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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.
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4. J. Lambek. The mathematics of sentence structure. American Mathematical
Monthly, 65(3):154–170 (1958).
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