rule within the ﬁrst rule and get A→σB1. . . Bm→στ C1. . . CkB2. . . Bm. This

completes the proof.

Lemma 4 (checking property a). ∀n∀T1, . . . , Tn∈T pExt JT1K. . . JTnK6=

ε.

Proof. Induction on n.

If n= 1, then T1=A/Bm/ . . . /B1, and there is a composition of rules of

G2which has the form A→σB1. . . Bm, σ ∈Σ+. Because there are no right

recursions in G2,A6=Bm. Then JT1K=AB−1

m· · · 6=ε.

The induction step is proved as follows: if there can be no reductions within

JT1K. . . JTnKthen it does not already equal ε. If Ti=A/Bm/ . . . /B1,Ti+1 =

C/Dk/ . . . /D1and JTiKJTi+1 Kcan be reduced, then C=B1and

A/Bm/ . . . /B2/Dk/ . . . /D1∈T pExt. The induction hypothesis completes the

proof.

Lemma 5 (checking property b). ∀n∀T1, . . . , Tn∈T pExt JT1K. . . JTnK=

S⇒L`T1, . . . , Tn→S.

Proof. Induction on n.

If n= 1, then T1=Sand L`S→S.

The induction step is proved as follows: if there can be no reductions within

JT1K. . . JTnK, then there are at least two positive occurences of primitive types

in this product and therefore it cannot be equal to S. Consequently, there is a

reduction between some JTiKand JTi+1K. Let Ti=A/Bm/ . . . /B1and Ti+1 =

B1/Dk/ . . . /D1. Then T=A/Bm/ . . . /B2/Dk/ . . . /D1∈T pExt. Applying the

induction hypothesis, one gets L`T1, . . . , Ti−1, T, Ti+2, . . . , Tn→S. Finally,

Ti, Ti+1 →T T1, . . . , Ti−1, T, Ti+2, . . . , Tn→S

T1, . . . , Tn→S(cut)

Now we present G4=hΣ, e

S, e

i, where e

Sis a new primitive type and ae

T iﬀ

a T or T=e

S/S •T0and a T 0. If some sequent T1, . . . , Tp→e

S,a1e

T1,. . . ,

ape

Tpis derived, then there is exactly one type in the antecedent of the form

e

S/S •T0— and it is the ﬁrst one. Hence T1=e

S/S •T0

1,a1 T 0

1, ai Ti(i≥2)

and L`T0

1, T2, . . . , Tp→S. This proves L(G4)⊆L(G3). The reverse inclusion

is obvious: if ai Ti(i≥1) and L`T1, . . . , Tp→S, then one can derive

L`e

S/S •T1, T2, . . . , Tp`e

S. Therefore, L=L(G1) = L(G2) = L(G3) = L(G4).

It is easy to see that G4inherits properties (a) and (b) of interpretable

grammars from G3, and that it also has the property (c). This completes the

proof. ut

8 Tikhon Pshenitsyn

3.4 How to model conjunction with optional divisions

The following construction, as it will be seen further, can be a substitute for

conjunction when proving theorems about ODLC-grammars.

Deﬁnition 9. T1×. . . ×Tn:= T1•(T1

∠

T2)•(T2

∠

T3)• · · · • (Tn−1

∠

Tn).

It works like a telescope: we can use each of optional divisions to consume a

type standing on the left-hand side of it. Consequently, we can reduce such type

as it shows

Lemma 6. If 1≤i<n, then ODLC `T1×. . . ×Tn→Ti•Tn.

Lemma 7. JT1×. . . ×TnK={JTi1K. . . JTikK|k≥1,1≤i1<· · · < ik=n}.

Proof. Induction on n. The case n= 1 is obvious. The general case is proven by

such equalities:

JT1•T1

∠

T2•T2

∠

T3• · · · • Tn−1

∠

TnK=JT1KJT1

∠

T2KJT2

∠

T3• · · · • Tn−1

∠

TnK=

=JT1K{JT−1

1K, ε}JT2•T2

∠

T3• · · · • Tn−1

∠

TnK={ε, JT1K}. . . {ε, JTn−1K}JTnK.

We apply induction hypothesis to obtain the last equality. ut

3.5 Proof of Theorem 1

Proof. Let’s consider an interpretable Lambek grammar G=hΣ, s, iwhich

generates a given language. Let’s ﬁx a new primitive type ξ∈P r. We construct

an ODLC-grammar e

G=hΣ, s, e

ias follows. Given a symbol a, let Ta

1, ..., T a

kabe

all types Tfor which aT holds. Then

ae

Pa:Pa= (Ta

1/ξ)× · · · × (Ta

ka/ξ)×ξ.

The role of ξis to restrict undesirable variants of using optional divisions,

e.g. when we omit all of them in a complex type.

First, using Lemma 6, we obtain ODLC `Pa→Ti/ξ •ξ. In the Lambek

calculus (and consequently in the ODLC) one can derive Ti/ξ •ξ→Ti, so

ODLC `Pa→Ti. Therefore if aj Tj, j = 1, . . . , l and L`T1, . . . , Tl→s, then

one can derive Pa1, . . . , Pal→susing the cut rule. This means that L(G)⊆

L(e

G).

In order to prove that L(e

G)⊆L(G) it is enough to show that if s∈

JPa1. . . PalK,aj∈Σ, then ∃Tj:aj Tj, j = 1, . . . , l and L`T1, . . . , Tl→s(see

Lemma 2; here it is applied to the sequent Pa1, . . . , Pal→s).

By Lemma 7, the interpretation of Pais the following set:

JPaK={ξ}∪{JTa

jK}ka

j=1 ∪ {JTa

j1Kξ−1JTa

j2Kξ−1. . . ξ−1JTa

jmK|m≥2, j1<· · · < jm}.

Let gi∈JPaiK, i = 1, . . . , l and Π:= g1. . . gl=s. Note that only ξ,ξ−1

and JTa

iKmay appear in Π, and there always is a nonempty sequence of type

Optional division Lambek calculus grammars 9

interpetations JTa

iKbetween any two ξand ξ−1standing nearby each other in Π.

Because Gis interpretable, sequence of JTa

iKcannot reduce to ε(by the property

(a)). This means that, if some primitive type ξor ξ−1appears in one of gi, then

it cannot reduce with some other primitive type ξ−1, ξ resp. Hence gican be

only of the form JTai

jiK. Using the property (b) of interpretable grammars, we

obtain

L`Ta1

j1, . . . , T al

jl→s,

which completes the proof. ut

3.6 Proof of Theorem 2

The main idea of the proof is to combine types from diﬀerent grammars with

the above construction.

The following denotation is used:

Deﬁnition 10.

A/Bδ=(Aif δ= 0,

A/B if δ= 1.

Proof. Let Gi=hΣ, si, ii, i = 1, . . . , N be interpretable Lambek grammars

generating context-free languages which we want to intersect. It can be assumed

without loss of generality that, for i6=j, types in T p(Gi)∪ {si}and types in

T p(Gj)∪ {sj}are built of disjoint sets of primitive types.

We ﬁx a new primitive type ξ∈P r and present a new grammar

G=hΣ, s1∧. . . ∧sN, i.

Given a symbol alet Ta,i

1, ..., T a,i

ka,i be all types Tfor which a iTholds. Then

aT iﬀ T= (Ta,1

j1/ξ)× · · · × (Ta,N

jN/ξ)×ξ , 1≤jl≤ka,l,1≤l≤N.

Clearly

N

T

i=1

L(Gi)⊆L(G): this follows from the fact that

∀l= 1, . . . , N (Ta,1

j1/ξ)× · · · × (Ta,N

jN/ξ)×ξ→Ta,l

jl.

Now let us assume that ai Ti, i = 1, . . . , n and JT1. . . TnK⊇ {s1, . . . , sN}.

As in the previous proof, one can notice that if a T , then the following holds

for some indices j1, . . . , jN:

JTK={ξ}∪{JTa,i

jiK}N

i=1∪{JTa,i1

ji1Kξ−1JTa,i2

ji2Kξ−1. . . ξ−1JTa,im

jimK|m≥2, i1<· · · < im}.

Let g1∈JT1K, . . . , gn∈JTnK, n ≥1 and let

Π=g1. . . gn=sM(2)

hold in F G for some ﬁxed M.

10 Tikhon Pshenitsyn

Can one of giequal ξor JTa,i1

ji1Kξ−1JTa,i2

ji2Kξ−1. . . ξ−1JTa,im

jimK? No, it cannot.

Otherwise there would be a positive number of occurrences of ξand ξ−1on the

left-hand side of (2). If ξappears, it has to reduce with some ξ−1. If this happens,

let’s consider the sequence of type interpretations between them, which has to

reduce to ε. This sequence is composed of several slices each of which corresponds

to some speciﬁc grammar:

Π=· · · ξJTa1

1,m1

j1

1

K. . . JTa1

M1,m1

j1

M1

K

| {z }

slice 1

. . . JTap

1,mp

jp

1

K. . . JTap

Mp,mp

jp

Mp

K

| {z }

slice p

ξ−1· · ·

Here slice 1 consists of interpretations of types in Gm1, slice 2 consists of inter-

pretations of types in Gm2and so on. One notices that none of such slices can

reduce to ε(because every Giis interpretable); besides reductions cannot hap-

pen between diﬀerent slices (because they consist of diﬀerent primitive types).

So the whole sequence of interpretations between ξand ξ−1cannot be reduced

to ε, whence ξor ξ−1appeared once shall not disappear.

Now it is quite obvious that gihave to be of the form JTa,M

lKfor some a, l:

otherwise, if gt=JTa,j

lKfor some tand j6=M, then the slice containing gt

shall not completely reduce (because every considered grammar is interpretable)

and the whole product g1. . . gncannot be equal to sM. This means that gi=

JT0

iK, such that aiMT0

i, i = 1, . . . , n, and ﬁnally, due to the property (b) of

interpretable grammars, which holds for GM,

L`T0

1, . . . , T 0

n→sM.

Because the above reasoning holds for each M, one can derive L(∧)`T1, . . . , Tn→

s1∧ · · · ∧ sNwhich ﬁnally proves TN

i=1 L(Gi) = L(G).

In order to construct the required ODLC-grammar, we have to change a

distinguished type. Let us suppose that N= 2u, u ∈N(proving under this

assumption suﬃces to prove the theorem for each N). Then we encode integers

k= 1, . . . , N by u-tuples (1(k); . . . ;u(k)), i(k)∈ {0,1},i= 1, . . . , u. Then we

present new primitive types s, σ1, . . . , σuand change each of Giby replacing si

by s/σ1(i)

1. . . /σu(i)

ueverywhere in T p(Gi) and in the distinguished type si. It

is easily seen that the language generated by Giis not changed in the result of

such substitution (this follows from interpetability of Gi).

Now we construct e

Gthe same way as it was done above for Gexcept for the

distinguished type: we shall take s∠σ1. . . ∠σuinstead of s1∧ · · · ∧ sN. Again

the only diﬃcult part is to prove that L(e

G)⊆TN

i=1 L(Gi). It is proven the

same way as for Gbut additionally we have to answer why, when considering a

sequence of type interpretations between ξand ξ−1, there cannot be reductions

between diﬀerent slices (note that now there are some common primitive types

contained in types of diﬀerent grammars Gi: these are sand σ1, . . . , σu). Because

of the property (c), siwas not in denominators of types in T p(Gi) before it was

replaced by s/σ1(i)

1. . . /σu(i)

u. Therefore s−1and σi,i= 1, . . . , u do not appear

in interpretations of types in T p(e

G), and there cannot be more reductions within

Πthan before replacing siby s/σ1(i)

1. . . /σu(i)

u.ut

Optional division Lambek calculus grammars 11

3.7 Examples

Consider the grammar hΣ, s, i:

a (s/t)/p, q/p;

b p, q\p;

c t, u, u\t, u/t.

This grammar generates the language Lc={anbnck|n, k ≥1}. Moreover,

this grammar is interpretable. Theorem 1 says that we can generate Lcby the

following ODLC-grammar where only one type is assigned to each symbol:

a q/p/ξ •(q/p/ξ)

∠

(((s/t)/p)/ξ)•(((s/t)/p)/ξ)

∠

ξ

b p/ξ •(p/ξ)

∠

(q\p/ξ)

c t/ξ •(t/ξ)

∠

(u/ξ)•(u/ξ)

∠

(u\t/ξ)•(u\t/ξ)

∠

((u/t)/ξ)•((u/t)/ξ)

∠

ξ

Now let us consider the grammar hΣ, s2, 0i:

a 0t2, u2, u2\t2, u2/t2.

b 0(t2\s2)/p2, q2/p2;

c 0p2, q2\p2;

This grammar is also interpretable, and it generates La={akbncn|n, k ≥

1}. Theorem 2 allows us to construct an ODLC-grammar hΣ, s∠σ1,e

iwhich

generates La∩Lc={anbncn|n≥1}:

ae

((s/t)/p)/ξ •(((s/t)/p)/ξ)

∠

(t2/ξ)•(t2/ξ)

∠

ξ,

(q/p)/ξ •((q/p)/ξ)

∠

(t2/ξ)•(t2/ξ)

∠

ξ,

. . . (8 types at all);

be

p/ξ •(p/ξ)

∠

(((t2\s/σ1)/p2)/ξ)•(((t2\s/σ1)/p2)/ξ)

∠

ξ,

p/ξ •(p/ξ)

∠

((q2/p2)/ξ)•((q2/p2)/ξ)

∠

ξ,

(q\p)/ξ •((q\p)/ξ)

∠

(((t2\s/σ1)/p2)/ξ)•(((t2\s/σ1)/p2)/ξ)

∠

ξ,

(q\p)/ξ •((q\p)/ξ)

∠

((q2/p2)/ξ)•((q2/p2)/ξ)

∠

ξ;

ce

t/ξ •(t/ξ)

∠

(p2/ξ)•(p2/ξ)

∠

ξ,

t/ξ •(t/ξ)

∠

((q2\p2)/ξ)•((q2\p2)/ξ)

∠

ξ,

. . . (8 types at all).

4 Conclusions and future work

An ability to generate ﬁnite intersections of context-free languages is quite pow-

erful. It is interesting that ODLC-grammars have it, because optional divisions

seem to be much weaker than conjunction. Note that we used only the left op-

tional division in proofs, and moreover it does not appear in denominators of

types — so may be we can generate many more languages when using both

divisions and more complex types? However, our proof of Theorem 2 does not

look pretty, and it is hard to generalize it: there are some technical tricks such

as a special symbol ξ, or a distinguished type of the form s∠σ1. . . ∠σu. It would

be nice to ﬁnd more elegant way to work with optional divisions.

There is a series of questions which naturally appear in view of Theorem 2:

12 Tikhon Pshenitsyn

1. Is the set of languages generated by ODLC-grammars closed under intersec-

tion?

2. Does this set contain the set of languages generated by conjunctive gram-

mars?

3. Does this set equal the set of languages generated by L(∧)-grammars?

Questions regarding algorithmical complexity are also interesting: are there

any beneﬁts in parsing sequents with optional divisions in comparison with con-

junction?

Another interesting question is connected with linguistics. Optional divisions

appear only in numerators of types in given examples in Section 1.2. Is there

any example in English or some other natural language when optional divisions

have to appear in a denominator of a type?

We hope that this theme shall lead to some nontrivial observations and re-

sults, so it is interesting for us to continiue this research.

5 Acknowledgments

Thanks to my teacher and scientiﬁc advisor Dr. Mati Pentus for guiding me and

for paying careful attention to my work. Thanks to my friend Alexey Starchenko

for discussing with me linguistic aspects of optional arguments.

References

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2. M. Kanazawa. The Lambek calculus enriched with additional connectives. Journal

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Michigan State University, East Lansing (2003).

4. J. Lambek. The mathematics of sentence structure. American Mathematical

Monthly, 65(3):154–170 (1958).

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