The Lambek Calculus with Iteration: Two Variants

Abstract

Formulae of the Lambek calculus are constructed using three binary connectives, multiplication and two divisions. We extend it using a unary connective, positive Kleene iteration. For this new operation, following its natural interpretation, we present two lines of calculi. The first one is a fragment of infinitary action logic and includes an omega-rule for introducing iteration to the antecedent. We also consider a version with infinite (but finitely branching) derivations and prove equivalence of these two versions. In Kleene algebras, this line of calculi corresponds to the *-continuous case. For the second line, we restrict our infinite derivations to cyclic (regular) ones. We show that this system is equivalent to a variant of action logic that corresponds to general residuated Kleene algebras, not necessarily *-continuous. Finally, we show that, in contrast with the case without division operations (considered by Kozen), the first system is strictly stronger than the second one. To prove this, we use a complexity argument. Namely, we show, using methods of Buszkowski and Palka, that the first system is $\varPi _1^0$-hard, and therefore is not recursively enumerable and cannot be described by a calculus with finite derivations.

This work is supported by the Russian Science Foundation under grant 16-11-10252.

Appendix: Safiullin’s Construction Revisited

Theorem 1 by Safiullin is a crucial component of our $\varPi _1^0$-hardness proof for $\mathbf {L}^{\!+}_{\omega }$. Unfortunately, Safiullin’s paper [22] is very brief and, moreover, includes this theorem (which is probably the most interesting result of that paper) as a side-effect of a more complicated construction. This makes it very hard to follow Safiullin’s ideas and arrive at a complete proof. Therefore, in this Appendix we present Safiullin’s proof clearly and in detail.

In this Appendix, the ${}^+$ connective is never used, and $\mathrm {Tp}$ stands for the set of types constructed from primitive ones using $\cdot$, $\mathop {\backslash }$, and $\mathop {/}$.

Define the top of a Lambek type in the following way: ${\mathrm {top}(q) = q \text { for } q \in \mathrm {Pr}}$; ${\mathrm {top}(A \mathop {\backslash }B)= \mathrm {top}(B{\mathop {/}}A) = \mathrm {top}(B).}$ Note that the $A \cdot B$ case is missing. Thus, not every type has a top.

For types with tops, the $(\rightarrow \cdot )$ rule is invertible (proof by induction):

Lemma 3

If all types of $\varPi$ have tops and $\varPi \rightarrow A_1 \cdot \ldots \cdot A_n$ is derivable in $\mathbf {L}$, then $\varPi = \varPi _1, \dots , \varPi _n$ and $\varPi _i \rightarrow A_i$ is derivable for every $i = 1, \ldots , n$.

If a sequent of the form $\varPi \rightarrow q$, $q \in \mathrm {Pr}$, has a cut-free derivation in $\mathbf {L}$, trace the occurrence of q back to the axiom of the form $q \rightarrow q$, and then trace the left q back to its occurrence in $\varPi$. This occurrence of q will be called the principal occurrence (for different derivations, the principal occurrences could differ).

Lemma 4

The principal occurrence has the following properties:

• 1. if all types in $\varPi$ have tops, then the principal occurrence is one of them;

• 2. if in a derivation of $\varPi , q, \varPhi \rightarrow q$ the occurrence of q between $\varPi$ and $\varPhi$ is principal, then $\varPi$ and $\varPhi$ are empty;

• 3. if in a derivation of $\varPi , q{\mathop {/}}A, \varPhi \rightarrow q$ the occurrence of q in $q{\mathop {/}}A$ is principal, then $\varPi$ is empty;

• 4. if in a derivation of $\varPi , A \mathop {\backslash }q, \varPhi \rightarrow q$ the occurrence of q in $A \mathop {\backslash }q$ is principal, then $\varPhi$ is empty.

Proof

For statement 1, proceed by induction on derivation. For statements 2–4, suppose the contrary and also proceed by induction on derivation.

Lemma 5

If all types of $\varPi$ have tops, and these tops are not q, then $\varPi \rightarrow q{\mathop {/}}q$ is not derivable in $\mathbf {L}$.

Proof

Since $(\rightarrow \mathop {/})$ is invertible, we get $\varPi , q \rightarrow q$, and by Lemma 4 $\varPi$ should be empty. But $\rightarrow q{\mathop {/}}q$ is not derivable due to Lambek’s restriction.

Lemma 6

If in a derivation of $q{\mathop {/}}A, \varPhi \rightarrow q$ the leftmost occurrence of q is principal or in a derivation of $\varPhi , A \mathop {\backslash }q \rightarrow q$ the rightmost occurrence of q is principal, then $\varPhi \rightarrow A$ is derivable.

Proof

Induction on the derivation.

Lemma 7

If $\varPi , q{\mathop {/}}q, \varPhi \rightarrow q{\mathop {/}}q$ is derivable in $\mathbf {L}$, all types from $\varPi$ and $\varPhi$ have tops, and these tops are not q, then $\varPi$ and $\varPhi$ are empty.

Proof

Again, by inverting $(\rightarrow \mathop {/})$ we get $\varPi , q{\mathop {/}}q, \varPhi , q \rightarrow q$. The rightmost q cannot be principal, because otherwise $\varPi , q{\mathop {/}}q, \varPhi$ is empty (Lemma 4). The second possibility is the top of $q{\mathop {/}}q$. Then, again by Lemma 4, $\varPi$ is empty, and by Lemma 6 $\varPhi , q \rightarrow q$ is derivable. Since tops of $\varPhi$ are not q, the rightmost occurrence of q is principal. By Lemma 4, $\varPhi$ is empty.

By $\mathbb {F}$ we denote the free group generated by the set of primitive types $\mathrm {Pr}$. For every $A \in \mathrm {Tp}$ we define its interpretation in this free group, $[\![A]\!]$, as follows: $[\![q]\!] = q \text { for } q \in \mathrm {Pr}$; $[\![A \cdot B]\!] = [\![A]\!] [\![B]\!]$; $[\![A \mathop {\backslash }B]\!] = [\![A]\!]^{-1} [\![B]\!]$; $[\![B{\mathop {/}}A]\!] = [\![B]\!] [\![A]\!]^{-1}$. If $[\![A]\!]$ is the unit of $\mathbb {F}$, A is called a zero-balance type.

The primitive type count, $\#_q(A)$, for $q \in \mathrm {Pr}$ and $A \in \mathrm {Tp}$, is defined as follows: $\#_q(q) = 1$; $\#_q(q') = 0$, if $q' \in \mathrm {Pr}$ and $q' \ne q$; $\#_q(A \cdot B) = \#_q(A) + \#_q(B)$; $\#_q(A \mathop {\backslash }B) = \#_q(B{\mathop {/}}A) = \#_q(B) - \#_q(A)$. Notice that if A is a zero-balance type, then $\#_q(A) = 0$ for every $q \in \mathrm {Pr}$.

If the sequent $A_1, \dots , A_n \rightarrow B$ is derivable in $\mathbf {L}$, then it is balanced, namely, $\#_q(B) = \#_q(A_1) + \ldots + \#_q(A_n)$ for every $q \in \mathrm {Pr}$, and $[\![A_1]\!] \ldots [\![A_n]\!] = [\![B]\!]$.

Theorem 4

(M. Pentus, 1994). If $[\![A_1]\!] = [\![A_2]\!] = \ldots = [\![A_n]\!]$, then there exists such $B \in \mathrm {Tp}$, that all sequents $A_1 \rightarrow B$, $A_2 \rightarrow B$, ..., $A_n \rightarrow B$ are derivable in $\mathbf {L}$ [16].

For a set of zero-balance types $\mathcal {U}= \{ A_1, \dots , A_n \}$, we construct an ersatz of their additive disjunction, $A_1 \vee \ldots \vee A_n$, in the following way. In the notations for types, we sometimes omit the multiplication sign, $\cdot$, if this doesn’t lead to misunderstanding. Let u, t, and s be fresh primitive types, not occurring in $A_i$. By Theorem 4, there exist such types F and G that the following sequents are derivable for all $i = 1, \ldots , n$:

$$ (t{\mathop {/}}t) A_i (t{\mathop {/}}t) \ldots (t{\mathop {/}}t) A_n (t{\mathop {/}}t) \rightarrow F, \qquad (t{\mathop {/}}t) A_1 (t{\mathop {/}}t) \ldots (t{\mathop {/}}t) A_i (t{\mathop {/}}t) \rightarrow G. $$

Now let

$$ E = (t{\mathop {/}}t) A_1 (t{\mathop {/}}t) A_2 (t{\mathop {/}}t) \ldots (t{\mathop {/}}t) A_n (t{\mathop {/}}t), $$

$$ B = E\, (((u{\mathop {/}}F) \mathop {\backslash }u) \mathop {\backslash }(t{\mathop {/}}t)), \qquad C = ((t{\mathop {/}}t) \mathop {/}(u{\mathop {/}}(G \mathop {\backslash }u))) \, E. $$

We omit the multiplication sign, $\cdot$, if this doesn’t lead to misunderstanding.

Finally, $\mathrm {is}(\mathcal {U}) = ((s{\mathop {/}}E) \cdot B) \mathop {\backslash }s{\mathop {/}}C.$

Lemma 8

For each $A_i \in \mathcal {U}$, the sequent $A_i \rightarrow \mathrm {is}(\mathcal {U})$ is derivable in $\mathbf {L}$.

Proof

The derivation is straightforward.

Lemma 9

If the sequent $\varPi \rightarrow \mathrm {is}(\mathcal {U})$ is derivable in $\mathbf {L}$, all types in $\varPi$ have tops, and these tops are not s or t, then for some $A_j \in \mathcal {U}$ the sequent $B_2, \varPi , C_1 \rightarrow A_j$, where $B_2$ is either empty or is a type such that $B = B_2$ or $B = B_1 \cdot B_2$ for some $B_1$, and $C_1$ is either empty or is a type such that $C = C_1$ or $C = C_1 \cdot C_2$ for some $C_2$ (up to associativity of $\cdot$).

Using the invertibility of $(\cdot \rightarrow )$, we replace $\cdot$’s in $B_2$ and $C_1$ by commas, and thus consider them as sequences of types that have tops. Actually, we want them to be empty, and it will be so in our final construction.

Proof

Let $\varPi \rightarrow \mathrm {is}(\mathcal {U})$ be derivable. Then one can derive $(s{\mathop {/}}E), B, \varPi , C \rightarrow s$, and then by Lemma 6 we get $B, \varPi , C \rightarrow E$ (since the leftmost s is the only top s, and it is the principal occurrence). Recall that $E = (t{\mathop {/}}t) A_1 \dots (t{\mathop {/}}t) A_n (t{\mathop {/}}t)$ and apply Lemma 3. It is sufficient so show that, after decompositon, the whole $\varPi$ comes to one part of the left-hand side of the sequent. Suppose the contrary, then locate the principal occurrence of t (it should be in B). Then proceed by induction: finally we run out of t’s in B and get a contradiction.

Proof

(of Theorem 1 ). Given a context-free grammar $\mathcal {G}$ without $\varepsilon$-rules, we need to construct an equivalent Lambek grammar with unique type assignment. Let $\varSigma = \{ a_1, \dots , a_\mu \}$ be the alphabet, $\mathcal {N}= \{ N_0, N_1, N_2, \dots , N_\nu \}$ be the set of non-terminal symbols of $\mathcal {G}$, $N_0$ is the starting symbol.

First we algorithmically transform $\mathcal {G}$ into Greibach normal form [6] with rules of the following three forms: $N_i \Rightarrow a_j N_k N_\ell$, $N_i \Rightarrow a_j N_k$, or $N_i \Rightarrow a_j$.

Now we construct the Lambek grammar. Let $\mathrm {Pr}$ include distinct primitive types p, $p_1, \dots , p_\nu$, r, u, t, and s. For each $i = 0, \dots , \nu$ let $H_i = p{\mathop {/}}((p_i{\mathop {/}}p_i) \cdot p)$ (this type corresponds to the non-terminal $N_i$). Next, for each $j = 1, \dots , \mu$, we form a set $\mathcal {U}_j$ in the following way:

$$\begin{aligned} \text {add } K_{i,k,\ell } = r{\mathop {/}}\bigl ( ( H_k \cdot H_\ell \cdot (p_i{\mathop {/}}p_i)) \mathop {\backslash }r \bigr )&\text { for each rule } N_i \Rightarrow a_j N_k N_\ell ,\\ \text {add } K_{i,k} = r{\mathop {/}}\bigl ( ( H_k \cdot (p_i{\mathop {/}}p_i)) \mathop {\backslash }r \bigr )&\text { for each rule } N_i \Rightarrow a_j N_k,\\ \text {add } K_i = r{\mathop {/}}\bigl ( (p_i{\mathop {/}}p_i) \mathop {\backslash }r \bigr )&\text { for each rule } N_i \Rightarrow a_j. \end{aligned}$$

Now let $D_j = \mathrm {is}(\mathcal {U}_j)$ and $A_j = p{\mathop {/}}(D_j \cdot p)$ be the type corresponding to $a_j$. For the target type H we take $H_0$. Our claim is that $a_{i_1} \dots a_{i_n} \in \mathcal {L}(\mathcal {G})$ iff the sequent $A_{i_1}, \dots , A_{i_n} \rightarrow H_0$ is derivable in $\mathbf {L}$.

For the easier “only if” part, we prove a more general statement: if $\gamma \in (\mathcal {N}\cup \varSigma )^+$ can be generated from $N_m$ in $\mathcal {G}$, then the sequent $\varGamma \rightarrow H_m$ is derivable in $\mathbf {L}$, where $\varGamma$ is a sequence of types corresponding to letters of $\gamma$, $A_j$ for $a_j \in \varSigma$ and $H_i$ for $N_i \in \mathcal {N}$. To establish this, it is sufficient to prove that the following sequents are derivable (each step of the context-free generation maps to a $(\mathrm {cut})$ with the corresponding sequent):

$$\begin{aligned} A_j, H_k, H_\ell \rightarrow H_i&\text { for each rule } N_i \Rightarrow a_j N_k N_\ell ,\\ A_j, H_k \rightarrow H_i&\text { for each rule } N_i \Rightarrow a_j N_k,\\ A_j \rightarrow H_i&\text { for each rule } N_i \Rightarrow a_j. \end{aligned}$$

Consider sequents of the first type (the second and the third types are handled similarly). Since $D_j = \mathrm {is}(\mathcal {U}_j)$ and $K_{i,k,\ell } = r{\mathop {/}}\bigl ( ( H_k \cdot H_\ell \cdot (p_i{\mathop {/}}p_i)) \mathop {\backslash }r \bigr ) \in \mathcal {U}_j$, we have $K_{i,k,\ell } \rightarrow D_j$ by Lemma 8. Then the derivation is as follows:

For the “if” part, let $A_{i_1}, \dots , A_{i_n} \rightarrow H_i$ be derivable and proceed by induction on the cut-free derivation (i is arbitrary here for induction; in the end $i = 0$). Since $H_i = p{\mathop {/}}((p_i{\mathop {/}}p_i) \cdot p)$ and $(\rightarrow \mathop {/})$ and $(\cdot \rightarrow )$ are invertible, we get $A_{i_1}, \dots , A_{i_n}, p_i{\mathop {/}}p_i, p \rightarrow p$. Locate the principal occurrence of p. By Lemma 4, it is the p in $A_{i_1} = p{\mathop {/}}(D_{i_1} \cdot p)$, and by Lemma 6 the sequent $A_{i_2}, \dots , A_{i_n}, p_i{\mathop {/}}p_i, p \rightarrow D_{i_1} \cdot p$ is derivable. Let $j = i_1$. Since all our types have tops, apply Lemma 3.

Case 1 (good). The sequent decomposes into $A_{i_2}, \dots , A_{i_n}, p_i{\mathop {/}}p_i \rightarrow D_j$ and $p \rightarrow p$. Consider the former sequent. Tops on the left side are p and $p_i$, we can apply Lemma 9 and get $B_2, A_{i_2}, \dots , A_{i_n}, p_i{\mathop {/}}p_i, C_1 \rightarrow K$ for some $K \in \mathcal {U}_j$.

Let’s prove that $B_2$ and $C_1$ in this case are empty. Suppose $K = K_{i',k,\ell }$ (the cases of $K_{i',k}$ and $K_{i'}$ are handled similarly). Then, by invertibility of $(\rightarrow \mathop {/})$, we get $B_2, A_{i_2}, \dots , A_{i_n}, p_i{\mathop {/}}p_i, C_1, (H_k \cdot H_\ell \cdot (p_{i'}{\mathop {/}}p_{i'})) \mathop {\backslash }r \rightarrow r$. Now locate the principal occurrence of r.

Subcase 1.1. The principal occurrence of r is the rightmost one. By Lemma 6, we get $B_2, A_{i_2}, \dots , A_{i_n}, p_i{\mathop {/}}p_i, C_1 \rightarrow H_k \cdot H_\ell \cdot (p_{i'}{\mathop {/}}p_{i'})$. Apply Lemma 3. First, by Lemma 5, $i = i'$, otherwise there’s no $p_i$ in tops of the left-hand side. Next, the part of the left-hand side that yields $(p_i{\mathop {/}}p_i)$, by Lemma 7, contains only $(p_i{\mathop {/}}p_i)$. Therefore, $C_1$ is empty. Now, for some part $\varPi$ we have $\varPi \rightarrow H_k$, and, decomposing $H_k$, we get $\varPi , p_i{\mathop {/}}p_i, p \rightarrow p$. By Lemma 4, the principal occurrence of p is not the rightmost one. Since $B_2$ doesn’t have p in tops, $\varPi$ should include also some of the $A_{i_2}, \dots$, and the principal p is the top of one of them. But, since $A_m$ is of the form $p{\mathop {/}}\dots$, by Lemma 4 the part to the left of this $A_m$, and, therefore, $B_2$ should be empty.

Subcase 1.2. The principal occurrence of r is somewhere in $B_2$ or $C_1$, in a type $K \in \mathcal {U}_j$. By Lemma 4, it is then the leftmost occurrence of r, because K has the form $r\mathop {/}\ldots$. This rules out the possibility of it being in $C_1$ (we definitely have $p_i{\mathop {/}}p_i$ to the left of it). If it is in $B_2$, again by Lemma 6, we get $B'_2, A_{i_2}, \dots , A_{i_n}, p_i{\mathop {/}}p_i, C_1 \rightarrow H_{k'} \cdot H_{\ell '} \cdot (p_i{\mathop {/}}p_i)$ ($H_{k'}$ and $H_{\ell '}$ are optional), and we’re in the same situation, as Subcase 1.1. Thus, $B'_2$ and $C_1$ should be empty. However, $B'_2$ now should contain the last type of B, $(((u{\mathop {/}}F)\mathop {\backslash }u){\mathop {/}}(t \mathop {/}t))$. Contradiction. Subcase 1.2 impossible.

Now we have $A_{i_2}, \dots , A_{i_n}, p_i{\mathop {/}}p_i \rightarrow H_k \cdot H_\ell \cdot (p_i{\mathop {/}}p_i)$ (the only choice for the principal r now is the rightmost one, and we’ve applied Lemma 6). By Lemma 3, we get $A_{i_2}, \dots , A_{i_z} \rightarrow H_k$, $A_{i_{z+1}}, \dots , A_{i_n} \rightarrow H_\ell$, $p_i{\mathop {/}}p_i \rightarrow p_i{\mathop {/}}p_i$ (the last left side is $p_i{\mathop {/}}p_i$ alone by Lemma 7).

Apply induction hypothesis. In the context-free grammar, we now have $N_k \Rightarrow ^* a_{i_2} \dots a_{i_z}$ and $N_\ell \Rightarrow ^* a_{i_{z+1}} \dots a_{i_n}$. Since $i' = i$, we also have the rule $N_i \Rightarrow a_j N_k N_\ell$ ($N_k$ and $N_\ell$ are optional) in the grammar (since the corresponding K type was in $\mathcal {U}_j$). Thus, $N_i \Rightarrow a_j a_{i_2} \dots a_{i_z} a_{i_{z+1}} \dots a_{i_n}$. Recall that $j = i_1$.

Case 2 (bad). The sequent decomposes in another way, yielding $A_{i_2}, \dots , A_{i_z} \rightarrow D_j$ and $\dots , p_i{\mathop {/}}p_i, p \rightarrow p$. Again, by Lemma 9, we get $B_2, A_{i_2}, \dots , A_{i_z}, C_1 \rightarrow K$ for some $K \in \mathcal {U}_j$, and further $B_2, A_{i_2}, \dots , A_{i_z}, C_1, (H_k \cdot H_\ell \cdot (p_{i'}{\mathop {/}}p_{i'})) \mathop {\backslash }r \rightarrow r$. Now we locate the principal occurrence of r and proceed as in Case 1. The only difference, however, is that now there is no $p_i{\mathop {/}}p_i$ in the left-hand side, and for that reason derivation fails by Lemma 5. Thus, Case 2 is impossible.