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The Review of Symbolic Logic
Volume 0, Number 0, — 2020
Complexity of the Infinitary Lambek Calculus with Kleene Star
STEPAN KUZNETSOVSteklov Mathematical Institute of the RAS
Abstract.
We consider the Lambek calculus, or non-commutative multiplicative in-tuitionistic linear logic, extended with iteration, or Kleene star, axiomatised by means ofan ω -rule, and prove that the derivability problem in this calculus is Π -hard. This solvesa problem left open by Buszkowski (2007), who obtained the same complexity bound forinfinitary action logic, which additionally includes additive conjunction and disjunction.As a by-product, we prove that any context-free language without the empty word canbe generated by a Lambek grammar with unique type assignment, without Lambek’snon-emptiness restriction imposed (cf. Safiullin 2007). Keywords:
Kleene star, Lambek calculus, non-commutative linear logic, infinitaryaction logic, complexity, categorial grammars. § Residuated structures play an important rˆole in abstractalgebra and substructural logic (Krull, 1924; Ward & Dilworth, 1939; Ono, 1993;Jipsen & Tsinakis, 2002; Galatos et al., 2007; Abramsky & Tzevelekos, 2010). Weintroduce the Lambek calculus with the unit (Lambek, 1969) as the algebraic logic(inequational theory) of residuated monoids . A residuated monoid is a partiallyordered algebraic structure h A ; (cid:22) , · , , \ , / i , where: • h A ; · , i is a monoid; • (cid:22) is a preorder; • \ and / are residuals of · w.r.t. (cid:22) , i.e. A (cid:22) C / B ⇐⇒ A · B (cid:22) C ⇐⇒ B (cid:22) A \ C. Notice that in the presence of residuals we do not need to postulate monotonicityof · w.r.t. (cid:22) explicitly: it follows from the rules for \ and / (Lambek, 1958).The Lambek calculus with the unit, L , axiomatises the set of atomic sentencesof the form A (cid:22) B (where A and B are formulae constructed from variablesand constant using three binary operations: · , \ , / ) which are generally truein residuated monoids.We formulate L in the form of a Gentzen-style sequent calculus. Formulae of L are built from a countable set of variables Var = { p , p , p , . . . } and constant using three binary connectives: \ , / , · . Sequents are expressions of the form Π → A ,where A (succedent) is a formula and Π (antecedent) is a finite sequence of formulae.The antecedent Π is allowed to be empty; the empty sequence of formulae is denotedby Λ.Sequent B , . . . , B n → A are interpreted as B · . . . · B n (cid:22) A ; sequent Λ → A means (cid:22) A .Axioms of L are sequents of the form A → A , and Λ → . Rules of inferenceare as follows: c (cid:13) doi:10.1017/S1755020300000000 U064-05-FPR Lstar˙Pi1˙final 4 May 2020 0:27 stepan kuznetsov Π → A Γ , B, ∆ → C Γ , Π , A \ B, ∆ → C ( \ → ) A, Π → B Π → A \ B ( → \ )Π → A Γ , B, ∆ → C Γ , B / A, Π , ∆ → C ( / → ) Π , A → B Π → B / A ( → / )Γ , A, B, ∆ → C Γ , A · B, ∆ → C ( · → ) Γ → A ∆ → B Γ , ∆ → A · B ( → · ) Γ , ∆ → C Γ , , ∆ → C ( → )Π → A Γ , A, ∆ → C Γ , Π , ∆ → C (cut)Completeness is proved by standard Lindenbaum – Tarski construction.One of the natural examples of residuated monoids is the algebra P (Σ ∗ ) offormal languages over an alphabet Σ. The preorder on P (Σ ∗ ) is the subset relation;multiplication is pairwise concatenation: A · B = { uv | u ∈ A, v ∈ B } and divisions are defined as follows: A \ B = { u ∈ Σ ∗ | ( ∀ v ∈ A ) vu ∈ B } ; B / A = { u ∈ Σ ∗ | ( ∀ v ∈ A ) uv ∈ B } . This interpretation of the Lambek calculus on formal languages corresponds tothe original idea of Lambek (1958) to use the Lambek calculus as a basis for categorial grammars.
The concept of categorial grammars goes back to Ajdukiewicz(1935) and Bar-Hillel (1953). Nowadays categorial grammars based on the Lambekcalculus and its extensions are used to describe fragments of natural languagein the type-logical linguistic paradigm. In this article we need Lambek categorialgrammars only as a technical gadget for our complexity proofs. Thus, in § ∧ ) and join ( ∨ ), which impose a lattice structure on thegiven preorder: a ∧ b = inf (cid:22) { a, b } , a ∨ b = sup (cid:22) { a, b } , and we postulate that thesesuprema and infima exist for any a, b . A residuated monoid with meets and joins isa residuated lattice . On the algebra of formal languages, meet and join correspondto set-theoretic intersection and union respectively.On the logical side, meet and join are called, respectively, additive conjunctionand disjunction. This terminology follows Girard’s linear logic (Girard, 1987). Themonoidal product, · , is multiplicative conjunction, and two divisions, \ and / , areleft and right linear implications. The rules for ∧ and ∨ are as follows:Γ , A i , ∆ → C Γ , A ∧ A , ∆ → C ( ∧ → ) , i = 1 , → A Π → A Π → A ∧ A ( → ∧ )Γ , A , ∆ → C Γ , A , ∆ → C Γ , A ∨ A , ∆ → C ( ∨ → ) Π → A i Π → A ∨ A ( → ∨ ) , i = 1 , U064-05-FPR Lstar˙Pi1˙final 4 May 2020 0:27
Lambek Calculus with Kleene Star split, and we have to consider two implications (left and right) and two versions ofconjunction (multiplicative and additive). For more discussion of substructuralitywe refer to Restall (2000).Another, more sophisticated operation to be added to residuated monoids is iteration, or Kleene star, first introduced by Kleene (1956). Residuated latticesextended with Kleene star are called residuated Kleene lattices (RKLs), or actionlattices (Pratt, 1991; Kozen, 1994b). Throughout this article, we consider only *-continuous RKLs, in which Kleene star is defined as follows: a ∗ = sup (cid:22) { a n | n ≥ } (where a n = a · . . . · a , n times, and a = ). In particular, the definition of a*-continuous RKL postulates existence of all these suprema.Axioms and rules for Kleene star naturally come from its definition: (cid:0) Γ , A n , ∆ → C (cid:1) ∞ n =0 Γ , A ∗ , ∆ → C ( ∗ → ) ω Π → A . . . Π n → A Π , . . . , Π n → A ∗ ( → ∗ ) , n ≥ A n means A, . . . , A , n times; A = Λ.The left rule for Kleene star is an ω -rule, i.e., has countably many premises. In thepresence of ω -rule, the notion of derivation should be formulated more accurately.A valid derivation is allowed to be infinite, but still should be well-founded, thatis, should not include infinite paths. Thus, the set of theorems of ACT ω is thesmallest set of sequents which includes all axioms and is closed under applicationof rules.Notice that meets and joins are not necessary for defining the Kleene star; thuswe can consider residuated monoids with iteration .The logic presented above axiomatises the inequational theory of *-continousRKLs. It is called infinitary action logic (Buszkowski & Palka, 2008) and denoted by ACT ω . Syntactically, the set of theorems of ACT ω is the smallest set that includesall axioms and is closed under inference rules presented above; completeness is againby Lindenbaum – Tarski construction.Palka (2007) proved cut elimination for ACT ω . As usual, cut elimination yieldssubformula property (if a formula appears somewhere in a cut-free derivation, thenit is a subformula of the goal sequent). Therefore, elementary fragments of ACT ω with restricted sets of connectives are obtained by simply taking the correspondingsubsets of inference rules. These fragments are denoted by listing the connectivesin parentheses after the name of the calculus, like ACT ω ( \ , ∨ , ∗ ), for example.Some of these fragments also have their specific names: ACT ω ( · , \ , /, , ∗ ) is theLambek calculus with the unit and iteration and is denoted by L ∗ ω ; its fragmentwithout the unit, ACT ω ( · , \ , /, ∗ ), is denoted by L ∗ ω . The multiplicative-additiveLambek calculus, MALC , is
ACT ω ( · , \ , /, ∨ , ∧ ). The Lambek calculus with theunit, L , is ACT ω ( · , \ , /, ).Finally, ACT ω ( · , \ , / ) is the Lambek calculus allowing empty antecedents, butwithout the unit constant. This calculus is usually denoted by L ∗ (Lambek, 1961).Unfortunately, this yields a notation clash with the Kleene star, which is also In the presence of divisions, the usual definition of *-continuity, b · a ∗ · c = sup (cid:22) { b · a n · c | n ≥ } , can be simplified by removing the context b , c .U064-05-FPR Lstar˙Pi1˙final 4 May 2020 0:27 stepan kuznetsov denoted by ∗ . Just “ L ” is reserved for the system with Lambek’s restriction (seethe next section). Therefore, we introduce a new name for this calculus: L Λ , whichmeans that empty antecedents are allowed in this calculus. By L Λ ( \ , / ) we denotethe product-free fragment of L Λ (which is the same as ACT ω ( \ , / )).Buszkowski (2007) and Palka (2007) show that the derivability problem in ACT ω is Π -complete—in particular, the set of all theorems of ACT ω is not recursivelyenumerable. Thus, the usage of an infinitary proof system becomes inevitable.Buszkowski (2007) also shows Π -hardness for fragments of ACT ω where one of theadditive connectives ( ∨ or ∧ , but not both) is removed. Complexity of the Lambekcalculus with Kleene star, but without both additives, i.e., the logic of residuatedmonoids with iteration, which we denote by L ∗ ω , however, is left by Buszkowski asan open problem. In this article we prove Π -hardness of this problem (the upperΠ bound is inherited by conservativity).The rest of this article is organised as follows. In § § § §
5. In §
4, we representcomplexity results by Buszkowski and Palka for
ACT ω and show how to strengthenthe lower bound and prove Π -completeness for the system without ∨ and ∧ . Thiscomplexity proof uses the version of Safiullin’s theorem for the Lambek calculuswithout Lambek’s restriction. We prove it in §
5, which is the most technically hardsection. Finally, § § § A preliminary version of this article was presentedat WoLLIC 2017 and published in its lecture notes (Kuznetsov, 2017). In theWoLLIC paper, we show Π -completeness of a system closely related to L ∗ ω —namely, the logic of residuated semigroups with positive iteration, denoted by L + ω .From the logical point of view, in L + ω there is no unit constant, and there shouldbe always something on the left-hand side.This constraint is called Lambek’s restriction . In the presence of Lambek’s re-striction one cannot add Kleene star, and it gets replaced by positive iteration, or“Kleene plus,” with the following rules: (cid:0) Γ , A n , ∆ → C (cid:1) ∞ n =1 Γ , A + , ∆ → C ( + → ) ω Π → A . . . Π n → A Π , . . . , Π n → A + ( → + ) n , n ≥ A + = A · A ∗ .Lambek’s restriction was imposed on the calculus L in Lambek’s original pa-per (Lambek, 1958) and is motivated linguistically (Moot & Retor´e, 2012, Sect. 2.5).Unfortunately, there are no conservativity relations between L + ω and L ∗ ω . For exam-ple, ( p \ p ) \ q → q is derivable in L Λ (thus in L ∗ ω ), but not in L (thus not in L + ω ),though the antecedent here is not empty. Hence, Π -hardness of the latter is notobtained automatically as a corollary.Moreover, the proof of Π -hardness of L + ω crucially depends on the followingresult by Safiullin (2007): any context-free language without the empty word canbe generated by a Lambek grammar with unique type assignment, and Safiullin’sproof essentially uses Lambek’s restriction. In this article, we feature a new result, U064-05-FPR Lstar˙Pi1˙final 4 May 2020 0:27
Lambek Calculus with Kleene Star namely, Π -completeness of L ∗ ω itself. We modify Safiullin’s construction and extendhis result to L Λ (which is already interesting on its own). Next, we use this newresult in order to prove Π -hardness of L ∗ ω . § In this section weintroduce Lambek categorial grammars and formulate equivalence results connect-ing them with a more widely known formalism, context-free grammars. The notionof
Lambek grammar is defined as follows:
Definition
A Lambek grammar over an alphabet Σ is a triple G = h Σ , H, ✄ i ,where H is a designated Lambek formula, called goal type, and ✄ is a finite binarycorrespondence between letters of Σ and Lambek formulae. Definition
A word a . . . a n over Σ is accepted by a grammar G , if there existformulae A , . . . , A n such that a i ✄ A i ( i = 1 , . . . , n ) and the sequent A , . . . , A n → H is derivable in the Lambek calculus. Definition
The language generated by grammar G is the set of all wordsaccepted by this grammar. In view of the previous section, one should distinguish grammars with and with-out Lambek’s restriction: the same grammar could generate different languages,depending on whether Lambek’s restriction is imposed or not.We also recall the more well-known notion of context-free grammars.
Definition
A context-free grammar over alphabet Σ is a quadruple G = hN , Σ , P, S i , where N is an auxiliary alphabet of non-terminal symbols, not inter-secting with Σ , S is a designated non-terminal symbol called the starting symbol,and P ⊂ N × ( N ∪ Σ) ∗ is a finite set of production rules. Production rules arewritten in the form A ⇒ α , where A is a non-terminal symbol and α is a word(possibly empty) over alphabet N ∪ Σ . Production rules with an empty α (of theform A ⇒ ε ) are called ε -rules. Definition
A word ηαθ over
N ∪ Σ is immediately derivable from ηAθ incontext-free grammar G (notation: ηAθ ⇒ G ηαθ ), if ( A ⇒ α ) ∈ P . The derivabilityrelation ⇒ ∗G is the reflexive-transitive closure of ⇒ G . Definition
The language generated by context-free grammar G is the set ofall words over Σ (i.e., without non-terminals) which are derivable from the startingsymbol S : { w ∈ Σ ∗ | S ⇒ ∗G w } . Such languages are called context-free. Two grammars (for example, a Lambek grammar and a context-free one) arecalled equivalent, if they generate the same language.Buszkowski’s proof of Π -hardness of ACT ω uses the following translation ofcontext-free grammars into Lambek grammars: Theorem
Any context-free grammar without ε -rules can be algorithmically transformed into an equivalent Lambek grammar, nomatter with or without Lambek’s restriction. We use Λ for the empty sequence of formulae and ε for the empty word over an alphabet.U064-05-FPR Lstar˙Pi1˙final 4 May 2020 0:27 stepan kuznetsov This theorem was proved by Gaifman (Bar-Hillel et al., 1960), but with basiccategorial grammars instead of Lambek grammars. Buszkowski (1985) noticed thatGaifman’s construction works for Lambek grammars also.The reverse translation is also available (in this article we do not need it):
Theorem
Any Lambek grammar, no matter with or withoutLambek’s restriction, can be algorithmically transformed into an equivalent context-free grammar. (Pentus, 1993)For our purposes we shall need a refined version of Theorem 3.7.
Definition
A Lambek grammar is a grammar with unique type assignment,if for any a ∈ Σ there exists exactly one formula A such that a ✄ A . Theorem
Any context-free grammar without ε -rules can bealgorithmically transformed into an equivalent Lambek grammar with unique typeassignment with Lambek’s restriction. (Safiullin, 2007)Notice that Theorem 3.10, as formulated and proved by Safiullin, needs Lambek’srestriction. If one applied Safiullin’s transformation and then abolished Lambek’srestriction, the resulting grammar could generate a different language, and the newgrammar would not be equivalent to the original context-free one. § We start with the known results on algorithmic complexity of
ACT ω , infinitaryaction logic with additive connectives. Theorem
The derivability problem in
ACT ω is Π -complete. (Buszkowski, 2007; Palka, 2007)An algorithmic decision problem, presented as a set A , belongs to Π , if thereexists a decidable set R of pairs h x, y i , such that x ∈ A if and only if h x, y i ∈ R for all y . The Π complexity class is dual to Σ , the class of recursively enumerablesets : a set is Π if and only if its complement is recursively enumerable. The “mostcomplex” sets in Π are called Π -complete sets: a set A is Π -complete, if (1)it belongs to Π (upper bound); (2) it is Π -hard, that is, any other Π set B ism-reducible to A (lower bound). The latter means that there exists a computablefunction f such that x ∈ A iff f ( x ) ∈ B . By duality, a set is Π -complete iffits complement is Σ -complete. For example, since the halting problem for Turingmachines is Σ -complete, the non-halting problem is Π -complete.Notice that a Π -complete set cannot belong to Σ : otherwise it would be decid-able by Post’s theorem, and this would lead to decidability of all sets in Π , which isnot the case. Thus, Theorem 4.11 implies the fact that the set of sequents provablein ACT ω is not recursively enumerable, and ACT ω itself cannot be reformulatedas a system with finite derivations.Division operations ( / and \ ) are essential for the Π lower complexity bound.For the fragment ACT ω ( · , ∨ , ∗ ) a famous result by Kozen (1994a) provides com-pleteness of an inductive axiomatization for Kleene star and establishes PSPACE We suppose that sequents of
ACT ω , as well as context-free grammars, are encoded aswords over a fixed finite alphabet, or as natural numbers.U064-05-FPR Lstar˙Pi1˙final 4 May 2020 0:27 Lambek Calculus with Kleene Star complexity. As we prove in this article, however, the fragment with divisions andwithout additives, ACT ω ( \ , /, · , ∗ ), is still Π -complete.The upper bound in Theorem 4.11 was proved by Palka (2007) using the following*-elimination technique. For each sequent define its n -th approximation by replacingall negative occurrences of A ∗ by A ≤ n = ∨ A ∨ A ∨ . . . ∨ A n . Formally this isdone by the following mutually recursive definitions: P n ( p i ) = p i N n ( p i ) = p i P n ( ) = N n ( ) = P n ( A \ B ) = N n ( A ) \ P n ( B ) N n ( A \ B ) = P n ( A ) \ N n ( B ) P n ( B / A ) = P n ( B ) / N n ( A ) N n ( B / A ) = N n ( B ) / P n ( A ) P n ( A · B ) = P n ( A ) · P n ( B ) N n ( A · B ) = N n ( A ) · N n ( B ) P n ( A ∨ B ) = P n ( A ) ∨ P n ( B ) N n ( A ∨ B ) = N n ( A ) ∨ N n ( B ) P n ( A ∧ B ) = P n ( A ) ∧ P n ( B ) N n ( A ∧ B ) = N n ( A ) ∧ N n ( B ) P n ( A ∗ ) = (cid:0) P n ( A ) (cid:1) ∗ N n ( A ∗ ) = (cid:0) N n ( A ) (cid:1) ≤ n Now the n -th approximation of a sequent A , . . . , A k → B can be defined as N n ( A ) , . . . , N n ( A n ) → P n ( B ). Palka (2007) proves that a sequent is derivablein ACT ω if and only if all its approximations are derivable. In a cut-free derivationof an approximation, however, the ( ∗ → ) ω rule could never be applied, since thereare no more negative occurrences of ∗ -formulae. Proof search without the ω -rule isdecidable. Thus, we get Π upper bound for ACT ω . By conservativity, this upperbound is valid for all elementary fragments of ACT ω , in particular, for L ∗ ω .For the lower bound (Π -hardness), Buszkowski (2007) presents a reduction of awell-known Π -complete problem, the totality problem for context-free grammars,to derivability in ACT ω . Buszkowski’s reduction is as follows. Let total + denotethe following algorithmic problem: given a context-free grammar without ε -rules,determine whether it generates all non-empty words over Σ. It is widely knownthat total + is Π -complete (Sipser, 2012; Du & Ko, 2001). The reduction of thisproblem to derivability in ACT ω is performed as follows: given a context-freegrammar, transform it (by Theorem 3.7) into an equivalent Lambek grammar G = h Σ , ✄ , H i . Suppose that Σ = { a , . . . , a m } and for each i = 1 , . . . , m the formulaethat are in ✄ correspondence with a i are A i, , . . . , A i,n i . Let A i = A i, ∧ . . . ∧ A i,n i ( i = 1 , . . . , n ) and E = A ∨ . . . ∨ A m . Then the grammar generates all non-emptywords if and only if E + → H is derivable in ACT ω (Buszkowski, 2007, Lm. 5).Recall that E + is E ∗ , E . Thus, we have a reduction of total + to the derivabilityproblem in ACT ω , and therefore the latter is Π -hard.This construction essentially uses additive connectives, ∨ and ∧ . As shown byBuszkowski (2007), one can easily get rid of ∨ by the following trick. First, noticethat total + is already Π -hard if we consider only languages over a two-letteralphabet { a , a } (Du & Ko, 2001). Second, E ∗ = ( A ∨ A ) ∗ can be equivalentlyreplaced by ( A ∗ · A ) ∗ · A ∗ (this equivalence is a law of Kleene algebra, thus provablein ACT ω ). The sequent E ∗ , E → H can be now replaced by a sequent without ∨ U064-05-FPR Lstar˙Pi1˙final 4 May 2020 0:27 stepan kuznetsov by the following chain of equivalent sequents:( A ∨ A ) ∗ , A ∨ A → H ( A ∗ · A ) ∗ · A ∗ , A ∨ A → H ( A ∗ · A ) ∗ · A ∗ → H / ( A ∨ A )( A ∗ · A ) ∗ · A ∗ → ( H / A ) ∧ ( H / A )(The last step is due to the equivalence A / ( B ∨ C ) ↔ ( A / B ) ∧ ( A / C ), which isprovable in
MALC and thus in
ACT ω .)Buszkowski also shows how to prove Π -hardness for ACT ω without ∧ —but then ∨ becomes irremovable. We show how to get rid of both ∨ and ∧ at once. Theorem
The derivability problem in L ∗ ω is Π -complete. The upper bound follows from Palka’s result by conservativity.For the lower bound (Π -hardness) we use the following alternation problem,denoted by alt , instead of total + . A context-free grammar over a two-letteralphabet { a , a } belongs to alt , if the language it generates includes all wordsbeginning with a and ending with a . (Other words can also belong to this lan-guage.) The alternation problem alt is also Π -hard, by the following reductionfrom total + : take a context-free grammar with starting symbol S and append anew starting symbol S ′ with rules S ′ ⇒ a Sa and S ′ ⇒ a a ; the new grammarbelongs to alt if and only if the original grammar belongs to total + .Now we translate our context-free grammar to a Lambek grammar, as Buszkowskidoes. It is easy to see that the grammar belongs to alt if and only if the sequent( A +1 · A +2 ) + → H is derivable in ACT ω : Lemma
The sequent ( A +1 · A +2 ) + → H is derivable in ACT ω if and only ifso are all sequents A n , A m , A n , A m , . . . , A n k , A n k → H for any k , n , m , n , m , . . . , n k , m k ≥ .Proof. By cut, one can easily establish that the ω -rule is invertible: A → A . . . A → AA n → A ∗ ( → ∗ ) n Γ , A ∗ , ∆ → C Γ , A n , ∆ → C (cut)and so is ( · → ): A → A B → BA, B → A · B ( → · ) Γ , A · B, ∆ → C Γ , A, B, ∆ → C (cut)Now the “if” part goes by direct application of ( · → ) and ( ∗ → ) ω and the “only if”one by their inversion. (cid:3) Now our proof of Theorem 4.12 will be finished if we manage to formulate A and A without ∧ . Recall that A i = A i, ∧ . . . ∧ A i,n i , where A i, , . . . , A i,n i arethe formulae which are in the ✄ correspondence with a i . For a Lambek grammarwith unique type assignment we have n i = 1, and A i does not contain ∧ . Thus,Theorem 4.12 now follows from the fact that any context-free grammar with-out ε -rules can be equivalently transformed to a Lambek grammar with uniquetype assignment, without Lambek’s restriction. Indeed, our language belongs to U064-05-FPR Lstar˙Pi1˙final 4 May 2020 0:27
Lambek Calculus with Kleene Star alt if and only if any word of the form a n a m . . . a n k a n k , for arbitrary k , n , m , n , m , . . . , n k , m k ≥
1, belong to the language. By definition of Lambekgrammar with unique type assignment, this happens exactly when all sequents A n , A m , A n , A m , . . . , A n k , A n k → H are derivable in L Λ , and, by conservativity,in ACT ω .In other words, we need an L Λ -variant of Safiullin’s Theorem 3.10, which we aregoing to prove in the next section (Theorem 5.14). § Λ In this section, we consider only theproduct-free fragment L Λ ( \ , / ). By “ ⊢ Π → B ” we mean that Π → B is derivable inthis calculus. By an L Λ ( \ , / )-grammar we mean a Lambek grammar without Lam-bek’s restriction, where all formulae do not include the multiplication operation. Theorem
Any context-free grammar without ε -rules can be algorithmicallytransformed to an equivalent L Λ ( \ , / ) -grammar with unique type assignment. Before presenting the construction of the grammar itself, we introduce Safiullin’stechnique of proof analysis for the product-free Lambek calculus, adapted for thecase without Lambek’s restriction. This technique has something in common withproof nets and focusing for non-commutative linear logic; however, due to thesimplicity of L Λ ( \ , / ), this technique works directly with Gentzen-style cut-freederivations.We start with a simple and well-known fact: Lemma
The ( → \ ) and ( → / ) rules are reversible, i.e., if ⊢ Π → A \ B ,then ⊢ A, Π → B , and if ⊢ Π → B / A , then ⊢ Π , A → B .Proof. Derivability is established using cut (and then, if we want a cut-free deriva-tion, eliminate cut).Π → A \ B A → A B → BA, A \ B → B ( \ → ) A, Π → B (cut) Π → B / A A → A B → BB / A, A → B ( / → )Π , A → B (cut) (cid:3) Definition
For any L Λ ( \ , / ) -formula let its top be a variable occurrencedefined recursively as follows: the top of a variable is this variable occurrence itself; the top of ( A \ B ) and, symmetrically, of ( B / A ) is the top of B . For convenience we consider cut-free derivations with axioms of the form q → q ,where q is a variable. Axioms A → A with complex formulae A are derivable(induction on A ). Definition
In a cut-free derivation of Π → q , the principal occurrence of q in Π is the one that comes from the same axiom as the q in the succedent. Wedenote the principal occurrence by q . In proof nets, the principal occurrence is the one connected by an axiom link to the q in the succedent.U064-05-FPR Lstar˙Pi1˙final 4 May 2020 0:27 stepan kuznetsov Notice that the notion of principal occurrence depends on a concrete derivation,not just on the fact of derivability of Π → q . Lemma
The principal occurrence is always a top.Proof.
Induction on derivation. (cid:3)
Using this lemma, one can locate possible principal occurrences by searching fortops with the same variable as the succedent (which has been reduced to its topvariable by Lemma 5.15).Introduce the following shortcut for curried passing of denominators: if Γ = X , . . . , X n and ∆ = Y , . . . , Y m are sequences of formulae, then letΓ \ q / ∆ = X n \ . . . \ X \ q / Y m / . . . / Y . ( / associates to the left and \ associates to the right). If Γ or ∆ is empty, we justomit the corresponding divisions:Λ \ q / ∆ = q / Y m / . . . / Y ;Γ \ q / Λ = X n \ . . . \ X \ q ;Λ \ q / Λ = q. (Since ( A \ B ) / C and A \ ( B / C ) are equivalent, we do not need parentheses here.)Next follows the decomposition lemma, which allows reverse-engineering of ( / → )and ( \ → ), once we have located the principal occurrence. Lemma If ⊢ Φ , ( X , . . . , X n ) \ q / ( Y , . . . , Y m ) , Ψ → q , then Φ = Φ , . . . , Φ n (some of Φ i may be empty), ⊢ Φ i → X i for any i = 1 , . . . , n ; Ψ = Ψ , . . . , Ψ m (some of Ψ j may be empty), ⊢ Ψ j → Y j for any j = 1 , . . . , m .Proof. Induction on derivation. By definition of principal occurrence, q always goesto the right branch in applications of ( / → ) and ( \ → ). (cid:3) In particular, Lemma 5.19 yields the following corollary: if the principal occur-rence is the top of a formula of the form p /
Γ, then it should be the leftmost formulain the antecedent.The next ingredient of Safiullin’s construction is the sentinel formula. Thisformula is used to delimit parts of the sequent and force them to behave indepen-dently. Safiullin uses a very simple formula, p / p , as a sentinel. Without Lambek’srestriction, however, this will not work, because Λ → p / p is now derivable, makingthe sentinel practically useless. We use a more complicated sentinel, using thetechnique of raising . A formula A raised using q is q / ( A \ q ). Our sentinel is asfollows: S p,q,r = ( r / ( p \ r )) / ( q / ( p \ q )) . Variables p , q , and r are parameters of the sentinel. We shall take fresh variablesfor them.The top of S p,q,r is r . In the notation of Lemma 5.19, S p,q,r = r / ( q / ( p \ q ) , p \ r ),thus, it is of the form r / Γ.The following lemma trivialises the analysis of sequents of the form S p,q,r , Π → S p,q,r , if the principal occurrence happens to be the top of the leftmost S p,q,r . U064-05-FPR Lstar˙Pi1˙final 4 May 2020 0:27
Lambek Calculus with Kleene Star Lemma If Π does not have p or q in tops (in particular, Π could include S p,q,r , since the top of the sentinel is r ) and ⊢ r / ( q / ( p \ q ) , p \ r ) , Π , q / ( p \ q ) , p \ r → r, then Π = Λ .Proof.
By Lemma 5.19, Π , q / ( p \ q ) , p \ r = Ψ , Ψ ; ⊢ Ψ → q / ( p \ q ); ⊢ Ψ → p \ r. Inverting ( → / ) in the first sequent yields ⊢ Ψ , p \ q → q . If Ψ does not include q / ( p \ q ), then the only principal occurrence is the q in p \ q , and by Lemma 5.19 ⊢ Ψ → p , which contradicts Lemma 5.18: there are no p tops in Ψ (which is asubsequence of Π).On the other hand, Ψ = Λ, since Λ → p \ r . Thus, Ψ = Π , q / ( p \ q ) and ⊢ Π , q / ( p \ q ) , p \ q → q or ⊢ Π , q / ( p \ q ) , p \ q → q. In the first case, Π = Λ by Lemma 5.19, q.e.d.
In the second case, ⊢ Π , q / ( p \ q ) → p fails, since there is no p top in the antecedent. (cid:3) Lemma If Φ , Φ , . . . , Φ n do not have p , q , or r in tops and ⊢ Φ , S p,q,r , Φ , S p,q,r , . . . , Φ n − , S p,q,r , Φ n → S p,q,r , then n = 1 and all Φ i are empty. (In other words, the only derivable sequent of thisform is the trivial S p,q,r → S p,q,r .) In particular, Λ → S p,q,r and S p,q,r , S p,q,r → S p,q,r .Proof. Inverting ( → / ) by Lemma 5.15, we get ⊢ Φ , S p,q,r , Φ , S p,q,r , . . . , Φ n − , S p,q,r , Φ n , q / ( p \ q ) , p \ r → r. Let us locate the principal occurrence of r . This should be a top, therefore it iseither the rightmost r in p \ r or r in one of the sentinels. In the first case we get,by Lemma 5.19, ⊢ Φ , S p,q,r , Φ , S p,q,r , . . . , Φ n − , S p,q,r , Φ n , q / ( p \ q ) → p, which immediately fails to be derivable, since there is no p top in the antecedentto become the principal occurrence.In the second case, since all sentinels are of the form r / Γ, the principal occurrenceshould be the leftmost one. Thus, Φ = Λ and the principal occurrence is the topof the leftmost S p,q,r . (In particular, at least one sentinel should exist, i.e., n ≥ ⊢ r / ( q / ( p \ q ) , p \ r ) , Φ , S p,q,r , . . . , Φ n − , S p,q,r , Φ n , q / ( p \ q ) , p \ r → r. Now by Lemma 5.20 Φ , S p,q,r , . . . , Φ n − , S p,q,r , Φ n should be empty. (cid:3) Next, we are going to need the construction of joining formulae (Pentus, 1994).A joining formula for A , . . . , A n is a formula B such that A i → B is derivablefor all i = 1 , . . . , n . The joining formula is a substitute for A ∨ . . . ∨ A n in thelanguage without ∨ and ∧ . One can also consider a joining formula for a set of U064-05-FPR Lstar˙Pi1˙final 4 May 2020 0:27 stepan kuznetsov sequences of formulae Γ , . . . , Γ n as such B that Γ i → B is derivable for all i . Dueto the substructural nature of the Lambek calculus, a joining formula does notalways exist. For example, two different variables, p and q , are not joinable. Pentus’criterion of joinability is based on the free group interpretation of the Lambekcalculus. Definition
Let FG be the free group generated by Var ; ε (the empty word)is its unit. Then [[ A ]] , the interpretation of Lambek formula A in FG , is definedrecursively as follows: • [[ p ]] = p for p ∈ Var ; • [[ ]] = ε ; • [[ A · B ]] = [[ A ]][[ B ]] ; • [[ A \ B ]] = [[ A ]] − [[ B ]] ; • [[ B / A ]] = [[ B ]][[ A ]] − .For a sequence Γ of formulae its interpretation in FG is defined as follows: • if Γ = A , . . . , A n , then [[Γ]] = [[ A ]] . . . [[ A n ]] ; • [[Λ]] = ε . One can easily see (by induction on derivation) that if A → B is derivable, then[[ A ]] = [[ B ]]. Thus, if there exists a joining formula for a set of formulae (or sequencesof formulae), then they should have the same free group interpretation. In fact, thisgives a criterion on the existence of a joining formula.
Theorem If [[Γ ]] = [[Γ ]] = . . . = [[Γ n ]] then there exists a formula B of thelanguage of \ , / , such that ⊢ Γ i → B for all i = 1 , . . . , n . This theorem is an easy corollary of the results of Pentus (1994). Return tothe calculus with multiplication, L Λ . In this calculus, for each Γ i = A i, , . . . , A i,n i consider its product G i = A i, · . . . · A i,n i . If Γ i = Λ, take G i = q / q for an arbitraryvariable q . Sequents Γ i → G i are derivable. Then apply the main result from Pentus’article (Pentus, 1994, Thm. 1) which yields a joining formula F for { G , . . . , G n } .This formula could include the multiplication connective. However, for each formula F , possibly with multiplication, there exists a formula B in the language of \ , / ,such that F → B is derivable (Pentus, 1994, Lm. 13(i)). By cut, we get ⊢ Γ i → B .A formula A is called zero-balanced if [[ A ]] = ε . By Theorem 5.23, if all formulaein all Γ i are zero-balanced, then { Γ , . . . , Γ n } has a joining formula B . Lemma
The sentinel formula S p,q,r is zero-balanced.Proof. Raising does not change the free group interpretation of a formula:[[ q / ( A \ q )]] = [[ q ]][[ A \ q ]] − = q ([[ A ]] − [[ q ]]) − = qq − [[ A ]] = [[ A ]] . Now [[ S p,q,r ]] = [[ r / ( p \ r )]][[ q / ( p \ q )]] − = pp − = ε. The free group interpretation can be seen as a special case of the interpretation onresiduated monoids, with the equality relation (=) taken as the preorder ( (cid:22) ). Fromthis perspective, the fact that derivability of A → B implies [[ A ]] = [[ B ]] follows fromthe general soundness statement.U064-05-FPR Lstar˙Pi1˙final 4 May 2020 0:27 Lambek Calculus with Kleene Star (cid:3) Now consider a set U = { A , . . . , A n } of zero-balance formulae and let u, t, v, w, s be fresh variables, not occurring in A i . Consider the following two sets of sequencesof formulae: (cid:8) E i + = S t,v,w , A i , S t,v,w , A i +1 , S t,v,w , . . . , S t,v,w , A n , S t,v,w | i = 1 , . . . , n (cid:9) ; (cid:8) E i − = S t,v,w , A , S t,v,w , A , S t,v,w , . . . , S t,v,w , A i , S t,v,w | i = 1 , . . . , n (cid:9) . All formulae here are zero-balanced. Therefore, Theorem 5.23 yields a joiningformula for each set: there exist formulae F and G (in the language of \ and / )such that for all i = 1 , . . . , n we have ⊢ E i + → F and ⊢ E i − → G. Let E = S t,v,w , A , S t,v,w , A , S t,v,w , . . . , S t,v,w , A n , S t,v,w ; B = E, ((( u / F ) \ u ) \ S t,v,w ); C = ( S t,v,w / ( u / ( G \ u ))) , E ;is( U ) = ( s / E, B ) \ s / C. The formula is( U ), in a sense, would play the rˆole of A ∨ . . . ∨ A n .We also define versions of E j + and E j − , for j = 1 , . . . , n , which lack the sentinelon one edge: E ′ j + = A j , S t,v,w , . . . , A n , S t,v,w ; E ′ j − = S t,v,w , A , . . . , S t,v,w , A j . For convenience, we also define E ′ − = E ′ ( n +1)+ = Λ. Now for any i = 1 , . . . , n wehave E = E i − , E ′ ( i +1)+ = E ′ ( i − − , E i + = E ′ ( i − − , S t,v,w , A i , S t,v,w , E ′ ( i +1)+ . Lemma ⊢ A i → is( U ) for any A i ∈ U .Proof. By ( → / ), ( → \ ), ( · → ), A i → is( U ) is derivable from s / E, B, A i , C, → s .By definition of B, C , the latter is the same as s / E, E ′ ( i − − , E i + , (( u / F ) \ u ) \ S t,v,w , A i , S t,v,w / ( u / ( G \ u )) , E i − , E ′ ( i +1)+ → s. By construction, we have ⊢ E i + → F → ( u / F ) \ u and ⊢ E i − → G → u / ( G \ u ) . Thus, applying ( \ → ) and ( / → ), we reduce to s / E, E ′ ( i − − , S t,v,w , A i , S t,v,w , E ′ ( i +1)+ → s. This sequent is exactly s / E, E → s , which is derivable by several applications of( / → ) (recall that E is a sequence of formulae). (cid:3) Lemma
Let Π be a non-empty sequence of types whose tops are not s , t , v , w , and let ⊢ Π → is( U ) . Then for some A j ∈ U we have ⊢ B , Π , C → A j , where B is a suffix of B and C is a prefix of C . U064-05-FPR Lstar˙Pi1˙final 4 May 2020 0:27 stepan kuznetsov Proof.
Inverting ( → / ) and ( → \ ) gives s / E, B, Π , C → s ( B , Π, and C donot have s in tops). Now apply Lemma 5.19. The sequence B, Π , C gets split intoΨ , . . . , Ψ n +1 , and we have ⊢ Ψ → S t,v,w , ⊢ Ψ → A , ⊢ Ψ → S t,v,w , . . . , ⊢ Ψ n → A n , ⊢ Ψ n +1 → S t,v,w .Here we consider the interesting case of a non-empty Π. The case of Π = Λ issimilar and is handled in Lemma 5.27 below. If Π as a whole comes into one of Ψ j , ⊢ Ψ j → A j , this is exactly what we want (Ψ j = B , Π , C ).In the other case, there are two possibilities: either a non-empty part of Π, denotedby Π ′ , comes to a part Ψ j +1 → S t,v,w , where Ψ j +1 = B , Π ′ , C ( B is a suffix of B , C is a prefix of C ; if Π ′ is not the whole Π, one of them is empty), or for some j we have Ψ j +1 = Λ, and parts of Π come to Ψ j and Ψ j +2 . The latter case isimpossible, since Λ → S t,v,w (Lemma 5.21) . In the former case we also wish toobtain contradiction.Inverting ( → / ) gives ⊢ B , Π ′ , C , v / ( t \ v ) , t \ w → w . We prove that such asequent cannot be derivable, proceeding by induction on the number of formulae in B . Suppose the sequent is derivable. Locate the principal occurrence of w . Sinceall tops in Π ′ are not w , and so are all A j , this principal occurrence is either therightmost one, or located in S t,v,w . Moreover, formulae with S t,v,w are of the form w / Γ, except for the last formula in B , which is (( u / F ) \ u ) \ S t,v,w . Thus, theprincipal occurrence is either in the leftmost S t,v,w of B , or in (( u / F ) \ u ) \ S t,v,w in the end of B . The non-emptiness of Π ′ prevents it from being in the first formulaof C , which is of the form S t,v,w / ( u / ( G \ u )). Consider these three possible cases. Case 1: ⊢ B , Π ′ , C , v / ( t \ v ) , t \ w → w . By Lemma 5.19, we obtain derivabilityof B , Π ′ , C , v / ( t \ v ) → t , which immediately fails, since there are no t tops in theantecedent. Case 2: the principal occurrence is in the leftmost S t,v,w in B : ⊢ w / ( v / ( t \ v ) , t \ w ) , B ′ , Π ′ , C , v / ( t \ v ) , t \ w → w. By Lemma 5.20, B ′ , Π ′ , C = Λ, which contradicts the non-emptiness of Π ′ . Case 3: the principal occurrence is in the last formula of B : ⊢ B ′′ , (( u / F ) \ u ) \ w / ( v / ( t \ v ) , t \ w ) , Π ′ , C , v / ( t \ v ) , t \ w → w. In this case essentially the same happens: by Lemma 5.19 we have ⊢ B ′′ → ( u / F ) \ u ;Π ′ , C , v / ( t \ v ) , t \ w = Ψ , Ψ ; ⊢ Ψ → v / ( t \ v ); ⊢ Ψ → t \ w. Again, as in the proof of Lemma 5.20, Ψ should be Π ′ , C , v / ( t \ v ), the principaloccurrence is located as follows: ⊢ Π ′ , C , v / ( t \ v ) , t \ v → v , and Π ′ , C = Λ, whichcontradicts with Π ′ = Λ. (cid:3) Lemma If ⊢ Λ → is( U ) , then we have ⊢ B , C → A j , where B is a suffixof B and C is a prefix of C . This is the difference from Safiullin’s sentinel!U064-05-FPR Lstar˙Pi1˙final 4 May 2020 0:27
Lambek Calculus with Kleene Star Notice that B and C are allowed to be empty. In particular, we could get just ⊢ Λ → A j . Proof.
As in the proof of the previous lemma, we get
B, C = Ψ , . . . , Ψ n +1 , where ⊢ Ψ j +1 → S t,v,w for j = 0 , , . . . , n and ⊢ Ψ i → A i for i = 1 , . . . , n . The only badcase is when both the last formula of B and the first formula of C come to Ψ j +1 :otherwise, for some i , Ψ i = B , C ( B , or C , or even both could be empty),which is what we need.In the bad case, we have ⊢ B , C → S t,v,w , where both B and C are non-empty.Inverting ( → / ) yields ⊢ B , C , v / ( t \ v ) , t \ w → w, and there are again three possibilities for the principal w , exactly as in the proofof the previous lemma. Notice that the usage of the first formula in C for theprincipal occurrence is now prohibited by non-emptiness of B , not Π ′ . Case 1: ⊢ B , C , v / ( t \ v ) , t \ w → w . Exactly as Case 1 in the previous lemma. Case 2: the principal occurrence is in the leftmost S t,v,w in B . As in Case 2 ofthe previous lemma, we get B ′ , C = Λ, which contradicts non-emptiness of C . Case 3: the principal occurrence is in the last formula of B : ⊢ B ′′ , (( u / F ) \ u ) \ w / ( v / ( t \ v ) , t \ w ) , C , v / ( t \ v ) , t \ w → w. This boils out into C = Λ (see proof of the previous lemma), which is false. (cid:3) Now we come to the main part of the construction.
Consider a languagewithout the empty word generated by a context-free grammar G in Greibach normalform (Greibach, 1965). Let Σ = { a , . . . , a µ } be its terminal alphabet and N = { N , N , . . . , N ν } be the non-terminal one; N is the starting symbol.Production rules of G are of the form N i ⇒ a j N k N ℓ , or N i ⇒ a j N k , or N i ⇒ a j . In order to simplify the proof, we write all these rules uniformly: N i ⇒ a j N ? k N ? ℓ , where ? means optionality.Our variables (all distinct) will be x , z , p i , q i r i (0 ≤ i ≤ ν ), plus u, t, v, w, s forthe construction described above. Introduce the following formulae: H i = ( z / z ) / S p i ,q i ,r i for each N i ∈ N (0 ≤ i ≤ ν ); A j ; i,k ? ,ℓ ? = x / (( H ? k , H ? ℓ , S p i ,q i ,r i ) \ x ) for each production rule N i ⇒ a j N ? k N ? ℓ . Lemma A j ; i,k ? ,ℓ ? is a zero-balance formula.Proof. [[ A j ; i,k ? ,ℓ ? ]] = xx − [[ H k ]] ? [[ H ℓ ]] ? [[ S p i ,q i ,r i ]].[[ S p i ,q i ,r i ]] = ε by Lemma 5.24.[[ H k ]] = zz − [[ S p k ,q k ,r k ]] − = ε . (cid:3) Let U j be the set of all A j ; i,k ? ,ℓ ? for each a j (1 ≤ j ≤ µ ). Since all formulae in U j are zero-balanced, we can construct is( U j ) obeying Lemma 5.25, Lemma 5.26, andLemma 5.27. Finally, let K j = ( z / z ) / is( U j ).In our derivation analysis, we shall frequently use Lemma 5.19 (decompositionof division in the antecedent). This requires locating the principal occurrence of aspecific variable. Since the principal occurrence is always a top, we need to recall U064-05-FPR Lstar˙Pi1˙final 4 May 2020 0:27 stepan kuznetsov the tops of formulae used in our construction. For the convenience of the reader,we gather all of them in one table: formula top variable A j ; i,k ? ,l ? xK j zH i zS p i ,q i ,r i r i formulae of B and C x or w Now we are ready to formulate the key lemma:
Lemma
For any non-empty word a i a i . . . a i n and any N m ∈ N , the word a i . . . a i n is derivable from N m in G if and only if ⊢ K i , K i , . . . , K i n → H m . Before proving it, we state and prove yet two technical statements.
Lemma
No sequent of the form e B , K i , . . . , K i m → is( U j ) , where e B is asuffix of B (maybe empty), is derivable. In particular, K i , . . . , K i m → is( U j ) and Λ → is( U j ) .Proof. In order to make the notation shorter, let ~K = K i , . . . , K i m Supposethat the sequent in question is derivable and apply Lemma 5.26 or Lemma 5.27,depending on whether e B , ~K is empty: ⊢ B , e B , ~K, C → x / (( H ? k , H ? ℓ , S p i ,q i ,r i ) \ x )for some A j ; i,k ? ,ℓ ? = x / (( H ? k , H ? ℓ , S p i ,q i ,r i ) \ x ) ∈ U j .Here B , or C , or both can be empty. In particular, Lemma 5.27 could yield ⊢ Λ → x / (( H ? k , H ? ℓ , S p i ,q i ,r i ) \ x ). In this case, however, we get ⊢ ( H ? k , H ? ℓ , S p i ,q i ,r i ) \ x → x , and by Lemma 5.19 ⊢ Λ → S p i ,q i ,r i , which is not true.Let b B = B , e B . This sequence is not always a suffix of B ; however, it keeps theproperty we shall need: all formulae in b B have tops x or w , and a formula withtop x is always of the form x / Γ and could not be the last formula of b B .Inverting ( → / ) (Lemma 5.15) yields: ⊢ b B , ~K, C , ( H ? k , H ? ℓ , S p i ,q i ,r i ) \ x → x. Locate the principal occurrence of x . Recall that tops of all K i ’s are z , thus this topis either in one of the A ’s in b B or C , or the rightmost occurrence of x . Since all A ’s are of the form x / Γ, the former is possible only if it is the leftmost occurrence.Moreover, this leftmost occurrence should be in b B , because, even if b B and ~K happen to be empty, the leftmost formula of C (and, thus, of its prefix C ) is notone of the A ’s.Thus, we have to consider two cases. Case 1.
The principal x is in the leftmost formula of the form A j ′ ; i ′ ,k ′ ? ,ℓ ′ ? in b B : ⊢ x / (( H ? k ′ , H ? ℓ ′ , S p i ′ ,q i ′ ,r i ′ ) \ x ) , b B ′ , ~K, C , ( H ? k , H ? ℓ , S p i ,q i ,r i ) \ x → x. Notice that A j ′ ; i ′ ,k ′ ? ,ℓ ′ ? , as a formula with top x , is not the rightmost formula of b B , thus b B ′ = Λ. Lemma 5.19 gives ⊢ b B ′ , ~K, C , ( H ? k , H ? ℓ , S p i ,q i ,r i ) \ x → ( H ? k ′ , H ? ℓ ′ , S p i ′ ,q i ′ ,r i ′ ) \ x. U064-05-FPR Lstar˙Pi1˙final 4 May 2020 0:27
Lambek Calculus with Kleene Star Inverting ( → \ ) (Lemma 5.15) yields ⊢ H ? k ′ , H ? ℓ ′ , S p i ′ ,q i ′ ,r i ′ , b B ′ , ~K, C , ( H ? k , H ? ℓ , S p i ,q i ,r i ′ ) \ x → x. This situation looks similar to the one in the beginning of the proof, but now thereis only one choice of the principal x . Indeed, tops of H i and K i are not x . The A ’sin b B ′ and C also could not include the principal x , since in this case it would bethe leftmost one. This is not true, since there is at least S p i ′ ,q i ′ ,r i ′ to the left. Thus, ⊢ H ? k ′ , H ? ℓ ′ , S p i ′ ,q i ′ ,r i ′ , b B ′ , ~K, C , ( H ? k , H ? ℓ , S p i ,q i ,r i ) \ x → x, and by Lemma 5.19 we have H ? k ′ , H ? ℓ ′ , S p i ′ ,q i ′ ,r i ′ , b B ′ , ~K, C = Ψ , Ψ , Ψ ; ⊢ Ψ → H ? k ; ⊢ Ψ → H ? ℓ ; ⊢ Ψ → S p i ,q i ,r i . Here and further we use the following convention about optionality: if ⊢ Γ → H ? k ,but there is no H k , then this statement means Γ = Λ; same for H ℓ . We also imposepriority of optionality: if there is H ℓ , there is also H k (otherwise we rename H ℓ to H k ).Notice that tops of H i , K i , and formulae of b B ′ and C are not p i , q i , or r i .Thus, by Lemma 5.21, i ′ should be equal to i and Ψ should be S p i ,q i ,r i . This yields b B ′ = Λ. Contradiction: we have shown above that b B ′ = Λ. Case 2.
The principal x is the rightmost one: ⊢ b B , ~K, C , ( H ? k , H ? ℓ , S p i ,q i ,r i ) \ x → x. Lemma 5.19 yields b B , ~K, C = Ψ , Ψ , Ψ ; ⊢ Ψ → H ? k ; ⊢ Ψ → H ? ℓ ; ⊢ Ψ → S p i ,q i ,r i . The third derivability, ⊢ Ψ → S p i ,q i ,r i , contradicts Lemma 5.21: there are no p i , q i , or r i tops in the left-hand side. (cid:3) Lemma
Let Φ be a non-empty sequence of formulae whose tops are not p i , q i , r i ; any formula with top z in Φ is of the form z / Γ ; the top of the leftmostformula in Φ is not z . Then no sequent of the form H ? k ′ , H ? ℓ ′ , S p i ′ ,q i ′ ,r i ′ , Φ → H k isderivable.Proof. Suppose the contrary. Recall that H k = ( z / z ) / S p k ,q k ,r k = z / ( S p k ,q k ,r k , z )and apply Lemma 5.15: ⊢ H ? k ′ , H ? ℓ ′ , S p i ′ ,q i ′ ,r i ′ , Φ , S p k ,q k ,r k , z → z. Since all z tops come in formulae of the form z / Γ, the principal occurrence shouldbe the leftmost one. (The standalone rightmost z could not be principal, becausethen by Lemma 5.19 it should have been the only formula in the antecedent.) This U064-05-FPR Lstar˙Pi1˙final 4 May 2020 0:27 stepan kuznetsov could happen only in H k ′ (thus, it should exist): the sentinel S p i ′ ,q i ′ ,r i ′ blocks otherpossibilities. Thus, we have ⊢ z / ( S p k ′ ,q k ′ ,r k ′ , z ) , H ? ℓ ′ , S p i ′ ,q i ′ ,r i ′ , Φ , S p k ,q k ,r k , z → z. Apply Lemma 5.19: H ? ℓ ′ , S p i ′ ,q i ′ ,r i ′ , Φ , S p k ,q k ,r k , z = Ψ , Ψ ; ⊢ Ψ → S p k ′ ,q k ′ ,r k ′ ; ⊢ Ψ → z. By Lemma 5.21 i ′ = k ′ and the first part, Ψ , should include only S p i ′ ,q i ′ ,r i ′ .Therefore, there is no H ℓ ′ , and Ψ = Φ , S p k ,q k ,r k , z .Now we have ⊢ Φ , S p k ,q k ,r k , z → z and we fail to locate the principal occurrence of z . It could be only in Φ, and sincethere z tops appear only in formulae of the form z / Γ, it should be the leftmostone. Contradiction: the leftmost formula of Φ does not have top z . (cid:3) Now we finally prove the key lemma.
Proof of Lemma 5.29.
For convenience let j = i .The “only if ” part is easier. Proceed by induction on the derivation of a j a i . . . a i n from N m in G . Consider the first production rule in this derivation: N m ⇒ a j N ? k N ? ℓ .Here N ? k derives a i . . . a i n ′ and N ? ℓ derives a i n ′ +1 . . . a i n . (If there is no N ℓ , then n ′ = n ; if there are no N k and N ℓ , then n = 0.) By induction hypothesis we havethe following: ⊢ K i , . . . , K i n ′ → H ? k ⊢ K i n ′ +1 , . . . , K i n → H ? ℓ Recall our convention about optionality: if ⊢ Γ → H ? k , but there is no H k , then thisstatement means Γ = Λ; same for H ℓ .Next, since the production rule N m ⇒ a j N ? k N ? ℓ is in G , the formula A j ; m,k ? ,ℓ ? belongs to U j , and by Lemma 5.25 we have ⊢ A j ; m,k ? ,ℓ ? → is( U j )or, explicitly, ⊢ x / (( H ? k , H ? ℓ , S p m ,q m ,r m ) \ x ) → is( U j ) . By cut with H ? k , H ? ℓ , S p m ,q m ,r m → x / (( H ? k , H ? ℓ , S p m ,q m ,r m ) \ x ) we get ⊢ H ? k , H ? ℓ , S p m ,q m ,r m → is( U j ) . Now the necessary sequent K j , K i , . . . , K i n ′ , K i n ′ +1 , . . . , K i n → H m is derived asfollows. Cut of K i , . . . , K i n ′ → H ? k , K i n ′ +1 , . . . , K i n → H ? ℓ , and H ? k , H ? ℓ , S p m ,q m ,r m → is( U j ) gives K i , . . . , K i n , S p m ,q m ,r m → is( U j ) (if there is no H k and/or H ℓ , we omitthe corresponding premise). Next, by ( / → ) and ( → / ) we get ⊢ ( z / z ) / is( U j ) , K i , . . . , K i n ′ , K i n ′ +1 , . . . , K i n → ( z / z ) / S p m ,q m ,r m . This is the necessary sequent, since K j = ( z / z ) / is( U j ) and H m = ( z / z ) / S p m ,q m ,r m . U064-05-FPR Lstar˙Pi1˙final 4 May 2020 0:27
Lambek Calculus with Kleene Star For the “if ” part, proceed by induction on a cut-free derivation of the sequent K j , K i , . . . , K i n → H m . Basically, we show that the only way this derivation couldgo is the one shown in the “only if” part of this proof.Recall that H m = ( z / z ) / S p m ,q m ,r m = z / ( S p m ,q m ,r m , z ) and reverse ( → / ) twice(Lemma 5.15). Thus we get ⊢ K j , K i , . . . , K i n , S p m ,q m ,r m , z → z. Each K i is of the form z / Γ. Thus, the principal occurrence of z is the top of theleftmost K i , that is, K j = ( z / z ) / is( U j ) = z / (is( U j ) , z ), and we have K i , . . . , K i n , S p m ,q m ,r m , z = Ψ , Ψ ; ⊢ Ψ → is( U j ); ⊢ Ψ → z. If Ψ contains only K ’s, then Ψ → is( U j ) is not derivable by Lemma 5.30.Moreover, Ψ = Λ, since Λ → z . Thus, Ψ = K i , . . . , K i n , S p m ,q m ,r m and ⊢ K i , . . . , K i n , S p m ,q m ,r m → is( U j ) . Let ~K = K i , . . . , K i n . By Lemma 5.26 (recall that ~K, S p m ,q m ,r m is definitely non-empty), ⊢ B , ~K, S p m ,q m ,r m , C → A j ; i,k ? ,ℓ ? for some A j ; i,k ? ,ℓ ? ∈ U j . Since A j ; i,k ? ,ℓ ? = x / (( H ? k , H ? ℓ , S p i ,q i ,r i ) \ x ), inversion of( → \ ) yields ⊢ B , ~K, S p m ,q m ,r m , C , ( H ? k , H ? ℓ , S p i ,q i ,r i ) \ x → x. Locate the principal occurrence of x . It is either in the leftmost formula in B , ofthe form A j ′ ; i ′ ,k ′ ? ,ℓ ′ ? , or the rightmost occurrence of x . Case 1. ⊢ x / (( H ? k ′ , H ? ℓ ′ , S p i ′ ,q i ′ ,r i ′ ) \ x ) , B ′ , ~K, S p m ,q m ,r m , C , ( H ? k , H ? ℓ , S p i ,q i ,r i ) \ x → x. The important notice here is that B ′ is not empty, since A j ′ ; i ′ ,k ′ ? ,ℓ ′ ? was not therightmost formula in B .Decomposition (Lemma 5.19) and inversion of ( → \ ) (Lemma 5.15) yields ⊢ H ? k ′ , H ? ℓ ′ , S p i ′ ,q i ′ ,r i ′ , B ′ , ~K, S p m ,q m ,r m , C , ( H ? k , H ? ℓ , S p i ,q i ,r i ) \ x → x. The sentinel S p i ′ ,q i ′ ,r i ′ , again, blocks other choices for the principal x . ApplyingLemma 5.19 once more: H ? k ′ , H ? ℓ ′ , S p i ′ ,q i ′ ,r i ′ , B ′ , ~K, S p m ,q m ,r m , C = Ψ , Ψ , Ψ ; ⊢ Ψ → H ? k ; ⊢ Ψ → H ? ℓ ; ⊢ Ψ → S p i ,q i ,r i . The only two formulae in the antecedents, whose tops could potentially be p i , q i ,or r i , are the sentinels S p i ′ ,q i ′ ,r i ′ and S p m ,q m ,r m . By Lemma 5.21, we get m = i andΨ = S p m ,q m ,r m . Therefore, C = Λ.Consider several subcases: U064-05-FPR Lstar˙Pi1˙final 4 May 2020 0:27 stepan kuznetsov Subcase 1.1: there are neither H k , nor H ℓ . Then Ψ = Ψ = Λ, which is not thecase, since Ψ or Ψ should include at least the other sentinel, S p i ′ ,q i ′ ,r i ′ . Subcase 1.2: there is only H k , but not H ℓ , thus, Ψ = Λ. Then we haveΨ = H ? k ′ , H ? ℓ ′ , S p i ′ ,q i ′ ,r i ′ , B ′ , ~K ; ⊢ Ψ → H k . Let us check that B ′ , ~K satisfies the conditions for Φ in Lemma 5.31. Indeed, topsof formulae in B ′ are w or x ; tops of K i are z , and they are of the form z / Γ.Finally, since B ′ is not empty, the first formula in our Φ is from B ′ , and thus doesnot have top z . By Lemma 5.31, H ? k ′ , H ? ℓ ′ , S p i ′ ,q i ′ ,r i ′ , Φ → H k . Contradiction. Subcase 1.3: there are both H k and H ℓ . ThenΨ , Ψ = H ? k ′ , H ? ℓ ′ , S p i ′ ,q i ′ ,r i ′ , B ′ , ~K. Take the one of Ψ and Ψ which includes S p i ′ ,q i ′ ,r i ′ . If it also includes the firstformula of B ′ , then the claim for it violates Lemma 5.31, exactly as in the previoussubcase. Otherwise Ψ = B ′ , ~K , and Ψ → H ℓ is not derivable by Lemma 5.30. Case 2.
This is the fruitful case. ⊢ B , K i , . . . , K i n , S p m ,q m ,r m , C , ( H ? k , H ? ℓ , S p i ,q i ,r i ) \ x → x. Decomposition (Lemma 5.19) yields B , K i , . . . , K i n , S p m ,q m ,r m , C = Ψ , Ψ , Ψ ; ⊢ Ψ → H ? k ; ⊢ Ψ → H ? ℓ ; ⊢ Ψ → S p i ,q i ,r i . By Lemma 5.21, m = i and Ψ should be S p m ,q m ,r m . Thus, C = Λ and B , K i , . . . , K i n =Ψ , Ψ .Let us show that B is empty. If there are no H k and H ℓ , this is trivial, since inthis case Ψ = Ψ = Λ. Let there be H k and suppose that B is not empty. Thenthe first formula of B is the first formula of Ψ (Ψ is not empty, since Λ → H k ).Recall that H k = z / ( S p k ,q k ,r k , z ) and invert ( → / ): ⊢ Ψ , S p k ,q k ,r k , z → z. The principal occurrence of z should be in one of the K i , and this formula, beingof the form z / Γ, should be the leftmost one. However, the leftmost formula of Ψ is from B and does not have top z . Contradiction.Thus, B = Λ, and we have Ψ = K i , . . . , K i n ′ , Ψ = K i n ′ +1 , . . . , K i n . Byinduction hypothesis, ⊢ Ψ → H k yields derivability of a i . . . a i n ′ from N k and ⊢ Ψ → H ℓ yields derivability of a i n ′ +1 . . . a i n from N ℓ in the context-free grammar G . Finally, since A j ; m,k ? ,ℓ ? ∈ U j (recall that m = i ), in G we have the productionrule needed to finish the derivation: N m ⇒ a j N ? k N ? ℓ . (cid:3) U064-05-FPR Lstar˙Pi1˙final 4 May 2020 0:27
Lambek Calculus with Kleene Star Theorem 5.14 immediately follows from Lemma 5.29. The necessary L Λ ( \ , / )-grammar is constructed as follows: for each a j ∈ Σ let a j ✄ K j (this type assignmentis unique) and let H = H .Notice that our construction also works with Lambek’s restriction, thus subsum-ing Safiullin’s original result (Theorem 3.10). § In this section we consider theoriesof three specific classes of residuated Kleene lattice, namely, language, regularlanguage, and relational models. Though, in the language with ∨ and ∧ , thesetheories are strictly greater than ACT ω , Buszkowski (2006) proves Π -hardnessfor them also. We propagate Buszkowski’s complexity results to the correspondingclasses of residuated monoids with iteration, getting rid of additive operations. Thecompleteness issue, i.e., the question whether these theories in the language without ∨ and ∧ coincide with L ∗ ω , is still open.Language models, or L-models for short, are interpretations of the Lambek cal-culus and its extensions on the algebra of formal languages over an alphabet Σ(see § W , i.e., subsets of W × W .Operations are defined as follows: R \ S = {h y, z i ∈ W × W | ( ∀h x, y i ∈ R ) h x, z i ∈ S } ; S / R = {h x, y i ∈ W × W | ( ∀h y, z i ∈ R ) h x, z i ∈ S } ; R · S = R ◦ S = {h x, z i ∈ W × W | ( ∃ y ∈ W ) h x, y i ∈ R and h y, z i ∈ S } . (For simplicity, we consider only “square” relational models, where any pair h x, y i could belong to a relation. There is also a broader class of “relativised” relationalmodels, where all relations are subsets of a “universal” relation U (Andr´eka & Mikul´as,1994), which is reflexive and transitive. Relativisation alters the definition of divi-sion operations. Relativised R-models are necessary for the Lambek calculus withLambek’s restriction ( § U .)In all these classes of models additive connectives are interpreted set-theoretically,as union and intersection.Unfortunately, no completeness results are known for L ∗ ω w.r.t. these specificclasses of models. Moreover, even adding only one additive connective, conjunction ∧ , yields incompleteness w.r.t. all three classes of models (Kuznetsov, 2018). Foraction logic in whole, incompleteness is connected with the distributivity principle,( A ∨ C ) ∧ ( B ∨ C ) → ( A ∧ B ) ∨ C . This principle is true in all models in which ∨ and ∧ are interpreted set-theoretically, as union and intersection, but is not derivablein ACT ω , cf. Ono & Komori (1985). Since A ∗ is essentially infinite disjunction( ∨ A ∨ A ∨ . . . ), incompleteness also propagates to the fragment without explicit ∨ . On the other hand, L Λ is complete w.r.t. L-models (Pentus, 1998) and R-models (Andr´eka & Mikul´as, 1994). For REGLAN-models, completeness is an openproblem, but there are some partial results which we shall use later. U064-05-FPR Lstar˙Pi1˙final 4 May 2020 0:27 stepan kuznetsov Completeness for L ∗ ω itself is an open problem, for any of these three interpre-tations. Thus, the inequational theories of L-models, R-models, and REGLAN-models, in the language with \ , / , · , and ∗ , could possibly be different from the setof theorems of L ∗ ω . We denote these theories by Th \ ,/, · , ∗ (L-Mod), Th \ ,/, · , ∗ (R-Mod),and Th \ ,/, · , ∗ (REGLAN-Mod), respectively.In the bigger language of ACT ω , the corresponding theories are denoted byTh \ ,/, · , ∗ , ∨ , ∧ ( M ), where M is L-Mod, R-Mod, or REGLAN-Mod. These theoriesare definitely different from ACT ω itself, due to the distributivity principle. How-ever, Buszkowski (2006) manages to propagate his Π -hardness result to these thesetheories also: Theorem
Theories Th \ ,/, · , ∗ , ∨ , ∧ (L - Mod) , Th \ ,/, · , ∗ , ∨ , ∧ (R - Mod) ,and Th \ ,/, · , ∗ , ∨ , ∧ (REGLAN - Mod) are Π -hard. Due to the semantic definition of these theories, however, the “out-of-the-box” up-per bound appears to be quite high. For Th \ ,/, · , ∗ , ∨ , ∧ (L-Mod) and Th \ ,/, · , ∗ , ∨ , ∧ (R-Mod)it is Π : indeed, the condition for a sequent to belong to one of these theoriesstarts with a second-order quantifier “for any model,” followed by an arithmeticallyformulated truth condition. No better upper complexity bounds are known. ForREGLAN-models, however, the situation is different (Buszkowski, 2006). Such amodel is essentially a bunch of regular expressions encoding the languages whichinterpret variables (for a given sequent, the set of its variables is finite), and thisbunch can be encoded by a natural number. This arithmetises the quantifier overall models and gives Π as the upper bound. Thus, the upper bound matches thelower one: Th \ ,/, · , ∗ , ∨ , ∧ (REGLAN-Mod) is Π -complete.Buszkowski’s method of proving Π -hardness is essentially based on the factthat the fragment which is really needed to prove Π -hardness of ACT ω is infact complete w.r.t. all three classes of models. Here we formulate this fragmentexplicitly, prove its completeness and thus obtain Π -hardness for the theorieswithout ∨ and ∧ , using our Theorem 4.12.Let us call a formula *-external, if no · or ∗ in it occurs within the scope of \ or / . More formally, the class of *-external formulae is defined recursively as follows: • any formula in the language of \ and / is *-external; • if A and B are *-external, then so is A · B ; • if A is *-external, then so is A ∗ .A sequent A , . . . , A n → B is called *-external, if all A i are *-external and B is aformula in the language of \ and / .By subformula property, all sequents in a cut-free derivation of a *-externalsequent are also *-external. Thus, the notion of *-externality induces a fragment of L ∗ ω —and we show that this fragment is simultaneously Π -hard and complete w.r.t.L-models, R-models, and REGLAN-models. The first claim immediately followsfrom the form of the sequent used in the proof of Theorem 4.12: it is *-external(recall that our construction of an L Λ -grammar with unique type assignment, The-orem 5.14, uses only two divisions, \ and / ).For the second claim, we present a version of Palka’s *-elimination via approx-imation, which reduces derivability of *-external sequents in L ∗ ω to derivabilityin L Λ ( \ , / ), without using ∨ . For each *-external formula A we define the set ofits instances, Inst( A ), where subformulae of the form B ∗ are replaced by concretenumbers of B ’s (since Kleene stars could be nested, these A ’s should also be replaced U064-05-FPR Lstar˙Pi1˙final 4 May 2020 0:27
Lambek Calculus with Kleene Star by instances, and therefore become different), and · ’s are replaced by metasyntacticcommas (thus, instances are sequences of formulae). The set of instances is definedby induction on the construction of a *-external formula: • if A is a formula in the language of \ and / , then Inst( A ) = { A } ; • Inst( A · B ) = { Γ , ∆ | Γ ∈ Inst( A ) , ∆ ∈ Inst( B ) } ; • Inst( A ∗ ) = { Γ , . . . , Γ n | n ≥ , Γ i ∈ Inst( A ) } .A typical example of the notion of the set of instances is given in the proof ofTheorem 4.12: for A = ( A +1 · A +2 ) + the set of instances is { A n , A m , A n , A m , . . . , A n k , A m k | k, n , m , n , m , . . . , n k , m k ≥ } . Lemma
A *-external sequent A , . . . , A n → B is derivable in L ∗ ω if andonly if Π , . . . , Π n → B is derivable in L Λ ( \ , / ) for any Π ∈ Inst( A ) , . . . , Π n ∈ Inst( A n ) , i.e., all its instances are derivable.Proof. Essentially the same as Lemma 4.13: the “if” part goes by applying ( · → )and ( ∗ → ) ω ; the “only if” one goes by their inversion. (cid:3) Lemma
For any *-external formula A and any Π ∈ Inst( A ) the sequent Π → A is derivable in L ∗ ω .Proof. Induction on the structure of A . If A is in the language of \ and / , thenΠ = A and Π → A is an axiom. If A = B · C , then Π = Γ , ∆, where Γ ∈ Inst( B )and ∆ ∈ Inst( C ). By induction hypothesis, Γ → B and ∆ → C are derivable; thenΠ → A is derivable by application of ( → · ). If A = B ∗ , then Π = Π , . . . , Π n , whereΠ i ∈ Inst( B ). By induction hypothesis, all sequents Π i → B are derivable, and thusΠ → A is derivable by application of ( → ∗ ) n . (cid:3) Now we are ready to prove completeness.
Theorem
A *-external sequent is derivable in L ∗ ω if and only if it is truein all L-models, or all REGLAN-models, or all R-models.Proof. The “only if” part follows from the general soundness theorem of L ∗ ω w.r.t.arbitrary RKLs.For the “if” part, we first recall completeness results for L Λ ( \ , / ): • for L-models, completeness of L Λ was proved by Pentus (1998); here we canactually use a simpler result by Buszkowski (1982a) for the product-freefragment; • for REGLAN-models, completeness follows from the fact that L Λ ( \ , / ) iscomplete even w.r.t. a narrower class of L-models, in which variables areinterpreted by cofinite languages (Buszkowski, 1982b; Sorokin, 2012); • for R-models, completeness was proved by Andr´eka & Mikul´as (1994).Now let a *-external sequent A , . . . , A n → B be true in all models of one of theclasses: L-Mod, REGLAN-Mod, or R-Mod. Let Π ∈ Inst( A ), . . . , Π n ∈ Inst( A n ).By Lemma 6.34, Π i → A i are derivable in L ∗ ω , and by soundness they are true in allmodels of the specified class. Thus, Π , . . . , Π n → B is also generally true. This isa sequent in the language of \ and / , and by completeness results mentioned aboveit is derivable in L Λ ( \ , / ). Now, since instances Π , . . . , Π n were taken arbitrarily, U064-05-FPR Lstar˙Pi1˙final 4 May 2020 0:27 stepan kuznetsov by Lemma 6.33 we conclude that the original sequent A , . . . , A n → B is derivablein L ∗ ω . (cid:3) Now we can prove the Π -hardness results. Theorem
Theories Th \ ,/, · , ∗ (L - Mod) , Th \ ,/, · , ∗ (REGLAN - Mod) , and Th \ ,/, · , ∗ (R - Mod) are Π -hard.Proof. If we consider only *-external sequents, the corresponding fragments of thesetheories are equal to the one of L ∗ ω . As noticed above, the latter is Π -hard. (cid:3) Like the system with additives, Th \ ,/, · , ∗ (REGLAN-Mod) also enjoys a Π uppercomplexity bound, and thus is Π -complete. For two other theories, Th \ ,/, · , ∗ (L-Mod)and Th \ ,/, · , ∗ (R-Mod), the best known upper bound is only Π . § We have proved Π -hardness for L ∗ ω , theLambek calculus with Kleene star, and the corresponding inequational theories ofthe algebras of languages, regular languages, and binary relations. These resultsstrengthen results by Buszkowski for the corresponding systems extended with ad-ditive connectives, ∨ and ∧ . The crucial component of our proof is the constructionof an L Λ ( \ , / )-grammar with unique type assignment for a context-free grammarwithout the empty word.Let us briefly survey the questions which are still open in this area.First, the completeness issue of L ∗ ω w.r.t. L-models, REGLAN-models, and R-models is still open.Second, it is interesting to characterise the class of languages that can be gen-erated by L ∗ ω -grammars—in particular, whether such a grammar could generatea Π -hard language. Notice that the complexity of the calculus in whole couldbe greater than that of concrete languages generated by grammars based on thiscalculus. For example, the original Lambek calculus is NP-complete (Pentus, 2006),while all languages generated by Lambek grammars are context-free (Pentus, 1993),and therefore decidable in polynomial time.Third, we do not yet know whether any context-free language without the emptyword can be generated by a Lambek grammar with unique type assignments withonly one division. (If one drops the uniqueness condition, then already Gaifman’sconstruction yields such a grammar.) A positive answer to this question would yieldΠ -hardness for the fragment of L ∗ ω in the language of · , / , and ∗ .Fourth, there is a small question on L Λ -grammars with unique type assignment.Grammars constructed in this article are not capable of generating the empty word(because context-free grammars in Greibach form cannot generate it), whereas ingeneral L Λ allows empty antecedents and therefore L Λ -grammars could potentiallygenerate grammars with the empty word. Moreover, in the case without uniquenesscondition there exists a construction that transforms context-free grammars withthe empty word into L Λ -grammars (Kuznetsov, 2012). The question whether allcontext-free languages with the empty word can be generated by L Λ -grammarswith unique type assignment is still open.Finally, as noticed by one of the referees, in our Π -hardness proof the product ( · )operation is used only once, in ( A +1 · A +2 ) + → H , while A , A , and H are product-free. The question is whether it is possible to get rid of the product completely andprove Π -hardness for ACT ω ( \ , /, ∗ ). We conjecture that this could be done by the U064-05-FPR Lstar˙Pi1˙final 4 May 2020 0:27
Lambek Calculus with Kleene Star “pseudo-double-negation” trick (Buszkowski, 2007; Kanovich et al., 2019). Namely,( A +1 · A +2 ) + → H is probably equiderivable with (cid:0) b / (( b / A +2 ) / A +1 ) (cid:1) + → b / ( b / H ),for a fresh variable b . Establishing this equiderivability (and, thus, Π -hardness ofthe product-free fragment of L ∗ ω ) is left for future research. Acknowledgments
The author is grateful to the organisers and participantsof WoLLIC 2017 and MIAN–POMI 2018 Winter Session on Mathematical Logicfor fruitful discussions and friendly atmosphere. The author is also indebted to theanonymous referees for valuable comments and suggestions. Being a Young RussianMathematics award winner, the author would like to thank its sponsors and juryfor this high honour.
Funding
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