On the Expressiveness of Parameterization in Process-Passing

Abstract

This paper studies higher-order processes with the capability of parameterization (or abstraction), which has been proven to be an effective measure of lifting the mere process-passing in expressiveness. We contribute to the understanding of two kinds of parameterization: name-parameterization and process-parameterization, particularly in a second-order setting (no currying of parameterization). Firstly, we show that in the expressiveness hierarchy of process-parameterization, n-ary parameterization can faithfully translate (n+1)-ary parameterization in a linear setting where each received process can be used only once. Secondly, the two kinds of parameterization are compared. We prove that name-parameterization is more basic than process-parameterization, i.e. the former can encode the union of them. As a result, name-parameterization can strictly promote the expressiveness of mere process-passing.

Xian Xu, Qiang Yin, and Huan Long are supported by the National Nature Science Foundation of China (61261130589, 61202023, 61173048), and the ANR project 12IS02001 (PACE).

Appendices

Appendix

Proofs

1.1 Proofs for Sect. 3

Proof of Lemma 1

Proof

$(1.1)$ and $(2.1)$ are trivial according to the encoding. $(3)$ becomes easy in a straightforward way from $(1.1)$, $(1.2)$, $(2.1)$, $(2.2)$. The most critical parts are $(1.2)$ and $(2.2)$. Since their proof strategies are somewhat similar, we focus on $(1.2)$. To prove that clause, we need to undergo an induction on the structure of $P$. Thus there are several cases: $0$, $a(Y).P_1$, $\overline{a}E.P_1$, $P_1|\,Q_1$, $(c)P_1$, $\langle X_1, X_2 \rangle P_1$, $P_1\langle B_1, B_2 \rangle$. We will expand the analysis on three of them, and the rest are similar or easier.

• $P\equiv a(Y).P_1$. In this case $P{\xrightarrow {a(G)}} P_1\{G/Y\}$ and $F\equiv P_1$. According to the encoding, $[\![P ]\!]^{}_{}\equiv a(Y). [\![P_1 ]\!]^{}_{} {\xrightarrow {a(H)}} [\![P_1 ]\!]^{}_{}\{H/Y\} \equiv T$, in which $H\equiv \langle Z \rangle \overline{m}Z$. Since $F\equiv P_1$, we immediately have $T = [\![F ]\!]^{}_{}\{H/Y\}$.

• $P\equiv P_1\langle B_1, B_2 \rangle$. The only countable case is $P\equiv Y\langle B_1, B_2 \rangle$, when $[\![P ]\!]^{}_{}\equiv (fgh)(\overline{g}|\,Y\langle g.(\overline{f}|\,\overline{h})+f.( [\![B_1 ]\!]^{}_{}|\,\overline{f}) \rangle |\,h.Y\langle [\![B_2 ]\!]^{}_{} \rangle )$. So this case is impossible.

• $P\equiv P_1|\,Q_1$. W.l.o.g., suppose $P_1{\xrightarrow {a(G)}} P_1'\equiv F_1\{G/Y\}$ for some $F_1$ and $Y$ is fresh. So $P\equiv P_1|\,Q_1 {\xrightarrow {a(G)}} P_1'|\,Q_1\equiv F_1\{G/Y\} |\,Q_1\equiv F\{G/Y\}$ in which $F\;{\mathop {=}\limits ^{def}}\; F_1|\,Q_1$. By induction hypothesis, $[\![P_1 ]\!]^{}_{}{\xrightarrow {a(H)}} T_1 = [\![F_1 ]\!]^{}_{}\{H/Y\}$ where $H\equiv \langle Z \rangle \overline{m}Z$. Thus

  $$\begin{aligned}{}[\![P ]\!]^{}_{}\equiv [\![P_1 ]\!]^{}_{}|\, [\![Q_1 ]\!]^{}_{} {\xrightarrow {a(H)}} T_1|\, [\![Q_1 ]\!]^{}_{}&= [\![F_1 ]\!]^{}_{}\{H/Y\} |\, [\![Q_1 ]\!]^{}_{} \\&= ( [\![F_1 ]\!]^{}_{}|\, [\![Q_1 ]\!]^{}_{})\{H/Y\} \\&= ( [\![F_1|\,Q_1 ]\!]^{}_{})\{H/Y\} \\&= ( [\![F ]\!]^{}_{})\{H/Y\} \end{aligned}$$

  $\square$

Proof of Lemma 3

Proof

Based on the operational correspondence, our target is to prove $P = Q$ implies $[\![P ]\!]^{}_{} = [\![Q ]\!]^{}_{}$, through showing

$$ \mathcal {R} \;{\mathop {=}\limits ^{def}}\; \{( [\![P ]\!]^{}_{}, [\![Q ]\!]^{}_{}) \;|\; P = Q\}\quad \text{ which } \text{ is } \text{ closed } \text{ under } \text{(variable) } \text{ substitution } $$

is a normal bisimulation up-to $=$ (see [19, 22] for up-to techniques). Remember “closed under (variable) substitution” means $P\{A/X\} \;\mathcal {R}\; Q\{A/X\}$ for every $A$ whenever $P\mathcal {R} Q$ and they are (open) processes with occurrence of variable $X$. Notice in the calculus $\varPi ^D_n$, normal bisimulation characterizes context bisimulation; see [17, 19, 29] for proof reference. In normal bisimulation, the communicated process only needs to be confined to triggers ($\langle Z \rangle \overline{m}Z.0$) and the testing environment in the output clause only needs to be $!n(X).[\,]\langle X \rangle$.

There are several cases needing analysis, among which the most crucial parts are concerning the communication of parameterized processes.

• 1. $[\![P ]\!]^{}_{}{\mathop {\Longrightarrow }\limits ^{a(H)}} T$ in which $H\equiv \langle Z \rangle \overline{m}Z$;

• 2. $[\![P ]\!]^{}_{}{\mathop {\Longrightarrow }\limits ^{(m)\overline{a}T_1}} T$ in which $T_1\equiv \langle Z \rangle \overline{m}Z$;

• 3. $[\![P ]\!]^{}_{}{\mathop {\Longrightarrow }\limits ^{\tau }} T$

We below proceed with their analysis.

• Case $1$. From the premise, $T\equiv S\{H/Y\}$ for some $S$, and by Lemma 2 $P{\mathop {\Longrightarrow }\limits ^{a(G)}} P'\equiv F\{G/Y\}$ for some $F$ and $G\equiv \langle X_1, X_2 \rangle E$, and also $S = [\![F ]\!]^{}_{}$. Because $P = Q$, $Q$ can simulate $P$ with $Q{\mathop {\Longrightarrow }\limits ^{a(G)}} Q'\equiv F'\{G/Y\}$ for some $F'$, and $P' = Q'$. Clearly we also have $F = F'$. By Lemma 1, $[\![Q ]\!]^{}_{}{\mathop {\Longrightarrow }\limits ^{a(H)}} T' = [\![F' ]\!]^{}_{}\{H/Y\}$, then

  $$ T = [\![F ]\!]^{}_{}\{H/Y\} \;\mathcal {R}\; [\![F' ]\!]^{}_{}\{H/Y\} = T' $$

  because $[\![F ]\!]^{}_{} \;\mathcal {R}\; [\![F' ]\!]^{}_{}$ due to $F = F'$.

• Case $2$. From the premise and Lemma 2, $P{\mathop {\Longrightarrow }\limits ^{(\widetilde{c})\overline{a}[\langle X_1, X_2 \rangle E]}} P'$ and

  $$ T = (\widetilde{c})( [\![P' ]\!]^{}_{}|\, [\![\langle X_1, X_2 \rangle E ]\!]^{}_{}). $$

  Since $P = Q$, $Q$ can simulate $P$ by $Q {\mathop {\Longrightarrow }\limits ^{(\widetilde{d})\overline{a}G}} Q'$ and

  $$\begin{aligned} (\widetilde{c})(P'|\,C[\langle X_1, X_2 \rangle E]) = (\widetilde{d})(Q'|\,C[G]) \end{aligned}$$

  (1)

  for every $\varPi ^D_2$ context $C$[ ]. There are two cases: whether $G$ is a parameterization or not. Actually they can be similarly dealt with, so we only consider the case $G\equiv \langle X_1, X_2 \rangle F$ which is a little more complicated. Then by Lemma 1,

  $$ [\![Q ]\!]^{}_{}{\mathop {\Longrightarrow }\limits ^{(m)\overline{a}[\langle Z \rangle \overline{m}Z]}} T' = (\widetilde{d})( [\![Q' ]\!]^{}_{}|\, [\![\langle X_1, X_2 \rangle F ]\!]^{}_{}) $$

  We intend to show

  $$ (m)(T|\,D[\langle Z \rangle \overline{m}Z]) \;\mathcal {R}\; (m)(T'|\,D[\langle Z \rangle \overline{m}Z]) $$

  for a $\varPi ^D_1$ context $D[\,]\equiv !n(X).[\,]\langle X \rangle$ in which $n$ is fresh. Or equivalently

  $$\begin{aligned}&\quad (m)((\widetilde{c})( [\![P' ]\!]^{}_{}|\, [\![\langle X_1, X_2 \rangle E ]\!]^{}_{})|\,!n(X).\overline{m}X) \nonumber \\&\mathcal {R}\; (m)((\widetilde{d})( [\![Q' ]\!]^{}_{}|\, [\![\langle X_1, X_2 \rangle F ]\!]^{}_{})|\,!n(X).\overline{m}X) \end{aligned}$$

  (2)

  In terms of (1), it is obvious that when fixing $C[\,]\equiv (m)([\,]|\,!n(X).\overline{m}X)$ we have

  $$ (\widetilde{c})(P'|\,(m)(\langle X_1, X_2 \rangle E|\,!n(X).\overline{m}X)) = (\widetilde{d})(Q'|\,(m)(\langle X_1, X_2 \rangle F|\,!n(X).\overline{m}X)) $$

  which becomes the following after applying some structural adjustment

  $$\begin{aligned} (m)((\widetilde{c})(P'|\,\langle X_1, X_2 \rangle E) |\,!n(X).\overline{m}X) = (m)((\widetilde{d})(Q'|\,\langle X_1, X_2 \rangle F)|\,!n(X).\overline{m}X) \end{aligned}$$

  (3)

  Now it is not hard to see that the left and right side of $=$ in (3) are respectively encoded into the left and right side of $\mathcal {R}$ in (2). That virtually proves the validity of (2) by the definition of $\mathcal {R}$.

• Case $3$. From the premise and Lemma 2, $P{\mathop {\Longrightarrow }\limits ^{\tau }} \text{ and } T{\mathop {\Longrightarrow }\limits ^{}} \cdot = [\![P' ]\!]^{}_{}$ or simply $T = [\![P' ]\!]^{}_{}$. Since $P = Q$, we have $Q{\mathop {\Longrightarrow }\limits ^{\tau }} Q' \text{ and } P' = Q'$. Then by Lemma 1, $[\![Q ]\!]^{}_{}{\mathop {\Longrightarrow }\limits ^{\tau }} T' = [\![Q' ]\!]^{}_{}$. Thus

  $$ T = [\![P' ]\!]^{}_{} \;\mathcal {R}\; [\![Q' ]\!]^{}_{} = T'. $$

Now the proof is completed. $\square$

1.2 Proofs for Sect. 4

Proof of Lemma 5

Proof

We only prove the strong transition correspondence part. The proof proceeds by induction on the structure of $P$, we consider the following cases.

• Case 1. $P \equiv a(X).P_1$. $P$ can only do an input action, suppose that $P {\xrightarrow {a(A)}} P'$ and $P'\equiv P_1\{A/X\}$, then we have

  $$ [\![P ]\!]^{}_{} \equiv a(X). [\![P_1 ]\!]^{}_{} {\xrightarrow {a( [\![A ]\!]^{}_{})}} \equiv [\![P_1 ]\!]^{}_{}\{ [\![A ]\!]^{}_{}/X\} = [\![P_1\{A/X\} ]\!]^{}_{}. $$

• Case 2. $P \equiv P_1 | P_2$. There are 3 subcases.

  • 1. $P {\xrightarrow {(\widetilde{c})\bar{a}A}} P'$. W.l.o.g. we can assume this is caused by $P_1 {\xrightarrow {(\widetilde{c})\bar{a}A}} P_1'$ and $P' \equiv P_1' | P_2$. Then by I.H. we have $[\![P_1 ]\!]^{}_{}{\mathop {\Longrightarrow }\limits ^{}} {\xrightarrow {(\widetilde{c})\bar{a} [\![A ]\!]^{}_{}}} T_1$ and $T_1 = [\![P_1' ]\!]^{}_{}$. As a result we have $[\![P ]\!]^{}_{} \equiv [\![P_1 ]\!]^{}_{} | [\![P_2 ]\!]^{}_{} {\mathop {\Longrightarrow }\limits ^{}} {\xrightarrow {(\widetilde{c})\bar{a} [\![A ]\!]^{}_{}}} T_1 | [\![P_2 ]\!]^{}_{}$ and $T_1 | [\![P_2 ]\!]^{}_{} = [\![P_1' ]\!]^{}_{}| [\![P_2 ]\!]^{}_{} = [\![P' ]\!]^{}_{}$.

  • 2. $P {\xrightarrow {a(A)}} P'$ this is case is similar to the above.

  • 3. $P {\xrightarrow {\tau }} P'$. W.l.o.g. we assume that this is caused by $P_1 {\xrightarrow {a(A)}} P_1$, $P_2 {\xrightarrow {(\widetilde{c})\bar{a}A}} P_2'$ and $P' \equiv (c)(P_1'|P_2')$. By I.H. we have $[\![P_1 ]\!]^{}_{} {\mathop {\Longrightarrow }\limits ^{}} {\xrightarrow {a( [\![A ]\!]^{}_{})}} T_1 = [\![P_1' ]\!]^{}_{}$ and $[\![P_2 ]\!]^{}_{} {\mathop {\Longrightarrow }\limits ^{}} {\xrightarrow {(\widetilde{c})\bar{a} [\![A ]\!]^{}_{}}} T_2 = [\![P_2' ]\!]^{}_{}$. As a result $[\![P ]\!]^{}_{} \equiv [\![P_1 ]\!]^{}_{}| [\![P_2 ]\!]^{}_{} {\mathop {\Longrightarrow }\limits ^{\tau }} T$ and

    $$T \equiv (\widetilde{c})(T_1|T_2) = (\widetilde{c})( [\![P_1' ]\!]^{}_{}| [\![P_2' ]\!]^{}_{}) \equiv [\![P' ]\!]^{}_{}.$$

• Case 3. $P \equiv \bar{a}A.P_1$ or $P \equiv (c)P_1$. This case is similar to case 2.

• Case 4. $P \equiv A\langle P_1 \rangle$. Suppose that $A \equiv \langle X \rangle Q$ and $P {\xrightarrow {\lambda }} P'$ there are three subcases to consider: $\lambda = a(B)$, $\lambda = (\widetilde{c})\bar{a}E$ and $\lambda = \tau$. We only deal with the case that $\lambda = a(B)$, the others are similar. Consider the different cause of $P {\xrightarrow {a(B)}} P'$.

  • 1. If $P {\xrightarrow {a(B)}} P'$ is caused by $Q$, then we can assume that $Q\{P_1 / X \} {\xrightarrow {a(A)}} P' \equiv Q_1\{P_1 /X \}\{ A / Y \}$ and $Q\{\bar{m}0 / X\} {\xrightarrow {a(A)}} Q_1\{\bar{m}0 / X\}\{A / Y \}$. By I.H. $[\![Q\{\bar{m}0 / X\} ]\!]^{}_{} {\mathop {\Longrightarrow }\limits ^{}} {\xrightarrow {a( [\![A ]\!]^{}_{})}} T'$ and $T' = [\![Q_1\{\bar{m}0 / X\}\{ A / Y \} ]\!]^{}_{}$. Thus we have $[\![P ]\!]^{}_{} \equiv (m)( [\![Q ]\!]^{}_{}\{\bar{m}0 / X \}| !m. [\![P_1 ]\!]^{}_{}) {\mathop {\Longrightarrow }\limits ^{}} {\xrightarrow {a( [\![A ]\!]^{}_{})}} T$ and

    $$\ T = (m)(T' | !m. [\![P_1 ]\!]^{}_{}) = (m)( [\![Q_1\{\bar{m}0 / X\}\{ A / Y \} ]\!]^{}_{} | !m. [\![P_1 ]\!]^{)}_{=} [\![P' ]\!]^{}_{}$$

  • 2. If $P {\xrightarrow {a(B)}} P'$ is caused by $P_1{\xrightarrow {a(A)}} P_2$ then we have $Q \equiv Q'| X$ for some $Q'$ , $Q\{P_1/X\} \equiv Q'\{P_1/X\} | P_1$ and $P' \equiv Q'\{P_1/X\} | P_2$. By I.H. $[\![P_1 ]\!]^{}_{} {\mathop {\Longrightarrow }\limits ^{}} {\xrightarrow {a( [\![A ]\!]^{}_{})}} T'$ and $T' = [\![P_2 ]\!]^{}_{}$. Then

    $$\begin{aligned}{}[\![P ]\!]^{}_{} \qquad \equiv \qquad&(m)( [\![Q ]\!]^{}_{}\{\bar{m}0 / X \}\langle m \rangle | !m. [\![P_1 ]\!]^{}_{}) \\ \equiv \qquad&(m)( [\![Q'|X ]\!]^{}_{}\{\bar{m}0 / X \}\langle m \rangle | !m. [\![P_1 ]\!]^{}_{}) \\ {\xrightarrow {\tau }} \qquad&(m)( [\![Q' ]\!]^{}_{}\{\bar{m}0 / X \}\langle m \rangle | [\![P_1 ]\!]^{}_{}| !m. [\![P_1 ]\!]^{}_{}) \\ {\mathop {\Longrightarrow }\limits ^{}} {\xrightarrow {a( [\![A ]\!]^{}_{})}} \;&(m)( [\![Q' ]\!]^{}_{}\{\bar{m}0 / X \}\langle m \rangle |T'| !m. [\![P_1 ]\!]^{}_{}) \\ = \qquad&(m)( [\![Q' ]\!]^{}_{}\{\bar{m}0 / X \}\langle m \rangle | [\![P_2 ]\!]^{}_{}| !m. [\![P_1 ]\!]^{}_{}) \\ \equiv \qquad&[\![P' ]\!]^{}_{} \end{aligned}$$

    $\square$

Proof of Lemma 7

Proof

We prove the relation $\mathcal {R}$ defined by

$$ \mathcal {R} \;{\mathop {=}\limits ^{def}}\; \{ ( [\![P ]\!]^{}_{}, [\![Q ]\!]^{)}_{~}|~ P = Q \} $$

is a context bisimulation up-to $=$. Suppose that $P {\mathop {\Longrightarrow }\limits ^{\lambda }} T$, there are three cases to consider.

• Case 1. $\lambda = a(A)$. By Lemma 6, $P {\mathop {\Longrightarrow }\limits ^{a(A)}} P'$ and $T = [\![P' ]\!]^{}_{}$. Since $P = Q$ , then $Q$ can simulate $P$ with $Q {\mathop {\Longrightarrow }\limits ^{a(A)}} Q' = P'$. By Lemma 5 we know that $[\![Q ]\!]^{}_{} {\mathop {\Longrightarrow }\limits ^{a( [\![A ]\!]^{}_{})}} T'$ and $T' = [\![Q' ]\!]^{}_{}$. As $A$ is an $\varPi ^d_1$ agent, $[\![A ]\!]^{}_{} \equiv A$. In conclusion we got $[\![Q ]\!]^{}_{} {\mathop {\Longrightarrow }\limits ^{a(A)}} T'$ and

  $$ T = [\![P' ]\!]^{}_{} \mathcal {R} [\![Q' ]\!]^{}_{} = T' $$

• Case 2. $\lambda = (\widetilde{c})\bar{a} A$. By Lemma 6, there exists some $\widetilde{c}$, $A'$ s.t. $P {\mathop {\Longrightarrow }\limits ^{(\widetilde{c})\bar{a}A'}} P'$ with $T = [\![P' ]\!]^{}_{}$ and $[\![A' ]\!]^{\equiv }_{A}$. Since $P = Q$, then $Q$ can simulate $P$ with $Q {\mathop {\Longrightarrow }\limits ^{(\widetilde{d})\bar{a}B'}} Q'$ for some $\widetilde{d}$ and $B'$ s.t. for every process $E[\,]$ with $\widetilde{c}\widetilde{d}\cap fn(E[\,])=\emptyset$ it holds that

  $$\begin{aligned} (\widetilde{c})(E[A']|\,P') = (\widetilde{d})(E[B']|\,Q') \end{aligned}$$

  (*)

  By Lemma 5 , there exist some $T'$ s.t. $Q{\mathop {\Longrightarrow }\limits ^{(\widetilde{d})\bar{a} [\![B' ]\!]^{}_{}}} T'$ and $T' = [\![Q' ]\!]^{}_{}$. And from (*) we know that for every process $F [\,]$ without higher-order abstraction and higher-order application s.t. $\widetilde{c}\widetilde{d}\cap fn(F[\,])=\emptyset$ we have

  $$ (\widetilde{c})(F [A]|\, [\![P' ]\!]^{}_{}) \,\mathcal {R}\, (\widetilde{d})(F [B]|\, [\![Q ]\!]^{}_{})\quad {\text {(let}}~{\text {B}} \equiv [\![B' ]\!]^{}_{}) $$

  In conclusion we have $Q{\mathop {\Longrightarrow }\limits ^{(\widetilde{d})\bar{a}B}} T'$ and for every process $F[\,]$ s.t. $\widetilde{c}\widetilde{d}\cap fn(F[\,])=\emptyset$ we have

  $$\begin{aligned} T = (\widetilde{c})(F [A]|\, [\![P' ]\!]^{}_{}) \,\mathcal {R}\, (\widetilde{d})(F [B]|\, [\![Q ]\!]^{}_{}) = T' \end{aligned}$$

• Case 3. $\lambda = \tau$. This case is similar to the case 1. $\square$