Regular and context-free nominal traces

Abstract

Two kinds of automata are presented, for recognising new classes of regular and context-free nominal languages. We compare their expressive power with analogous proposals in the literature, showing that they express novel classes of languages. Although many properties of classical languages hold no longer in the nominal case, we design a slight restriction of our models that preserve some interesting ones. In particular, we prove the emptiness problem decidable and we construct the intersection between (restricted) regular and context-free automata. By examples and walking through their properties we argue the relevance of our models in the context of the verification of resource usage patterns.

Appendix: Usages

To compare the expressive power of ${\text {PDNA}}_{\pm }$ against ${\text {Usages}}$ we need to extend the expressiveness theorems about ${\text {Usages}}$.

Theorem 9

The language ${\varSigma }^{*}$ is not generated by any ${\text {Usages}}$.

Proof

The proof has the following structure: a new transition system for ${\text {Usages}}$ is given, this is provably equivalent to the original one [7]. Then a few lemmata are proved to provide support for the final argument.

In the following we will use the word redex to identify the source of a transition deduced by only applying an (instance of an) axiom.

Lemma 5

If and $\mu h. U$ is its redex, then .

Proof

(By induction on the depth of the proof)

There are two exhaustive cases:

• Base case: $U_{0} = \mu h. U$ and the thesis follows easily.

• Inductive case: has been proved by

  

  by rule (seq) as last step. By the inductive hypothesis we have that and the thesis follows easily.

$\square$

Lemma 6

If and k is the least index such that $\mu h . U$ is the redex of and rule (rec) is never used in reducing $U_{i}, i < k$ then $U_{i} = C_{i}[\mu h. U]$ for some $C_{i}$ and $a_{k}=\varepsilon$.

Lemma 7

Let be a computation and let k be the least index such that $\mu h . U$ is the redex of and rule (rec) is never used in reducing $U_{i}, i < k$ then $\exists U_{i}' \equiv U_{i}, i < k$ such that and (rec) is never used reducing $U_{i}, 0 \le i \le k$.

Proof

By Lemma 6 $U_{i} = C_{i}[\mu h . U], i \le k$, also $\mu h . U$ is never the redex of $U_{i}, i < k$, hence also . By Lemma 5 . $\square$

Lemma 8

Let such that such that is deduced using (rec) rule.

Proof

By repeated application of Lemma 7. $\square$

Property 7

(of capture avoiding substitutions) Given $\mu h . U$, if n occurs in the scope of $\nu n$, then contains a term $\nu n'. U', n' \ne n$ and n does not occur in $U'$.

Proof

Follows because unfolding a recursion is capture-avoiding and all the bound names are different. $\square$

For simplicity , we write $a_{n}$ for the dynamic resource replacing a name n in U when the rule (new) is applied.

Let U with k occurrences of $\nu n_{i}$ be such that $\llbracket U \rrbracket = {\varSigma }^{*}$ and let $s = a_{n_{1}}a_{n_{2}}\dots a_{n_{k+1}}a_{n_{1}}a_{n_{2}}\dots a_{n_{k+1}}$. By Lemma 8, there exists $U' \equiv U$ such that with no transition deduced using rule (rec).

Since $\nu .n_{k+1}$ does not occur in U, then there exists a subterm of U of the form $\mu h. {\overline{U}}$, with $\nu n_{i}. U'', (0 < i \le k)$ in ${\overline{U}}$. Therefore, $U' = U\{\mu h . {\overline{U}} /h\}$ and the replacing term contains $\nu n_{k+1}. U''\{n_{k+1}/n_{i}\}, n_{k+1}\ne n_{i}$ for some i because our assumption of keeping bound names apart.

Therefore $\nu n_{k+1}. U''\{n_{k+1}/n_{i}\}$ occurs in $U'$ for some $U''$ (by Lemma 8) and $n_{i}$ does not occur in $U''\{n_{k+1}/n_{i}\}$. Also $n_{k+1}$ must occur at least twice in $U''\{n_{k+1}/n_{i}\}$ and nowhere else. Since $U''\{n_{k+1}/n_{i}\}$ can not generate $a_{n_{i}}$, it can not generate $a_{n_{k+1}}a_{n_{1}}\dots a_{n_{k}}a_{n_{k+1}}$. $\square$