Completeness

in

Discrete-Time

Process

Algebra

Michel

A.

Reniers

and

Jan

Joris

Vereijken

Department

of

Computing

Science

Eindhoven

University

of

Technology

P.O.

Box

MB

Eindhoven

The

Netherlands

michelr@win.tue.nl,

janjoris@acm.org

August

Abstract

We

prove

soundness

and

completeness

for

some

ACP-style

concrete

relative-time

discrete-time

process

algebras

We

treat

non-delayable

actions

delayable

actions

and

immediate

deadlock

Basic

process

algebras

are

examined

extensively

and

also

some

concurrent

process

algebras

are

considered

We

conclude

with

ACPdrt

which

combines

all

described

features

in

one

theory

Mathematics

Subject

Classification

CR

Categories

D.

D.

F.

F.

Note

The

investigations

of

the

first

author

were

supported

by

Philips

Research

Laboratories

Eindhoven

The

Netherlands

The

investigations

of

the

second

author

were

supported

by

the

Netherlands

Computer

Science

Research

Foundation

-LRB-

SION

-RRB-

with

financial

support

from

the

Netherlands

Organization

for

Scientific

Research

-LRB-

NWO

-RRB-

Contents

Introduction

Discrete-Time

Process

Algebras

Soundness

and

Completeness

Basic

Process

Algebras

Introduction

Soundness

and

Completeness

of

BPA

Soundness

and

Completeness

of

BPAffi

Soundness

and

Completeness

of

BPA?drt?ffi

Soundness

and

Completeness

of

BPA?drt?ID

Soundness

and

Completeness

of

BPA?drt?ID

Soundness

and

Completeness

of

BPA?drt

Soundness

and

Completeness

of

BPA?drt

Concurrent

Process

Algebras

Introduction

Soundness

and

Completeness

of

PA?drt?ID

Soundness

and

Completeness

of

Soundness

and

Completeness

of

ACP?drt?ID

Soundness

and

Completeness

of

Soundness

and

Completeness

of

PA?drt

Soundness

and

Completeness

of

ACP?drt

Conclusions

Acknowledgments

A

Nomenclature

and

notational

issues

A.

Signatures

A.

Comparison

with

other

notations

B

Overview

B.

Axioms

B.

Theories

B.

Theorems

List

of

Tables

Bibliography

Introduction

Discrete-Time

Process

Algebras

Since

-LSB-

-RSB-

appeared

the

ACP

framework

of

process

algebras

has

been

extended

with

among

many

other

things

discrete-time

extensions

Papers

describing

such

extensions

are

for

example

-LSB-

-RSB-

With

the

recent

appearance

of

-LSB-

-RSB-

discrete-time

process

algebra

seems

to

have

reached

a

decent

state

of

maturity

and

we

felt

now

was

the

time

to

write

a

paper

about

the

soundness

and

completeness

issues

involved

with

discrete

time

To

our

knowledge

no

such

results

have

been

published

in

the

context

of

ACP

before

And

although

we

never

really

doubted

the

soundness

and

completeness

of

the

respective

theories

we

felt

that

it

would

not

hurt

to

prove

these

beliefs

explicitly

And

rightly

so

it

turned

out

that

the

axiomatizations

we

started

out

with

were

neither

sound

nor

complete

Soundness

and

Completeness

In

this

paper

we

will

give

elimination

soundness

and

completeness

results

for

several

discrete-time

process

algebras

We

restrict

ourselves

to

concrete

process

algebras

-LRB-

i.e.

without

abstraction

without

a

silent

step

o

and

without

an

empty

step

-RRB-

relative

time

closed

terms

-LRB-

i.e.

no

completeness

-RRB-

and

mostly

basic

process

algebras

-LRB-

i.e.

without

merge

operators

-RRB-

We

do

treat

delayable

actions

and

immediate

deadlock

In

our

definitions

and

notations

we

try

to

conform

to

-LSB-

-RSB-

We

use

term

deduction

system

semantics

in

the

style

of

Section

of

-LSB-

-RSB-

In

the

proofs

of

elimination

theorems

we

make

abundant

use

of

term

rewriting

analysis

For

further

details

on

these

techniques

especially

the

lexicographical

path

ordering

see

-LSB-

-RSB-

In

proving

completeness

we

sometimes

make

use

of

Verhoef

's

General

Completeness

Theorem

For

more

information

on

this

see

-LSB-

-RSB-

As

the

definitions

and

notations

used

for

discrete-time

process

algebras

have

been

subject

to

vehement

revision

over

the

past

few

years

we

have

compiled

an

appendix

that

tries

to

shed

some

light

on

these

matters

Please

study

this

Appendix

A

if

you

are

not

completely

familiar

with

discrete-time

ACP

Then

in

Appendix

B

we

give

a

concise

summary

of

the

most

important

results

presented

in

this

paper

Finally

the

actual

theorems

and

proofs

on

soundness

and

completeness

are

given

in

Section

-LRB-

for

process

algebras

that

do

not

contain

a

merge

operator

-RRB-

and

Section

-LRB-

for

the

ones

that

do

-RRB-

Basic

Process

Algebras

Introduction

In

this

section

we

prove

soundness

and

completeness

for

some

discrete-time

basic

process

algebras

i.e.

not

containing

a

merge

operator

But

we

start

by

giving

soundness

and

completeness

theorems

for

two

untimed

basic

process

algebras

namely

BPA

and

BPAffi

The

purpose

of

giving

these

results

about

untimed

algebras

is

twofold

First

we

want

to

show

the

principles

our

proof

techniques

are

based

on

and

these

simple

process

algebras

are

more

suited

to

this

purpose

than

the

timed

versions

that

will

follow

Secondly

we

felt

that

soundness

and

completeness

proofs

given

in

the

literature

are

often

a

bit

twisted

and

sometimes

even

plain

wrong

After

these

untimed

basic

process

algebras

by

adding

the

time-unit

delay

we

proceed

to

the

most

simple

timed

one

BPA?drt?ffi

Then

we

add

to

this

algebra

undelayable

deadlock

delayable

actions

and

finally

immediate

deadlock

The

section

culminates

in

soundness

and

correctness

theorems

for

a

basic

process

algebra

that

combines

all

the

above

extensions

at

the

same

time

BPAdrt

Soundness

and

Completeness

of

BPA

Remark

-LRB-

Alphabet

-RRB-

For

this

section

and

all

sections

to

come

we

presume

the

existence

of

a

fixed

finite

alphabet

A

that

can

be

considered

a

parameter

of

the

respective

theories

Furthermore

we

define

Affi

as

A

-LSB-

fffig

and

Aoe

as

A

-LSB-

foeg

where

ffi

and

oe

are

still

to

be

treated

symbols

that

are

not

contained

in

A.

Definition

-LRB-

Signature

of

BPA

-RRB-

The

signature

of

BPA

consists

of

the

atomic

actions

faja

Ag

the

alternative

composition

operator

and

the

sequential

composition

operator

Remark

-LRB-

Symbol

versus

atom

-RRB-

Note

that

in

Definition

in

the

expression

faja

Ag

the

second

a

refers

to

the

symbol

a

while

the

first

one

refers

to

the

atom

a

This

distinction

should

be

clearly

made

and

it

can

be

considered

a

tragic

historical

incident

that

these

different

notions

have

received

the

same

notation

Remark

-LRB-

Range

of

a

-RRB-

When

we

write

a

-LRB-

or

b

or

c

-RRB-

in

the

context

of

an

equality

or

a

partial

ordering

we

mean

this

a

to

range

over

Affi

-LRB-

provided

of

course

deadlock

is

part

of

the

relevant

signature

-RRB-

When

we

write

it

in

the

context

of

a

deduction

rule

we

mean

it

to

range

over

A

In

all

other

cases

or

when

we

deviate

from

the

above

rule

we

explicitly

state

whether

it

ranges

over

A

or

Affi

Definition

-LRB-

Operator

precedence

-RRB-

Throughout

this

paper

we

adhere

to

the

following

operator

precedence

scheme

which

consist

of

four

categories

of

operators

The

four

categories

from

strongly

binding

to

weakly

binding

are

-LRB-

i

-RRB-

all

unary

operators

-LRB-

ii

-RRB-

the

sequential

composition

operator

-LRB-

iii

-RRB-

all

binary

operators

except

the

and

the

-LRB-

iv

-RRB-

the

alternative

composition

operator

Within

one

category

all

operators

bind

equally

strong

Definition

-LRB-

Axioms

of

BPA

-RRB-

The

process

algebra

BPA

is

axiomatized

by

Axioms

shown

in

Table

BPA

x

y

y

x

x

y

z

x

y

z

x

x

x

x

y

z

x

z

y

z

x

y

z

x

y

z

Table

Axioms

of

BPA

Definition

-LRB-

Notation

Regarding

Semantics

I

-RRB-

In

order

to

define

a

semantics

we

will

use

term

deduction

system

semantics

in

the

style

of

Section

of

-LSB-

-RSB-

-LRB-

also

called

Structured

Operational

Semantics

or

Plotkin-style

semantics

-RRB-

We

use

the

notation

x

a

to

denote

that

x

can

do

an

a-step

to

x

a

p

to

denote

that

x

can

do

an

a-step

and

then

terminate

x

to

denote

that

x

can

not

do

an

a-step

and

x

to

denote

that

x

can

not

do

any

step

at

all

For

each

process

algebra

we

define

we

will

give

a

term

deduction

system

By

using

the

concept

of

bisimulation

-LRB-

to

be

defined

in

Definition

on

the

following

page

-RRB-

we

then

turn

the

term

deduction

system

into

a

model

of

the

given

axioms

Definition

-LRB-

Semantics

of

BPA

-RRB-

The

semantics

of

BPA

are

given

by

the

term

deduction

system

T?BPA

induced

by

the

deduction

rules

given

in

Table

and

Table

on

the

following

page

a

a

p

Table

Deduction

rule

for

a.

x

a

x

y

a

x

a

p

x

y

a

p

x

a

x

y

a

y

y

a

x

y

a

y

a

p

x

y

a

p

x

a

p

x

y

a

y

Table

Deduction

rules

for

and

Definition

-LRB-

Symmetric

Closure

-RRB-

For

a

binary

relation

R

we

denote

its

symmetric

closure

by

RS

RS

R

-LSB-

f?y

x?j?x

y

Rg

Definition

-LRB-

Bisimulation

for

BPA

-RRB-

Bisimulation

for

BPA

is

defined

as

follows

a

binary

relation

R

on

closed

BPA

terms

is

a

bisimulation

if

the

following

transfer

conditions

hold

for

all

closed

BPA

terms

p

and

q

-LRB-

i

-RRB-

If

RS?p

q

and

T?BPA

p

a

where

a

A

then

there

exists

a

closed

term

such

that

T?BPA

q

a

and

-LRB-

ii

-RRB-

If

RS?p

q

and

T?BPA

p

a

p

where

a

A

then

T?BPA

q

a

p.

Two

BPA

terms

p

and

q

are

called

bisimilar

notation

p

BPA

q

if

there

exists

a

bisimulation

relation

R

such

that

R?p

q

Definition

-LRB-

Bisimulation

Model

for

BPA

-RRB-

Using

bisimulation

we

can

now

construct

a

model

of

the

axioms

of

BPA

In

order

to

do

this

we

first

need

to

know

that

bisimulation

is

a

congruence

with

respect

to

all

operators

In

-LSB-

-RSB-

it

is

proven

that

a

sufficient

condition

for

this

is

that

the

deduction

rules

satisfy

the

so

called

panth

format

It

is

easy

to

check

that

this

is

indeed

the

case

We

then

construct

the

bisimulation

model

for

BPA

by

taking

the

equivalence

classes

of

the

set

of

all

closed

BPA

terms

with

respect

to

bisimulation

equivalence

As

bisimulation

is

a

congruence

the

operators

can

be

trivially

defined

on

the

equivalence

classes

For

example

for

the

operator

x

BPA

y

BPA

x

y

BPA

Here

x

BPA

denotes

the

equivalence

class

of

x

with

respect

to

the

equivalence

relation

BPA

The

other

operators

are

defined

similarly

Definition

-LRB-

Basic

Terms

of

BPA

-RRB-

We

define

basic

terms

inductively

as

follows

-LRB-

i

-RRB-

Every

a

A

is

a

basic

term

-LRB-

ii

-RRB-

if

a

A

and

t

is

a

basic

term

then

a

t

is

a

basic

term

-LRB-

iii

-RRB-

if

t

and

s

are

basic

terms

then

t

s

is

a

basic

term

Definition

-LRB-

Number

of

Symbols

of

a

BPA

Term

-RRB-

We

define

n?x

the

number

of

symbols

of

x

inductively

as

follows

-LRB-

i

-RRB-

For

a

A

we

define

n?a

-LRB-

ii

-RRB-

for

closed

BPA

terms

x

and

y

we

define

n?x

y

n?x

y

n?x

n?y

Definition

-LRB-

Closed

Terms

of

BPA

-RRB-

We

denote

the

set

of

all

closed

terms

-LRB-

i.e.

terms

not

containing

free

variables

-RRB-

of

BPA

by

C?BPA

This

notation

extends

as

expected

to

other

theories

Remark

-LRB-

Results

from

-LSB-

-RSB-

-RRB-

In

-LSB-

-RSB-

several

results

are

given

about

elimination

soundness

and

completeness

with

respect

to

BPA

and

BPA

with

some

extensions

We

will

not

repeat

the

proofs

given

for

those

results

here

but

instead

refer

to

that

article

One

might

object

that

the

proofs

of

-LSB-

-RSB-

are

often

very

sketchy

-LRB-

Easy

by

induction

-RRB-

and

sometimes

even

incorrect

-LRB-

see

for

example

Remark

on

page

of

our

paper

-RRB-

Nevertheless

we

felt

that

no

useful

purpose

would

be

served

by

writing

out

those

sketchy

proofs

in

full

here

as

in

a

sense

they

are

encompassed

by

the

proofs

regarding

BPA?drt?ID

-LRB-

to

be

treated

in

Section

-RRB-

which

we

do

give

in

full

Proposition

-LRB-

Elimination

for

BPA

-RRB-

Let

t

be

a

closed

BPA

term

Then

there

is

a

basic

term

s

such

that

BPA

t

s.

Proof

This

is

Proposition

of

-LSB-

-RSB-

See

the

proof

given

there

Theorem

-LRB-

Soundness

of

BPA

-RRB-

The

set

of

closed

BPA

terms

modulo

bisimulation

equivalence

is

a

model

of

BPA

Proof

This

is

Theorem

of

-LSB-

-RSB-

See

the

proof

given

there

Remark

-LRB-

Proving

Completeness

using

the

Direct

Method

-RRB-

All

completeness

proofs

regarding

basic

process

algebras

-LRB-

i.e.

all

completeness

proofs

in

Section

-RRB-

follow

the

same

scheme

which

we

will

outline

in

this

remark

in

some

detail

so

we

do

not

have

to

go

over

these

details

again

and

again

in

the

actual

proofs

To

prove

completeness

of

process

theory

P

we

first

derive

an

auxiliary

lemma

Towards

Completeness

of

P

-LRB-

see

for

example

Lemma

on

the

following

page

-RRB-

that

contains

sublemmata

of

the

general

form

T?P

-RRB-

P

Typically

each

sublemma

relates

a

certain

transition

in

the

term

deduction

system

of

P

with

a

certain

equality

in

P

-LRB-

some

sublemmata

slightly

deviate

from

this

format

-RRB-

Armed

with

the

implications

proven

in

the

Towards

lemma

we

then

set

out

to

actually

prove

completeness

-LRB-

see

for

example

Theorem

on

page

-RRB-

This

is

done

using

Lemma

on

the

following

page

by

proving

that

for

all

basic

terms

x

and

y

of

P

we

have

that

x

y

P

y

-RRB-

P

x

y

y

This

part

of

the

proof

is

done

by

induction

on

the

number

of

symbols

in

x

using

case

distinction

on

the

form

of

-LRB-

basic

term

-RRB-

x

The

Towards

sublemmata

are

chosen

in

such

a

way

that

each

case

we

encounter

in

completing

our

proof

is

now

easily

handled

The

proof

method

outlined

in

this

remark

is

taken

from

-LSB-

-RSB-

Lemma

-LRB-

Towards

Completeness

of

BPA

-RRB-

Let

x

and

y

be

closed

BPA

terms

Then

we

have

-LRB-

i

-RRB-

T?BPA

x

a

p

-RRB-

BPA

x

a

x

-LRB-

ii

-RRB-

T?BPA

x

a

y

-RRB-

BPA

x

a

y

x

-LRB-

iii

-RRB-

T?BPA

x

a

y

-RRB-

n?x

n?y

Proof

For

part

-LRB-

i

-RRB-

and

-LRB-

ii

-RRB-

we

assume

by

Proposition

and

Theorem

without

loss

of

generality

that

x

is

a

basic

term

and

then

apply

induction

on

the

structure

of

basic

terms

For

part

-LRB-

iii

-RRB-

this

does

not

work

as

bisimulation

obviously

is

not

a

congruence

for

n?x

Therefore

in

proving

-LRB-

iii

-RRB-

we

use

induction

on

the

general

structure

of

terms

-LRB-

i

-RRB-

Case

x

is

an

atomic

action

Because

T?BPA

x

a

p

it

must

then

be

the

case

that

x

a

So

we

have

BPA

x

a

a?a

a?x

Case

x

is

of

the

form

atomic

action

followed

by

another

basic

term

This

is

in

contradiction

with

T?BPA

x

a

p

so

this

case

does

not

occur

Case

x

is

of

the

form

s

t

where

s

and

t

are

again

basic

terms

As

T?BPA

s

t

a

p

necessarily

T?BPA

s

a

p

or

T?BPA

t

a

p.

Therefore

by

the

induction

hypothesis

BPA

s

a

s

or

BPA

t

a

t

But

then

in

both

cases

BPA

x

s

t

a

s

t

a

x.

-LRB-

ii

-RRB-

Case

x

is

an

atomic

action

This

is

in

contradiction

with

T?BPA

x

a

y

so

this

case

does

not

occur

Case

x

is

of

the

form

atomic

action

followed

by

another

basic

term

Then

because

T?BPA

x

a

y

it

must

be

that

x

a

y.

So

BPA

x

a

y

a

y

a

y

a

y

x.

Case

x

is

of

the

form

s

t

where

s

and

t

are

again

basic

terms

As

T?BPA

s

t

a

y

necessarily

T?BPA

s

a

y

or

T?BPA

t

a

y.

Therefore

by

the

induction

hypothesis

BPA

s

a

y

s

or

BPA

t

a

y

t

So

in

both

cases

BPA

x

s

t

a

y

s

t

a

y

x.

-LRB-

iii

-RRB-

Case

x

is

an

atomic

action

This

is

in

contradiction

with

T?BPA

x

a

y

so

this

case

does

not

occur

Case

x

is

of

the

form

s

t

for

certain

terms

s

and

t.

Then

by

T?BPA

x

a

y

we

either

have

T?BPA

s

a

p

and

y

t

or

we

have

T?BPA

s

a

and

y

t

for

some

term

In

the

first

case

we

have

n?x

n?s?t

n?s

n?t

n?t

n?y

and

in

the

second

we

can

apply

the

induction

hypothesis

to

arrive

at

n?s

so

we

get

n?x

n?s

t

n?s

n?t

n?t

n?y

Case

x

is

of

the

form

s?t

for

certain

terms

s

and

t

As

T?BPA

s?t

a

y

necessarily

T?BPA

s

a

y

or

T?BPA

t

a

y.

Therefore

by

the

induction

hypothesis

n?s

n?y

or

n?t

n?y

As

n

ranges

over

the

positive

naturals

only

in

both

cases

n?x

n?s

t

n?s

n?t

n?y

Lemma

-LRB-

Towards

Completeness

of

BPA

-RRB-

In

order

to

prove

for

all

BPA

terms

x

and

y

that

x

BPA

y

-RRB-

BPA

x

y

-LRB-

-RRB-

it

is

sufficient

to

prove

that

x

y

BPA

y

-RRB-

BPA

x

y

y

-LRB-

y

-RRB-

Proof

Assume

-LRB-

y

-RRB-

and

the

left-hand

side

of

-LRB-

-RRB-

to

hold

Then

prove

the

right-hand

side

of

-LRB-

-RRB-

This

is

done

as

follows

by

Theorem

the

fact

that

BPA

is

a

congruence

and

Axiom

for

all

bisimilar

BPA

terms

x

and

y

we

have

x

y

BPA

y

y

BPA

y

and

y

x

BPA

x

x

BPA

x.

Therefore

by

-LRB-

y

-RRB-

also

BPA

x

y

y

and

BPA

y

x

x

But

that

gives

us

BPA

x

y

x

x

y

y.

Corollary

-LRB-

Generalization

of

Lemma

-RRB-

Lemma

generalizes

from

BPA

to

any

equational

process

theory

P

with

corresponding

term

deduction

system

T?P

provided

-LRB-

i

-RRB-

P

is

a

sound

axiomatization

of

T?P

-LRB-

ii

-RRB-

P

is

a

congruence

for

the

function

symbols

from

the

signature

of

P

and

-LRB-

iii

-RRB-

P

contains

the

axioms

of

BPA

Proof

As

the

proof

of

Lemma

only

depends

on

the

soundness

of

BPA

with

respect

to

T?BPA

the

fact

that

P

is

a

congruence

for

P

and

the

axioms

of

BPA

the

proof

is

trivially

valid

for

P

too

Theorem

-LRB-

Completeness

of

BPA

-RRB-

The

axiom

system

BPA

is

a

complete

axiomatization

of

the

set

of

closed

BPA

terms

modulo

bisimulation

equivalence

Proof

Let

x

and

y

be

bisimilar

closed

BPA

terms

We

have

to

prove

that

BPA

x

y.

With

the

aid

of

Proposition

and

Theorem

it

is

enough

to

prove

this

for

basic

terms

By

Lemma

it

is

even

enough

to

prove

for

all

basic

terms

x

and

y

that

x

y

BPA

y

-RRB-

BPA

x

y

y

We

will

prove

this

by

induction

on

n?x

using

-LRB-

iii

-RRB-

and

case

distinction

on

the

form

of

basic

term

x.

Case

x

is

of

the

form

a

for

a

A

Then

T?BPA

x

a

p

so

T?BPA

x

y

a

p

and

because

x

y

BPA

y

we

have

T?BPA

y

a

p

so

with

Lemma

-LRB-

i

-RRB-

we

find

that

BPA

x?y

y

This

proves

the

basis

of

our

induction

Case

x

is

of

the

form

a

s

where

a

A

and

s

again

a

basic

BPA

term

Then

T?BPA

x

a

s

and

therefore

T?BPA

x

y

a

s

so

because

x

y

BPA

y

there

is

an

with

T?BPA

y

a

and

s

BPA

But

then

by

Theorem

and

Axiom

also

BPA

and

s

BPA

s

and

with

induction

-LRB-

note

that

n?s

n?x

-RRB-

we

find

BPA

s

and

BPA

s

s

So

BPA

s

Now

BPA

x

y

a

s

y

a

y

y

with

Lemma

-LRB-

ii

-RRB-

Case

x

is

of

the

form

s

t

for

certain

basic

BPA

terms

s

and

t

Since

x

y

BPA

y

we

also

have

s

y

BPA

y

and

t

y

BPA

y.

By

induction

BPA

s

y

y

and

BPA

t

y

y

So

BPA

x

y

s

t

y

s

y

y.

Soundness

and

Completeness

of

BPAffi

Definition

-LRB-

Signature

of

BPAffi

-RRB-

The

signature

of

BPAffi

consists

of

the

atomic

actions

faja

Ag

the

deadlock

constant

ffi

the

alternative

composition

operator

and

the

sequential

composition

operator

x

ffi

x

ffi

x

ffi

Table

Axioms

for

ffi

Definition

-LRB-

Axioms

of

BPAffi

-RRB-

The

process

algebra

BPAffi

is

axiomatized

by

the

axioms

of

BPA

given

in

Definition

on

page

and

Axioms

shown

in

Table

BPAffi

Remark

-LRB-

Axiom

versus

Axiom

-RRB-

Note

that

in

the

presence

of

the

other

axioms

of

BPA

Axiom

given

in

Table

is

equivalent

for

closed

BPAffi

terms

with

Axiom

given

in

Table

Therefore

we

could

replace

in

BPA

by

without

affecting

the

soundness

or

completeness

of

the

resulting

theory

One

such

reason

to

do

so

could

be

the

fact

that

remains

valid

in

all

discretetime

process

algebras

we

will

describe

whereas

does

not

Still

for

historical

reasons

we

prefer

to

be

used

in

the

definition

of

BPAffi

We

will

later

return

to

this

subject

in

Remark

on

page

and

Remark

on

page

a

ffi

a

Table

Alternative

for

Axiom

Definition

-LRB-

Semantics

of

BPAffi

-RRB-

The

semantics

of

BPAffi

are

given

by

the

term

deduction

system

T?BPAffi

induced

by

the

deduction

rules

given

in

Table

on

page

and

Table

on

page

Note

that

this

term

deduction

system

T?BPAffi

is

practically

identical

to

the

term

deduction

system

T?BPA

given

in

Definition

on

page

as

there

are

no

deduction

rules

for

ffi

However

T?BPAffi

does

differ

from

T?BPA

in

the

fact

that

it

contains

the

symbol

ffi

in

its

signature

Definition

-LRB-

Bisimulation

and

Bisimulation

Model

for

BPAffi

-RRB-

Bisimulation

for

BPAffi

and

the

corresponding

bisimulation

model

are

defined

in

the

same

way

as

for

BPA

Replace

BPA

by

BPAffi

in

Definition

on

page

and

Definition

on

page

Definition

-LRB-

Basic

Terms

of

BPAffi

-RRB-

We

define

ffi-basic

terms

inductively

as

follows

-LRB-

i

-RRB-

Every

a

Affi

is

a

ffi-basic

term

-LRB-

ii

-RRB-

if

a

Affi

and

t

is

a

ffi-basic

term

then

a

t

is

a

ffi-basic

term

-LRB-

iii

-RRB-

if

t

and

s

are

ffi-basic

terms

then

t

s

is

a

ffi-basic

term

From

now

on

if

we

speak

of

basic

terms

in

the

context

of

BPAffi

we

mean

ffi-basic

terms

Remark

-LRB-

Definition

of

Basic

Terms

-RRB-

Usually

the

basic

terms

of

BPAffi

are

defined

a

bit

differently

with

respect

to

deadlock

ffi

t

for

some

basic

term

t

is

usually

not

considered

basic

We

chose

to

deviate

from

established

practice

because

it

made

our

proofs

quite

a

bit

shorter

The

reason

is

that

in

this

way

we

get

rid

of

a

nasty

case

distinction

that

would

otherwise

have

popped

up

in

just

about

every

other

line

of

our

proofs

Definition

-LRB-

Number

of

Symbols

of

a

BPAffi

Term

-RRB-

We

define

n?x

the

number

of

symbols

of

x

inductively

as

follows

-LRB-

i

-RRB-

For

a

Affi

we

define

n?a

-LRB-

ii

-RRB-

for

closed

BPAffi

terms

x

and

y

we

define

n?x

y

n?x

y

n?x

n?y

Proposition

-LRB-

Elimination

for

BPAffi

-RRB-

Let

t

be

a

closed

BPAffi

term

Then

there

is

a

basic

term

s

such

that

BPAffi

t

s.

Proof

This

is

Proposition

of

-LSB-

-RSB-

see

the

proof

given

there

Note

that

although

-LSB-

-RSB-

uses

a

slightly

different

definition

of

basic

terms

their

definition

is

narrower

so

their

elimination

result

also

holds

for

our

definition

of

basic

terms

Theorem

-LRB-

Soundness

of

BPAffi

-RRB-

The

set

of

closed

BPAffi

terms

modulo

bisimulation

equivalence

is

a

model

of

BPAffi

Proof

This

is

Theorem

of

-LSB-

-RSB-

See

the

proof

given

there

Theorem

-LRB-

Completeness

of

BPAffi

-RRB-

The

axiom

system

BPAffi

is

a

complete

axiomatization

of

the

set

of

closed

BPAffi

terms

modulo

bisimulation

equivalence

Proof

Since

there

are

no

transitions

for

the

new

constant

ffi

this

is

proven

in

the

same

way

as

Theorem

Soundness

and

Completeness

of

BPA?drt?ffi

Definition

-LRB-

Signature

of

BPA?drt?ffi

-RRB-

The

signature

of

BPA?drt?ffi

consists

of

the

undelayable

atomic

actions

faja

Ag

the

alternative

composition

operator

the

sequential

composition

operator

and

the

time

unit

delay

operator

oerel

Definition

-LRB-

Axioms

of

BPA?drt?ffi

-RRB-

The

process

algebra

BPA?drt?ffi

is

axiomatized

by

the

axioms

of

BPA

given

in

Definition

on

page

and

Axioms

shown

in

Table

on

the

next

page

BPA?drt?ffi

oerel?x

oerel?y

oerel?x

y

oerel?x

y

oerel?x

y

Table

Axioms

for

oerel

Definition

-LRB-

Notation

Regarding

Semantics

II

-RRB-

Next

to

the

deduction

rule

notations

x

a

x

a

p

and

x

and

x

introduced

in

Definition

on

page

we

now

also

use

x

oe

to

denote

that

x

can

do

a

oe-step

to

-LRB-

i.e.

move

to

the

following

time-slice

and

become

-RRB-

and

x

to

denote

that

x

can

not

do

an

oe-step

Definition

-LRB-

Semantics

of

BPA?drt?ffi

-RRB-

The

semantics

of

BPA?drt?ffi

are

given

by

the

term

deduction

system

T?BPA?drt?ffi

induced

by

the

deduction

rules

given

in

Table

on

page

and

Table

a

a

p

oerel?x

oe

x

x

oe

x

y

oe

y

x

oe

y

oe

x

y

oe

x

oe

y

x

y

oe

x

y

oe

x

y

oe

Table

Deduction

rules

for

a

and

oerel

Definition

-LRB-

Bisimulation

for

BPA?drt?ffi

-RRB-

Bisimulation

for

BPA?drt?ffi

is

defined

as

follows

a

binary

relation

R

on

closed

BPA?drt?ffi

terms

is

a

bisimulation

if

the

following

transfer

conditions

hold

for

all

closed

BPA?drt?ffi

terms

p

and

q

-LRB-

i

-RRB-

If

RS?p

q

and

T?BPA?drt?ffi

p

a

where

a

A

then

there

exists

a

closed

term

such

that

T?BPA?drt?ffi

q

a

and

-LRB-

ii

-RRB-

If

RS?p

q

and

T?BPA?drt?ffi

p

oe

then

there

exists

a

closed

term

such

that

T?BPA?drt?ffi

q

oe

and

-LRB-

iii

-RRB-

If

RS?p

q

and

T?BPA?drt?ffi

p

a

p

where

a

A

then

T?BPA?drt?ffi

q

a

p.

Two

BPA?drt?ffi

terms

p

and

q

are

bisimilar

notation

p

BPA?drt?ffi

q

if

there

exists

a

bisimulation

relation

R

such

that

R?p

q

Definition

-LRB-

Bisimulation

Model

for

BPA?drt?ffi

-RRB-

The

bisimulation

model

for

BPA?drt?ffi

is

defined

in

the

same

way

as

for

BPA

Replace

BPA

by

BPA?drt?ffi

in

Definition

on

page

Definition

-LRB-

Basic

Terms

of

BPA?drt?ffi

-RRB-

We

define

oe-basic

terms

inductively

as

follows

-LRB-

i

-RRB-

For

every

a

A

the

atomic

action

a

is

a

oe-basic

term

-LRB-

ii

-RRB-

if

a

A

and

t

is

a

oe-basic

term

then

a

t

is

a

oe-basic

term

-LRB-

iii

-RRB-

if

t

and

s

are

oe-basic

terms

then

t

s

is

a

oe-basic

term

-LRB-

iv

-RRB-

if

t

is

a

basic

term

then

oerel?t

is

a

oe-basic

term

From

now

on

if

we

speak

of

basic

terms

in

the

context

of

BPA?drt?ffi

we

mean

oe-basic

terms

Definition

-LRB-

Number

of

Symbols

of

a

BPA?drt?ffi

Term

-RRB-

We

define

n?x

the

number

of

symbols

of

x

inductively

as

follows

-LRB-

i

-RRB-

For

a

A

we

define

n?a

-LRB-

ii

-RRB-

for

closed

BPA?drt?ffi

terms

x

and

y

we

define

n?x?y

n?x

y

n?x

n?y

-LRB-

iii

-RRB-

for

a

closed

BPA?drt?ffi

term

x

we

define

n?oerel?x

n?x

Definition

-LRB-

Symbol

for

a

Chain

of

oe

's

-RRB-

We

will

write

x

oe

-RRB-

y

to

indicate

that

x

can

reach

y

by

doing

zero

or

more

oe-transitions

Formally

oe

-RRB-

denotes

the

transitive

reflexive

closure

of

oe

Remark

-LRB-

Proving

Elimination

using

Term

Rewriting

Analysis

-RRB-

To

prove

elimination

we

will

use

where

possible

a

method

that

is

based

on

term

rewriting

analysis

This

method

works

by

associating

a

term

rewriting

system

to

an

equational

specification

and

then

proving

that

this

term

rewriting

system

is

strongly

normalizing

and

its

normal

forms

are

basic

terms

See

Theorem

for

an

example

of

this

method

Theorem

-LRB-

Elimination

for

BPA?drt?ffi

-RRB-

Let

t

be

a

closed

BPA?drt?ffi

term

Then

there

is

a

basic

term

s

such

that

BPA?drt?ffi

s

t.

Proof

This

theorem

is

proven

as

follows

First

a

number

of

axioms

of

BPA?drt?ffi

are

selected

and

subsequently

oriented

as

rewriting

rules

This

gives

us

a

term

rewriting

system

Then

it

is

proven

that

this

term

rewriting

system

is

strongly

normalizing

and

that

every

normal

form

of

a

closed

BPA?drt?ffi

term

is

a

basic

term

In

this

way

a

recipe

is

obtained

for

transforming

a

closed

BPA?drt?ffi

term

into

a

basic

term

The

rewriting

rules

of

the

term

rewriting

system

for

BPA?drt?ffi

are

given

in

Table

on

the

next

page

The

proof

that

this

term

rewriting

system

is

strongly

normalizing

uses

the

method

of

the

lexicographical

path

ordering

The

well-founded

ordering

on

constants

and

function

symbols

is

the

following

oerel

a

To

we

assign

the

lexicographical

status

for

the

first

argument

Now

we

show

that

the

left-hand

side

of

every

rewriting

rule

is

bigger

than

the

right-hand

side

with

respect

to

the

ordering

Olpo

This

is

done

by

the

following

reductions

-LRB-

taken

from

-LSB-

-RSB-

-RRB-

x

y

z

Olpo

x

y

z

Olpo

x

y

z

x

y

z

Olpo

x

y

z

x

y

z

Olpo

x

z

y

z

x

y

z

x

z

y

z

x

y

z

x

y

z

oerel?x

y

oerel?x

y

Table

Term

Rewriting

System

for

BPA?drt?ffi

x

y

z

Olpo

x

y

z

Olpo

x

y

x

y

z

Olpo

x

x

y

z

Olpo

x

y

z

oerel?x

y

Olpo

oerel?x

y

Olpo

oerel?oerel?x

y

Olpo

oerel?oe?rel?x

y

Olpo

oerel?x

y

Next

we

will

prove

that

the

normal

forms

of

the

closed

BPA?drt?ffi

terms

are

basic

terms

Thereto

suppose

that

s

is

a

normal

form

of

some

closed

BPA?drt?ffi

term

Furthermore

suppose

that

s

is

not

a

basic

term

Let

denote

the

smallest

subterm

of

s

which

is

not

a

basic

term

Note

that

consequently

all

proper

subterms

of

are

basic

terms

Then

we

can

prove

that

is

not

a

normal

form

by

case

analysis

We

distinguish

all

possible

cases

-LRB-

i

-RRB-

is

an

atomic

action

But

then

is

a

basic

term

This

is

in

contradiction

with

the

assumption

that

is

not

a

basic

term

so

this

case

does

not

occur

-LRB-

ii

-RRB-

is

of

the

form

for

basic

terms

and

With

case

analysis

on

the

structure

of

basic

term

-LRB-

a

-RRB-

If

is

an

atomic

action

a

then

is

a

basic

term

and

so

is

a

basic

term

which

again

contradicts

the

assumption

that

is

not

a

basic

term

This

case

can

therefore

not

occur

-LRB-

b

-RRB-

If

is

of

the

form

a?t

for

some

atomic

action

a

and

basic

term

t

then

rewriting

rule

can

be

applied

So

is

not

a

normal

form

-LRB-

c

-RRB-

If

is

of

the

form

for

and

basic

terms

Then

rewriting

rule

is

applicable

Therefore

is

not

a

normal

form

-LRB-

d

-RRB-

If

is

of

the

form

oerel?t

for

some

basic

term

t

Then

rewriting

rule

is

applicable

So

is

not

a

normal

form

-LRB-

iii

-RRB-

is

of

the

form

for

basic

terms

and

In

this

case

would

be

a

basic

term

which

contradicts

the

assumption

that

is

not

a

basic

term

Therefore

this

case

can

not

occur

-LRB-

iv

-RRB-

is

of

the

form

for

some

basic

term

But

then

is

basic

term

too

so

the

case

does

not

occur

In

any

case

that

can

occur

it

follows

that

is

not

a

normal

form

Since

is

a

subterm

of

s

we

conclude

that

s

is

not

a

normal

form

This

contradicts

the

assumption

that

s

is

a

normal

form

From

this

contradiction

we

conclude

that

s

is

a

basic

term

which

completes

the

proof

Remark

-LRB-

Elimination

for

BPA?drt?ffi

-RRB-

Elimination

for

BPA?drt?ffi

is

also

claimed

-LRB-

without

proof

-RRB-

in

Theorem

of

-LSB-

-RSB-

-LRB-

where

BPA?drt?ffi

is

called

BPAdt

-RRB-

Remark

-LRB-

Proving

Soundness

using

the

Direct

Method

-RRB-

Most

of

the

soundness

proofs

given

in

this

paper

follow

the

same

scheme

which

we

will

outline

in

this

remark

in

some

detail

so

we

do

not

have

to

go

over

these

details

again

and

again

in

the

actual

proofs

To

prove

the

soundness

of

a

certain

axiom

of

the

general

form

xn

xn

where

tl

and

tr

are

process

expressions

in

the

free

variables

xn

for

some

n

with

respect

to

some

bisimulation

model

we

proceed

as

follows

First

we

give

a

relation

R

which

will

be

a

binary

relation

on

closed

terms

Then

we

show

that

this

R

is

a

bisimulation

relation

that

for

all

closed

instantiations

of

xn

relates

the

left-hand

and

right-hand

side

of

the

axiom

This

involves

two

steps

-LRB-

i

-RRB-

R

should

relate

both

sides

of

the

axiom

for

all

closed

terms

i.e.

for

all

closed

instantiations

of

xn

we

should

have

that

xn

xn

R

This

is

mostly

so

trivial

that

we

do

not

mention

it

at

all

-LRB-

ii

-RRB-

R

should

be

a

bisimulation

In

order

to

prove

that

we

show

that

for

all

closed

terms

s

t

in

the

relation

the

transfer

conditions

from

the

definition

of

bisimulation

are

satisfied

For

example

in

proving

soundness

of

an

axiom

of

BPA?drt?ffi

we

have

to

show

that

for

all

closed

term

s

t

such

that

s

t

R

we

have

that

for

any

transition

s

u

-LRB-

where

u

Aoe

-RRB-

there

is

a

corresponding

transition

t

u

such

that

R

and

vice

versa

for

any

transition

t

u

there

is

a

corresponding

transition

s

u

such

that

again

R

This

part

of

the

proof

is

done

using

case

distinction

on

the

different

kinds

of

steps

that

are

possible

-LRB-

an

action

a

time

step

termination

-RRB-

Note

that

the

vice

versa

part

of

proof

obligation

-LRB-

ii

-RRB-

results

from

the

fact

that

the

transfer

conditions

for

bisimulation

-LRB-

see

for

example

Definition

on

page

-RRB-

are

defined

with

respect

to

the

symmetric

closure

of

R

This

completes

the

general

outline

of

our

soundness

proofs

In

the

actual

proofs

we

will

sometimes

slightly

deviate

from

it

for

example

because

we

re-use

an

earlier

proof

or

because

we

do

not

faithfully

list

the

too

trivial

proof

obligations

The

above

described

general

outline

will

however

remain

visible

in

the

background

Finally

note

that

in

soundness

proofs

we

do

not

explicitly

indicate

the

term

reduction

system

in

which

a

transition

occurs

so

instead

of

T?BPA?drt?ffi

x

a

y

we

simply

write

x

a

y

This

is

no

problem

as

the

theory

in

which

we

are

working

is

always

clear

The

proof

method

outlined

in

this

remark

is

taken

from

-LSB-

-RSB-

Theorem

-LRB-

Soundness

of

BPA?drt?ffi

-RRB-

The

set

of

closed

BPA?drt?ffi

terms

modulo

bisimulation

equivalence

is

a

model

of

BPA?drt?ffi

Proof

Since

bisimulation

equivalence

is

a

congruence

-LRB-

see

for

example

Theorem

of

-LSB-

-RSB-

-RRB-

also

for

the

new

operators

we

only

need

to

verify

the

soundness

of

every

closed

instantiation

of

the

axioms

We

do

this

by

giving

a

relation

R

for

every

axiom

and

we

prove

that

this

relation

is

a

bisimulation

relation

for

every

closed

instantiation

of

the

left-hand

and

right-hand

sides

of

the

axiom

In

the

setting

of

BPA

soundness

of

Axioms

has

already

been

proven

-LRB-

see

for

example

Theorem

of

-LSB-

-RSB-

-RRB-

The

theory

BPA?drt?ffi

adds

to

this

the

possibility

to

perform

oe-transitions

However

any

term

headed

by

the

operator

added

to

BPA

is

not

capable

of

performing

a

transition

labeled

by

an

atomic

action

So

we

argue

that

the

left-hand

side

and

the

right-hand

side

of

Axioms

can

perform

exactly

the

same

transitions

labeled

by

an

atomic

action

in

the

theory

BPA?drt?ffi

as

in

the

theory

BPA

Therefore

we

only

consider

the

transitions

labeled

by

oe

for

Axioms

For

Axioms

we

of

course

have

to

consider

both

the

transitions

labeled

by

an

atomic

action

and

the

transitions

labeled

by

a

oe

Finally

note

that

BPA?drt?ffi

contains

atomic

actions

of

the

form

a

whereas

BPA

had

actions

of

the

form

a

This

is

however

not

relevant

for

the

purpose

of

extending

the

soundness

proofs

of

Axioms

from

BPA

to

BPA?drt?ffi

as

neither

a

nor

a

appears

in

Axioms

and

both

have

exactly

the

same

deduction

rules

-LRB-

see

Table

on

page

and

Table

on

page

-RRB-

Axiom

Take

the

relation

R

s

s

s

t

t

s

s

t

C?BPA?drt?ffi

First

we

look

at

the

transitions

of

the

left-hand

side

Suppose

s

t

oe

p

Then

one

of

the

following

situations

occurs

-LRB-

i

-RRB-

s

oe

p

and

t

then

also

t

s

oe

p

and

note

that

p

p

R.

-LRB-

ii

-RRB-

s

and

t

oe

p

then

also

t

s

oe

p

and

note

that

p

p

R.

-LRB-

iii

-RRB-

s

oe

and

t

oe

and

p

then

t

s

oe

and

note

that

R

The

proof

for

the

right-hand

side

is

analogous

Axiom

Take

the

relation

R

s

s

s

t

u

s

t

u

s

t

u

C?BPA?drt?ffi

First

we

look

at

the

transitions

of

the

left-hand

side

Suppose

s

t

u

oe

p

Then

one

of

the

following

situations

occurs

-LRB-

i

-RRB-

s

t

oe

p

and

u

then

the

transition

s

t

oe

p

must

be

due

to

one

of

the

following

-LRB-

a

-RRB-

s

oe

p

and

t

in

that

case

s

oe

p

and

t

u

So

s

t

u

oe

p

and

note

that

p

p

R.

-LRB-

b

-RRB-

s

and

t

oe

p

in

that

case

s

and

t

u

oe

p.

Therefore

s

t

u

oe

p.

-LRB-

c

-RRB-

s

oe

and

t

oe

and

p

in

that

case

s

oe

and

t?u

oe

Therefore

s

t

u

oe

and

note

that

p

p

R.

-LRB-

ii

-RRB-

s

t

and

u

oe

p

then

s

t

and

u

oe

p.

So

s

and

t

u

oe

p.

Therefore

s

t

u

oe

p

and

note

that

p

p

R.

-LRB-

iii

-RRB-

s

t

oe

and

u

oe

and

p

then

the

transition

s

t

oe

must

be

due

to

one

of

the

following

-LRB-

a

-RRB-

s

oe

and

t

in

that

case

s

oe

and

t

u

oe

So

s

t?u

oe

and

note

that

p

p

R.

-LRB-

b

-RRB-

s

and

t

oe

in

that

case

s

and

t

u

oe

Therefore

s

t

u

oe

and

note

that

p

p

R.

-LRB-

c

-RRB-

s

oe

and

t

oe

and

in

that

case

s

oe

and

t

u

oe

Therefore

s

t

u

oe

and

note

that

R

The

proof

for

the

right-hand

side

is

analogous

Axiom

Take

the

relation

R

s

s

s

s

s

s

C?BPA?drt?ffi

First

we

look

at

the

transitions

of

the

left-hand

side

Suppose

s?s

oe

p

Then

s

oe

and

p

Then

s

oe

and

note

that

R

The

proof

for

the

right-hand

side

is

analogous

Axiom

Take

the

relation

R

s

s

s

t

u

s

u

t

u

s

t

u

C?BPA?drt?ffi

First

we

look

at

the

transitions

of

the

left-hand

side

Suppose

s

t

u

oe

p

Then

s?t

oe

and

p

The

transition

s?t

oe

must

be

due

to

one

of

the

following

-LRB-

i

-RRB-

s

oe

and

t

then

s

u

t

u

oe

u

and

note

that

p

p

R.

-LRB-

ii

-RRB-

s

and

t

oe

analogous

to

the

previous

case

-LRB-

iii

-RRB-

s

oe

and

t

oe

and

then

s

u

t

u

oe

u

and

note

that

u

u

u

R.

Secondly

we

look

at

the

transitions

of

the

right-hand

side

Suppose

s

u

t

u

oe

p

This

must

be

due

to

one

of

the

following

-LRB-

i

-RRB-

s?u

oe

p

and

t?u

then

s

oe

and

p

Also

t

Therefore

s?t

oe

and

s

t

u

oe

u

and

note

that

p

p

R.

-LRB-

ii

-RRB-

s

u

and

t

u

oe

p

analogous

to

the

previous

case

-LRB-

iii

-RRB-

s

u

oe

and

t

u

oe

and

p

then

s

oe

and

u

and

t

oe

and

u.

Therefore

s

t

oe

and

s

t

u

oe

u

and

note

that

u

u

u

R.

Axiom

Take

the

relation

R

s

s

s

t

u

s

t

u

s

t

u

C?BPA?drt?ffi

First

we

look

at

the

transitions

of

the

left-hand

side

Suppose

s

t

u

oe

p

Then

this

must

be

due

to

s

oe

and

p

t

u.

So

s

t

u

oe

t

u

and

note

that

t

u

t

u

R

The

proof

for

the

right-hand

side

is

analogous

Axiom

Take

the

relation

R

s

s

oerel?s

oerel?t

oerel?s

t

s

t

C?BPA?drt?ffi

We

look

at

the

transitions

of

both

sides

at

the

same

time

First

note

that

oerel?s

oerel?t

and

oerel?s

t

Secondly

oerel?s

oerel?t

oe

p

iff

p

s

t

iff

oerel?s

t

oe

p

and

note

that

p

p

R.

Axiom

Take

the

relation

R

s

s

oerel?s

t

oerel?s

t

s

t

C?BPA?drt?ffi

We

look

at

the

transitions

of

both

sides

at

the

same

time

First

note

that

oerel?s

t

and

oerel?s

t

Secondly

oerel?s

t

oe

p

iff

p

s

t

iff

oerel?s

t

oe

p

and

note

that

p

p

R.

Remark

-LRB-

Soundness

of

BPA?drt?ffi

-RRB-

Soundness

of

BPA?drt?ffi

is

also

claimed

-LRB-

without

proof

-RRB-

in

Theorem

of

-LSB-

-RSB-

-LRB-

where

BPA?drt?ffi

is

called

BPAdt

-RRB-

Lemma

-LRB-

Towards

Completeness

of

BPA?drt?ffi

-RRB-

Let

x

and

y

be

closed

BPA?drt?ffi

terms

Then

we

have

-LRB-

i

-RRB-

T?BPA?drt?ffi

x

a

p

-RRB-

BPA?drt?ffi

x

a

x

-LRB-

ii

-RRB-

T?BPA?drt?ffi

x

a

y

-RRB-

BPA?drt?ffi

x

a

y

x

-LRB-

iii

-RRB-

T?BPA?drt?ffi

x

oe

y

-RRB-

BPA?drt?ffi

x

oerel?y

x

-LRB-

iv

-RRB-

T?BPA?drt?ffi

x

a

y

-RRB-

n?x

n?y

-LRB-

v

-RRB-

T?BPA?drt?ffi

x

oe

y

-RRB-

n?x

n?y

Proof

For

part

-LRB-

i

-RRB-

-LRB-

ii

-RRB-

and

-LRB-

iii

-RRB-

we

assume

by

Theorem

and

Theorem

without

loss

of

generality

that

x

is

a

basic

term

and

apply

induction

on

the

structure

of

basic

terms

For

part

-LRB-

iv

-RRB-

and

-LRB-

v

-RRB-

we

again

have

to

use

induction

on

the

general

structure

of

terms

-LRB-

i

-RRB-

Case

x

is

an

atomic

action

Because

T?BPA?drt?ffi

x

a

p

it

must

then

be

the

case

that

x

a

So

we

have

BPA?drt?ffi

x

a

a

a

a

x.

Case

x

is

of

the

form

atomic

action

followed

by

another

basic

term

This

is

in

contradiction

with

T?BPA?drt?ffi

x

a

p

so

this

case

does

not

occur

Case

x

is

of

the

form

s

t

where

s

and

t

are

again

basic

terms

As

T?BPA?drt?ffi

s

t

a

p

necessarily

T?BPA?drt?ffi

s

a

p

or

T?BPA?drt?ffi

t

a

p.

Therefore

by

the

induction

hypothesis

BPA?drt?ffi

s

a

s

or

BPA?drt?ffi

t

a

t

But

then

in

both

cases

BPA?drt?ffi

x

s

t

a

s

t

a

x.

Case

x

is

of

the

form

oerel?s

for

some

basic

term

s

As

we

know

that

T?BPA?drt?ffi

x

a

p

this

case

can

not

occur

-LRB-

ii

-RRB-

Case

x

is

an

atomic

action

This

is

in

contradiction

with

T?BPA?drt?ffi

x

a

y

so

this

case

does

not

occur

Case

x

is

of

the

form

atomic

action

followed

by

another

basic

term

Then

because

T?BPA?drt?ffi

x

a

y

it

must

be

that

x

a?y

So

BPA?drt?ffi

x

a?y

a?y?a?y

a?y?x

Case

x

is

of

the

form

s?t

where

s

and

t

are

again

basic

terms

As

T?BPA?drt?ffi

s

t

a

y

necessarily

T?BPA?drt?ffi

s

a

y

or

T?BPA?drt?ffi

t

a

y.

Therefore

by

the

induction

hypothesis

BPA?drt?ffi

s

a

y

s

or

BPA?drt?ffi

t

a

y

t

So

in

both

cases

BPA?drt?ffi

x

s

t

a

y

s

t

a

y

x.

Case

x

is

of

the

form

oerel?s

for

some

basic

term

s

As

we

know

that

T?BPA?drt?ffi

x

a

y

this

case

can

not

occur

-LRB-

iii

-RRB-

Case

x

is

an

atomic

action

This

is

in

contradiction

with

T?BPA?drt?ffi

x

oe

y

so

this

case

does

not

occur

Case

x

is

of

the

form

atomic

action

followed

by

another

basic

term

For

the

same

reason

this

case

can

not

occur

either

Case

x

is

of

the

form

s?t

where

s

and

t

are

again

basic

terms

As

T?BPA?drt?ffi

x

oe

y

weknow

that

either

T?BPA?drt?ffi

s

oe

y

or

T?BPA?drt?ffi

t

oe

y

or

both

So

by

the

induction

hypothesis

either

BPA?drt?ffi

t

oerel?y

t

or

BPA?drt?ffi

s

oerel?y

s

or

both

So

in

all

cases

BPA?drt?ffi

x

s

t

oerel?y

s

t

oerel?y

x.

Case

x

is

of

the

form

oerel?s

for

some

basic

term

s

Then

necessarily

s

y.

So

BPA?drt?ffi

x

x

x

oerel?y

x

oerel?s

x.

-LRB-

iv

-RRB-

Case

x

is

an

atomic

action

This

is

in

contradiction

with

T?BPA?drt?ffi

x

a

y

so

this

case

does

not

occur

Case

x

is

of

the

form

s

t

for

certain

terms

s

and

t.

Then

by

T?BPA?drt?ffi

x

a

y

we

either

have

T?BPA?drt?ffi

s

a

p

and

y

t

or

we

have

T?BPA?drt?ffi

s

a

and

y

t

for

some

term

In

the

first

case

we

have

n?x

n?s

t

n?s

n?t

n?t

n?y

and

in

the

second

we

can

apply

the

induction

hypothesis

to

arrive

at

n?s

so

we

get

n?x

n?s?t

n?s

n?t

n?t

t

n?y

Case

x

is

of

the

form

s?t

for

certain

terms

s

and

t

As

T?BPA?drt?ffi

s

t

a

y

necessarily

T?BPA?drt?ffi

s

a

y

or

T?BPA?drt?ffi

t

a

y.

Therefore

by

the

induction

hypothesis

n?s

n?y

or

n?t

n?y

As

n

ranges

over

the

positive

naturals

only

in

both

cases

n?x

n?s

t

n?s

n?t

n?y

Case

x

oerel?s

for

a

certain

term

s

does

not

occur

as

T?BPA?drt?ffi

oerel?s

-LRB-

v

-RRB-

Case

x

is

an

atomic

action

This

is

in

contradiction

with

T?BPA?drt?ffi

x

oe

y

so

this

case

does

not

occur

Case

x

is

of

the

form

s

t

for

certain

terms

s

and

t.

Then

necessarily

T?BPA?drt?ffi

s

oe

and

y

t

for

some

term

We

now

can

apply

the

induction

hypothesis

to

arrive

at

n?s

so

we

get

n?x

n?s

t

n?s

n?t

n?t

t

n?y

Case

x

is

of

the

form

s

t

for

certain

terms

s

and

t.

Now

by

T?BPA?drt?ffi

x

oe

y

we

know

that

either

T?BPA?drt?ffi

s

oe

y

or

T?BPA?drt?ffi

t

oe

y

or

both

So

by

the

induction

hypothesis

either

n?s

n?y

or

n?t

n?y

or

both

So

in

all

cases

n?x

n?s

t

n?s

n?t

n?y

Case

if

x

is

of

the

form

oerel?s

for

a

certain

term

s

it

must

be

the

case

that

s

y.

So

n?x

n?oerel?s

n?oerel?y

n?y

n?y

Theorem

-LRB-

Completeness

of

BPA?drt?ffi

-RRB-

The

axiom

system

BPA?drt?ffi

is

a

complete

axiomatization

of

the

set

of

closed

BPA?drt?ffi

terms

modulo

bisimulation

equivalence

Proof

This

proof

is

almost

identical

to

the

one

of

Theorem

with

the

exception

of

the

fourth

case

Let

x

and

y

be

bisimilar

closed

BPA?drt?ffi

terms

We

have

to

prove

that

BPA?drt?ffi

x

y.

With

the

aid

of

Theorem

and

Theorem

it

is

enough

to

prove

this

for

basic

terms

By

Corollary

it

is

even

enough

to

prove

for

all

basic

terms

x

and

y

that

x

y

BPA?drt?ffi

y

-RRB-

BPA?drt?ffi

x

y

y

We

will

prove

this

by

induction

on

n?x

using

Lemma

-LRB-

iv

-RRB-

-LRB-

v

-RRB-

and

case

distinction

on

the

form

of

basic

term

x.

Case

x

is

of

the

form

a

for

a

A

Then

T?BPA?drt?ffi

x

a

p

so

T?BPA?drt?ffi

x

y

a

p

and

because

x

y

BPA?drt?ffi

y

we

have

T?BPA?drt?ffi

y

a

p

so

with

Lemma

-LRB-

i

-RRB-

we

find

that

BPA?drt?ffi

x?y

y

This

proves

the

basis

of

our

induction

Case

x

is

of

the

form

a

s

when

s

is

again

a

basic

BPA?drt?ffi

term

Then

T?BPA?drt?ffi

x

a

s

and

therefore

T?BPA?drt?ffi

x

y

a

s

so

because

x

y

BPA?drt?ffi

y

there

is

an

with

T?BPA?drt?ffi

y

a

and

s

BPA?drt?ffi

But

then

by

Theorem

and

Axiom

also

s

BPA?drt?ffi

and

s

BPA?drt?ffi

s

and

with

induction

we

find

BPA?drt?ffi

and

BPA?drt?ffi

s

So

BPA?drt?ffi

s

Now

BPA?drt?ffi

x?y

a?s?y

y

with

Lemma

-LRB-

ii

-RRB-

Case

x

is

of

the

form

s?t

for

certain

basic

BPA?drt?ffi

terms

s

and

t

Since

x

y

BPA?drt?ffi

y

we

also

have

s

y

BPA?drt?ffi

y

and

t

y

BPA?drt?ffi

y

Then

by

the

induction

hypothesis

BPA?drt?ffi

s?y

y

and

BPA?drt?ffi

t

y

y

So

BPA?drt?ffi

x

y

s

t

y

s

y

y.

Case

x

is

of

the

form

for

a

certain

basic

BPA?drt?ffi

term

Now

T?BPA?drt?ffi

x

oe

and

since

x?y

BPA?drt?ffi

y

we

also

have

T?BPA?drt?ffi

y

oe

and

T?BPA?drt?ffi

x

y

oe

for

some

such

that

BPA?drt?ffi

By

Lemma

-LRB-

iii

-RRB-

we

have

BPA?drt?ffi

y

y.

By

the

induction

hypothesis

we

have

BPA?drt?ffi

So

BPA?drt?ffi

x?y

y

y

y

y

y.

Remark

-LRB-

Completeness

of

BPA?drt?ffi

-RRB-

Completeness

of

BPA?drt?ffi

is

also

claimed

in

Theorem

of

-LSB-

-RSB-

-LRB-

where

BPA?drt?ffi

is

called

BPAdt

-RRB-

The

proof

given

there

however

is

incorrect

the

-LRB-

supposedly

-RRB-

bijective

mapping

is

not

bijective

as

oe

is

undefined

Soundness

and

Completeness

of

BPA?drt?ID

Definition

-LRB-

Signature

of

BPA?drt?ID

-RRB-

The

signature

of

BPA?drt?ID

consists

of

the

undelayable

atomic

actions

faja

Ag

the

undelayable

deadlock

constant

ffi

the

alternative

composition

operator

the

sequential

composition

operator

the

time

unit

delay

operator

oerel

and

the

now

operator

rel

Definition

-LRB-

Axioms

of

BPA?drt?ID

-RRB-

The

process

algebra

BPA?drt?ID

is

axiomatized

by

the

axioms

of

BPA?drt?ffi

given

in

Definition

on

page

Axioms

shown

in

Table

on

the

following

page

and

Axioms

shown

in

Table

on

the

next

page

BPA?drt?ID

Remark

-LRB-

and

versus

-RRB-

Note

that

for

closed

BPA?drt?ID

terms

Axioms

given

in

Table

on

the

following

page

are

equivalent

with

Axiom

given

in

Table

on

the

next

page

Therefore

we

could

replace

Axioms

in

BPA?drt?ID

by

without

affecting

the

soundness

or

completeness

of

the

resulting

theory

This

is

for

example

done

in

-LSB-

-RSB-

ffi

x

ffi

a

ffi

a

oerel?x

ffi

oerel?x

Table

Axioms

for

ffi

One

such

reason

to

do

so

could

be

the

fact

that

is

a

straightforward

reformulation

of

Axiom

from

BPAffi

in

a

setting

with

undelayable

actions

However

we

prefer

Axioms

over

because

the

latter

does

not

extend

to

the

discrete-time

theories

with

immediate

deadlock

we

will

describe

in

the

sections

to

follow

If

we

need

the

equality

of

we

can

derive

it

as

a

lemma

-LRB-

which

we

do

in

Lemma

-LRB-

iv

-RRB-

-RRB-

We

will

return

to

this

subject

in

Remark

on

page

x

ffi

x

Table

Alternative

for

Axioms

rel?a

a

rel?x

y

rel?x

rel?y

rel?x

y

rel?x

y

rel?oerel?x

ffi

Table

Axioms

for

rel

Definition

-LRB-

Semantics

of

BPA?drt?ID

-RRB-

The

semantics

of

BPA?drt?ID

are

given

by

the

term

deduction

system

T?BPA?drt?ID

induced

by

the

deduction

rules

for

BPA?drt?ffi

given

in

Definition

on

page

and

the

deduction

rules

for

rel

shown

in

Table

on

the

following

page

Definition

-LRB-

Bisimulation

and

Bisimulation

Model

for

BPA?drt?ID

-RRB-

Bisimulation

for

BPA?drt?ID

and

the

corresponding

bisimulation

model

are

defined

in

the

same

way

as

for

BPA?drt?ffi

and

BPA

respectively

Replace

BPA?drt?ffi

by

BPA?drt?ID

in

Definition

on

page

and

BPA

by

BPA?drt?ID

in

Definition

on

page

Definition

-LRB-

Basic

Terms

of

BPA?drt?ID

-RRB-

We

define

oe

ffi

basic

terms

inductively

as

follows

x

a

rel?x

a

x

a

p

rel?x

a

p

Table

Deduction

rules

for

rel

-LRB-

i

-RRB-

For

every

a

Affi

a

is

a

oe

ffi

basic

term

-LRB-

ii

-RRB-

if

a

Affi

and

t

is

a

oe

ffi

basic

term

then

a

t

is

a

oe

ffi

basic

term

-LRB-

iii

-RRB-

if

t

and

s

are

oe

ffi

basic

terms

then

t

s

is

a

oe

ffi

basic

term

-LRB-

iv

-RRB-

if

t

is

a

oe

ffi

basic

term

then

oerel?t

is

a

oe

ffi

basic

term

From

now

on

if

we

speak

of

basic

terms

in

the

context

of

BPA?drt?ID

we

mean

oe

ffi

basic

terms

Definition

-LRB-

Number

of

Symbols

of

a

BPA?drt?ID

Term

-RRB-

We

define

n?x

the

number

of

symbols

of

x

inductively

as

follows

-LRB-

i

-RRB-

For

a

Affi

we

define

n?a

-LRB-

ii

-RRB-

for

closed

BPA?drt?ID

terms

x

and

y

we

define

n?x?y

n?x

y

n?x

n?y

-LRB-

iii

-RRB-

for

a

closed

BPA?drt?ID

term

x

we

define

n?oerel?x

n

rel?x

n?x

Definition

-LRB-

Summation

Convention

-RRB-

We

will

use

the

convention

that

a

summation

over

the

empty

set

yields

the

undelayable

deadlock

X

i

ti

ffi

Theorem

-LRB-

General

Form

of

Basic

Terms

of

BPA?drt?ID

-RRB-

Modulo

the

commutativity

and

associativity

of

the

all

basic

terms

t

of

BPA?drt?ID

are

of

the

form

t

X

i

m

ai

si

X

j

n

bj

X

k

p

oerel?uk

for

m

n

p

N

ai

bj

Affi

and

basic

terms

si

and

uk

Proof

Trivial

by

inspection

of

the

definition

of

basic

terms

Definition

Observe

that

the

general

form

of

basic

terms

is

closed

under

the

formation

rules

gives

in

Definition

See

also

-LSB-

-RSB-

Lemma

-LRB-

Representation

of

BPA?drt?ID

Terms

-RRB-

Let

t

be

a

basic

term

Then

either

BPA?drt?ID

t

rel?t

or

there

exists

a

basic

term

s

such

that

BPA?drt?ID

t

rel?t

oerel?s

and

n?s

n?t

Proof

Let

t

be

a

basic

term

By

Theorem

we

may

now

proceed

by

case

analysis

on

the

general

form

of

basic

terms

-LRB-

i

-RRB-

Either

we

have

no

oerel-summands

-LRB-

p

in

Definition

-RRB-

t

X

i

m

ai

si

X

j

n

bj

for

m

n

N

ai

bj

Affi

and

basic

terms

si

Then

we

have

the

following

computation

BPA?drt?ID

t

X

i

m

ai

si

X

j

n

bj

X

i

m

rel?ai

si

X

j

n

rel?bj

X

i

m

rel?ai

si

X

j

n

rel?bj

rel

@X

i

m

ai

si

X

j

n

bj

A

rel?t

-LRB-

ii

-RRB-

Or

we

have

at

least

one

oerel-summand

-LRB-

p

in

Definition

-RRB-

t

X

i

m

ai

si

X

j

n

bj

X

k

p

oerel?uk

for

m

n

p

N

ai

bj

Affi

and

basic

terms

si

and

uk

Then

we

have

the

following

computation

BPA?drt?ID

t

X

i

m

ai

si

X

j

n

bj

X

k

p

oerel?uk

X

i

m

rel?ai

si

X

j

n

rel?bj

X

k

p

oerel?uk

X

i

m

rel?ai

si

X

j

n

rel?bj

X

k

p

oerel?uk

rel

@X

i

m

ai

si

X

j

n

bj

A

oerel

@X

k

p

uk

A

rel

@X

i

m

ai

si

X

j

n

bj

X

k

p

oerel?uk

A

oerel

@X

k

p

uk

A

rel?t

oerel?s

Where

we

define

s

X

k

p

uk

Note

that

n?s

n?t

is

now

trivially

satisfied

as

for

every

summand

uk

of

s

there

is

a

corresponding

summand

oerel?uk

of

t

and

at

least

one

such

summand

exists

as

p

Remark

-LRB-

Representation

of

BPA?drt?ID

Terms

-RRB-

The

main

use

of

Lemma

will

be

in

induction

proofs

regarding

the

-LRB-

not

yet

treated

-RRB-

theories

PA?drt?ID

and

ACP?drt?ID

-LRB-

see

Sections

to

-RRB-

Theorem

-LRB-

Elimination

for

BPA?drt?ID

-RRB-

Let

t

be

a

closed

BPA?drt?ID

term

Then

there

is

a

basic

term

s

such

that

BPA?drt?ID

s

t.

Proof

This

theorem

is

proven

as

follows

First

a

number

of

axioms

of

BPA?drt?ID

are

selected

and

subsequently

oriented

as

rewriting

rules

This

gives

us

a

term

rewriting

system

Then

it

is

proven

that

this

term

rewriting

system

is

strongly

normalizing

and

that

every

normal

form

of

a

closed

BPA?drt?ID

term

is

a

basic

term

In

this

way

a

recipe

is

obtained

for

transforming

a

closed

BPA?drt?ID

term

into

a

basic

term

The

rewriting

rules

of

the

term

rewriting

system

for

BPA?drt?ID

are

given

in

Table

The

proof

that

this

term

rewriting

system

is

strongly

normalizing

uses

the

method

of

x

y

z

x

z

y

z

x

y

z

x

y

z

oerel?x

y

oerel?x

y

rel?a

a

rel?x

y

rel?x

rel?y

rel?x

y

rel?x

y

rel?oerel?x

ffi

Table

Term

Rewriting

System

for

BPA?drt?ID

the

lexicographical

path

ordering

The

well-founded

ordering

on

constants

and

function

symbols

is

the

following

rel

oerel

a

Moreover

has

the

lexicographical

status

of

the

first

argument

Now

we

show

that

the

left-hand

side

of

every

rewriting

rule

is

bigger

than

the

right-hand

side

with

respect

to

the

ordering

Olpo

This

is

done

by

the

following

reductions

rel?a

Olpo

rel

a

Olpo

a

rel?x

y

Olpo

rel

x

y

Olpo

rel

x

y

rel

x

y

Olpo

rel?x

y

rel?x

y

Olpo

rel?x

rel?y

rel?x

y

Olpo

rel

x

y

Olpo

rel

x

y

rel

x

y

Olpo

rel?x

y

x

y

Olpo

rel?x

x

y

Olpo

rel?x

x

y

Olpo

rel?x

x

y

Olpo

rel?x

y

rel?oerel?x

Olpo

rel

oerel?x

Olpo

ffi

Note

that

we

do

not

give

reductions

for

and

as

these

already

have

been

given

in

the

proof

of

Theorem

and

since

the

new

ordering

is

a

proper

extension

of

the

old

one

these

proofs

remain

valid

Next

we

will

prove

that

the

normal

forms

of

the

closed

BPA?drt?ID

terms

are

basic

terms

Thereto

suppose

that

s

is

a

normal

form

of

some

closed

BPA?drt?ID

term

Furthermore

suppose

that

s

is

not

a

basic

term

Let

denote

the

smallest

subterm

of

s

which

is

not

a

basic

term

Then

we

can

prove

that

is

not

a

normal

form

by

case

analysis

We

distinguish

all

possible

cases

-LRB-

i

-RRB-

is

an

atomic

action

or

ffi

But

then

is

a

basic

term

This

is

in

contradiction

with

the

assumption

that

is

not

a

basic

term

so

this

case

does

not

occur

-LRB-

ii

-RRB-

is

of

the

form

for

basic

terms

and

With

case

analysis

on

the

structure

of

basic

term

-LRB-

a

-RRB-

If

is

an

atomic

action

or

ffi

then

is

a

basic

term

and

so

is

a

basic

term

which

again

contradicts

the

assumption

that

is

not

a

basic

term

This

case

can

therefore

not

occur

-LRB-

b

-RRB-

If

is

of

the

form

a

t

for

some

a

Affi

and

basic

term

t

then

rewriting

rule

can

be

applied

So

is

not

a

normal

form

-LRB-

c

-RRB-

If

is

of

the

form

for

and

basic

term

s

Then

rewriting

rule

is

applicable

Therefore

is

not

a

normal

form

-LRB-

d

-RRB-

If

is

of

the

form

oerel?t

for

some

basic

term

t

Then

rewriting

rule

is

applicable

So

is

not

a

normal

form

-LRB-

iii

-RRB-

is

of

the

form

for

basic

terms

and

In

this

case

would

be

a

basic

term

which

contradicts

the

assumption

that

is

not

a

basic

term

Therefore

this

case

can

not

occur

-LRB-

iv

-RRB-

is

of

the

form

oerel?t

for

some

basic

term

t

But

then

is

basic

term

too

so

the

case

does

not

occur

-LRB-

v

-RRB-

is

of

the

form

rel?t

for

some

basic

term

t

But

then

one

of

is

applicable

so

is

not

a

normal

form

In

any

case

that

can

occur

it

follows

that

is

not

a

normal

form

Since

is

a

subterm

of

s

we

conclude

that

s

is

not

a

normal

form

This

contradicts

the

assumption

that

s

is

a

normal

form

From

this

contradiction

we

conclude

that

s

is

a

basic

term

which

completes

the

proof

Remark

-LRB-

Elimination

for

BPA?drt?ID

-RRB-

Elimination

for

BPA?drt?ID

is

also

claimed

-LRB-

without

proof

-RRB-

in

Theorem

of

-LSB-

-RSB-

Theorem

-LRB-

Soundness

of

BPA?drt?ID

-RRB-

The

set

of

closed

BPA?drt?ID

terms

modulo

bisimulation

equivalence

is

a

model

of

BPA?drt?ID

Proof

In

Theorem

we

already

proved

the

soundness

of

Axioms

and

with

respect

to

the

term

deduction

system

T?BPA?drt?ffi

Since

the

term

deduction

system

T?BPA?drt?ID

uses

the

same

underlying

model

as

the

term

deduction

system

T?BPA?drt?ffi

these

proofs

remain

valid

in

the

setting

of

BPA?drt?ID

Therefore

we

only

have

to

prove

soundness

of

the

additional

Axioms

and

Axiom

Take

the

relation

R

s

s

s

ffi

s

s

C?BPA?drt?ID

We

look

at

the

transitions

of

both

sides

at

the

same

time

First

s

ffi

a

p

iff

s

a

p

and

note

that

p

p

R.

Secondly

s

ffi

a

p

iff

s

a

p.

Thirdly

s

ffi

oe

p

iff

s

oe

p

and

note

that

p

p

R.

Axiom

Take

the

relation

R

ffi

s

ffi

s

C?BPA?drt?ID

We

look

at

the

transitions

of

both

sides

at

the

same

time

Observe

that

neither

the

left-hand

side

nor

the

right-hand

side

of

the

axiom

can

perform

any

transition

ffi

s

ffi

s

and

ffi

ffi

Axiom

Take

the

relation

R

s

s

oerel?s

ffi

oerel?s

s

C?BPA?drt?ID

We

look

at

the

transitions

of

both

sides

at

the

same

time

Observe

that

neither

the

left-hand

side

nor

the

right-hand

side

of

the

axiom

can

perform

an

a-transition

oerel?s

ffi

and

oerel?s

Furthermore

the

only

oe-transitions

are

oerel?s

ffi

oe

s

and

oerel?s

oe

s

and

note

that

s

s

R.

Axiom

Take

the

relation

R

f

rel?a

a

g

We

look

at

the

transitions

of

both

sides

at

the

same

time

Observe

that

either

side

of

the

axiom

can

only

do

an

a-transition

to

p

rel?a

a

p

and

a

a

p.

No

other

transitions

are

possible

Axiom

Take

the

relation

R

s

s

rel?s

t

rel?s

rel?t

s

t

C?BPA?drt?ID

We

look

at

the

transitions

of

both

sides

at

the

same

time

Observe

that

neither

side

of

the

axiom

can

do

a

oe-transition

rel?s

t

and

rel?s

rel?t

Furthermore

rel?s

t

a

p

iff

s

t

a

p

iff

s

a

p

or

t

a

p

iff

rel?s

a

p

or

rel?t

a

p

iff

rel?s

rel?t

a

p

and

note

that

p

p

R.

Finally

rel?s

t

a

p

iff

s

t

a

p

iff

s

a

p

or

t

a

p

iff

rel?s

a

p

or

rel?t

a

p

iff

rel?s

rel?t

a

p.

Axiom

Take

the

relation

R

s

s

rel?s

t

rel?s

t

s

t

C?BPA?drt?ID

We

look

at

the

transitions

of

both

sides

at

the

same

time

Observe

that

neither

side

of

the

axiom

can

do

a

oe-transition

rel?s

t

and

rel?s

t

Furthermore

rel?s

t

a

p

iff

s

t

a

p

iff

s

a

p

and

p

t

or

s

a

and

p

t

iff

rel?s

a

p

and

p

t

or

rel?s

a

and

p

t

iff

rel?s

t

a

p

and

note

that

p

p

R.

Finally

rel?s

t

p

and

rel?s

t

p.

Axiom

Take

the

relation

R

rel?oerel?s

ffi

s

C?BPA?drt?ID

We

look

at

the

transitions

of

both

sides

at

the

same

time

Observe

that

neither

the

left-hand

side

nor

the

right-hand

side

of

the

axiom

can

perform

any

a-transition

or

oe-transition

rel?oerel?s

rel?oerel?s

and

ffi

ffi

Remark

-LRB-

Soundness

of

BPA?drt?ID

-RRB-

Soundness

of

BPA?drt?ID

is

also

claimed

-LRB-

without

proof

-RRB-

in

Section

of

-LSB-

-RSB-

-LRB-

where

BPA?drt?ID

is

called

BPAffidt

-RRB-

and

in

Theorem

of

-LSB-

-RSB-

Lemma

-LRB-

Towards

Completeness

of

BPA?drt?ID

-RRB-

Let

x

be

a

closed

BPA?drt?ID

term

and

let

a

A

Then

we

have

-LRB-

i

-RRB-

T?BPA?drt?ID

x

a

p

-RRB-

BPA?drt?ID

x

a

x

-LRB-

ii

-RRB-

T?BPA?drt?ID

x

a

y

-RRB-

BPA?drt?ID

x

a

y

x

-LRB-

iii

-RRB-

T?BPA?drt?ID

x

-RRB-

BPA?drt?ID

x

rel?x

-LRB-

iv

-RRB-

BPA?drt?ID

x

ffi

x

-LRB-

v

-RRB-

T?BPA?drt?ID

x

oe

y

-RRB-

BPA?drt?ID

x

oerel?y

rel?x

-LRB-

vi

-RRB-

T?BPA?drt?ID

x

a

y

-RRB-

n?x

n?y

-LRB-

vii

-RRB-

T?BPA?drt?ID

x

oe

y

-RRB-

n?x

n?y

Proof

For

part

-LRB-

i

-RRB-

-LRB-

v

-RRB-

we

assume

by

Theorem

and

Theorem

without

loss

of

generality

that

x

is

a

basic

term

and

then

apply

induction

on

the

structure

of

basic

terms

For

part

-LRB-

vi

-RRB-

and

-LRB-

vii

-RRB-

we

again

have

to

use

induction

on

the

general

structure

of

terms

-LRB-

i

-RRB-

Suppose

that

T?BPA?drt?ID

x

a

p.

Case

x

b

where

b

Affi

Because

T?BPA?drt?ID

x

a

p

it

must

be

the

case

that

b

a

So

we

have

BPA?drt?ID

x

b

b

b

a

b

a

x.

Case

x

b

where

b

Affi

and

is

a

basic

term

This

is

in

contradiction

with

T?BPA?drt?ID

x

a

p

so

this

case

does

not

occur

Case

x

where

and

are

basic

terms

As

T?BPA?drt?ID

x

a

p

necessarily

T?BPA?drt?ID

a

p

or

T?BPA?drt?ID

a

p.

Therefore

by

the

induction

hypothesis

BPA?drt?ID

a

or

BPA?drt?ID

a

But

then

in

both

cases

BPA?drt?ID

x

a?x

Case

x

where

is

a

basic

term

This

is

in

contradiction

with

T?BPA?drt?ID

x

a

p

so

this

case

does

not

occur

-LRB-

ii

-RRB-

Suppose

that

T?BPA?drt?ID

x

a

y.

Case

x

b

where

b

Affi

This

is

in

contradiction

with

T?BPA?drt?ID

x

a

y

so

this

case

does

not

occur

Case

x

b

where

b

Affi

and

is

a

basic

term

Then

because

T?BPA?drt?ID

x

a

y

it

must

be

that

b

a

and

y.

So

BPA?drt?ID

x

x

x

b

x

a

y

x.

Case

x

where

and

are

basic

terms

As

T?BPA?drt?ID

x

a

y

necessarily

T?BPA?drt?ID

a

y

or

T?BPA?drt?ID

a

y.

Therefore

by

the

induction

hypothesis

BPA?drt?ID

a

y

or

BPA?drt?ID

a

y

But

then

in

both

cases

BPA?drt?ID

x

a

a

y?x

Case

x

where

is

a

basic

term

This

is

in

contradiction

with

T?BPA?drt?ID

x

a

y

so

this

case

does

not

occur

-LRB-

iii

-RRB-

Suppose

that

T?BPA?drt?ID

x

Case

x

a

where

a

Affi

We

have

BPA?drt?ID

x

a

rel?a

rel?x

Case

x

a

where

a

Affi

and

is

a

basic

term

We

have

BPA?drt?ID

x

a

rel?a

rel?a

rel?x

Case

x

where

and

are

basic

terms

As

T?BPA?drt?ID

x

necessarily

T?BPA?drt?ID

and

T?BPA?drt?ID

Therefore

by

the

induction

hypothesis

BPA?drt?ID

and

BPA?drt?ID

But

then

also

BPA?drt?ID

x

rel?x

Case

x

where

is

a

basic

term

This

is

in

contradiction

with

T?BPA?drt?ID

x

so

this

case

does

not

occur

-LRB-

iv

-RRB-

Case

x

a

where

a

Affi

Then

we

have

BPA?drt?ID

x?ffi

a?ffi

a

x.

Case

x

a

where

a

Affi

and

is

a

basic

term

Then

BPA?drt?ID

x

ffi

a

ffi

a

ffi

a

ffi

a

x.

Case

x

where

and

are

basic

terms

Then

by

the

induction

hypothesis

BPA?drt?ID

ffi

ffi

So

BPA?drt?ID

x

ffi

ffi

x.

Case

x

where

is

a

basic

term

Then

BPA?drt?ID

x

ffi

ffi

x.

-LRB-

v

-RRB-

Suppose

that

T?BPA?drt?ID

x

oe

y.

Case

x

a

where

a

Affi

This

is

in

contradiction

with

T?BPA?drt?ID

x

oe

y

so

this

case

does

not

occur

Case

x

a

where

a

Affi

and

is

a

basic

term

This

is

in

contradiction

with

T?BPA?drt?ID

x

oe

y

so

this

case

does

not

occur

Case

x

where

and

are

basic

terms

As

T?BPA?drt?ID

x

oe

y

necessarily

-LRB-

-RRB-

T?BPA?drt?ID

oe

y

or

-LRB-

-RRB-

T?BPA?drt?ID

oe

y

or

-LRB-

-RRB-

T?BPA?drt?ID

oe

oe

where

y

In

the

first

case

by

the

induction

hypothesis

we

have

BPA?drt?ID

oerel?y

and

by

-LRB-

iii

-RRB-

BPA?drt?ID

Therefore

BPA?drt?ID

x

oerel?y

oerel?y

oerel?y

rel?x

The

second

case

is

treated

analogously

In

the

third

case

we

have

by

the

induction

hypothesis

BPA?drt?ID

Therefore

we

have

BPA?drt?ID

x

oerel?y

rel?x

Case

x

where

is

a

basic

term

Because

T?BPA?drt?ID

x

oe

y

it

must

be

the

case

that

y

So

we

have

BPA?drt?ID

x

oerel?y

oerel?y

ffi

oerel?y

oerel?y

rel?x

-LRB-

vi

-RRB-

Suppose

that

T?BPA?drt?ID

x

a

y.

Case

x

b

where

b

Affi

This

is

in

contradiction

with

T?BPA?drt?ID

x

a

y

so

this

case

does

not

occur

Case

x

for

certain

terms

and

Then

because

T?BPA?drt?ID

x

a

y

we

either

have

T?BPA?drt?ID

a

p

and

y

or

we

have

T?BPA?drt?ID

a

and

y

for

some

term

In

the

first

case

we

have

n?x

n?y

and

in

the

second

we

can

apply

the

induction

hypothesis

to

arrive

at

so

we

get

n?x

n?y

Case

x

for

certain

terms

and

Since

T?BPA?drt?ID

x

a

y

necessarily

T?BPA?drt?ID

a

y

or

T?BPA?drt?ID

a

y.

Therefore

by

the

induction

hypothesis

n?y

or

n?y

In

both

cases

n?x

n?y

Case

x

for

a

certain

term

This

is

in

contradiction

with

T?BPA?drt?ID

x

a

y

so

this

case

does

not

occur

Case

x

for

a

certain

term

Since

T?BPA?drt?ID

x

a

y

necessarily

T?BPA?drt?ID

a

y.

Therefore

by

the

induction

hypothesis

n?y

So

n?x

n

n?y

-LRB-

vii

-RRB-

Suppose

that

T?BPA?drt?ID

x

oe

y.

Case

x

a

where

a

Affi

This

is

in

contradiction

with

T?BPA?drt?ID

x

oe

y

so

this

case

does

not

occur

Case

x

for

certain

terms

and

Then

necessarily

oe

and

y

for

some

term

We

now

can

apply

the

induction

hypothesis

to

arrive

at

so

we

get

n?x

n?y

Case

x

for

certain

terms

and

As

T?BPA?drt?ID

x

oe

y

necessarily

-LRB-

-RRB-

T?BPA?drt?ID

oe

y

or

-LRB-

-RRB-

T?BPA?drt?ID

oe

y

or

-LRB-

-RRB-

T?BPA?drt?ID

oe

oe

where

y

In

the

first

case

by

the

induction

hypothesis

n?y

So

n?x

n?y

The

second

case

is

treated

analogously

In

the

third

case

by

the

induction

hypothesis

and

So

n?x

n?y

Case

x

for

a

certain

term

Because

T?BPA?drt?ID

x

oe

y

it

must

be

the

case

that

y

Then

we

have

n?x

n?y

n?y

Case

x

for

a

certain

term

This

is

in

contradiction

with

T?BPA?drt?ID

x

oe

y

so

this

case

does

not

occur

Theorem

-LRB-

Completeness

of

BPA?drt?ID

-RRB-

The

axiom

system

BPA?drt?ID

is

a

complete

axiomatization

of

the

set

of

closed

BPA?drt?ID

terms

modulo

-LRB-

strong

-RRB-

bisimulation

equivalence

Proof

Suppose

x?y

BPA?drt?ID

y

We

will

prove

with

induction

on

the

structure

of

basic

term

x

that

BPA?drt?ID

x

y

y.

By

Theorem

we

can

restrict

ourselves

to

basic

terms

without

loss

of

generality

The

proof

is

done

with

induction

on

n?x

using

Lemma

-LRB-

vi

-RRB-

-LRB-

vii

-RRB-

and

case

distinction

on

the

form

of

basic

term

x.

-LRB-

i

-RRB-

x

ffi

Then

using

Lemma

-LRB-

iv

-RRB-

we

have

BPA?drt?ID

x

y

ffi

y

y

ffi

y.

-LRB-

ii

-RRB-

x

a

where

a

A.

From

the

deduction

rules

we

have

T?BPA?drt?ID

x

a

p

and

T?BPA?drt?ID

x?y

a

p

Since

x?y

BPA?drt?ID

y

we

also

have

T?BPA?drt?ID

y

a

p.

By

Lemma

-LRB-

i

-RRB-

we

obtain

BPA?drt?ID

y

a?y

So

BPA?drt?ID

x?y

a?y

y.

-LRB-

iii

-RRB-

x

ffi

s

where

s

is

a

basic

term

Then

we

have

BPA?drt?ID

x

ffi

s

ffi

and

using

-LRB-

i

-RRB-

BPA?drt?ID

x

y

y.

-LRB-

iv

-RRB-

x

a

s

where

a

A

and

s

is

a

basic

term

From

the

deduction

rules

we

obtain

T?BPA?drt?ID

x

a

s

and

T?BPA?drt?ID

x

y

a

s

Since

x

y

BPA?drt?ID

y

we

then

also

have

T?BPA?drt?ID

y

a

t

for

some

t

such

that

s

BPA?drt?ID

t.

By

the

induction

hypothesis

we

have

BPA?drt?ID

s

t.

From

Lemma

-LRB-

ii

-RRB-

we

have

BPA?drt?ID

y

a

t

y.

So

BPA?drt?ID

x

y

a

s

y

a

t

y

y.

-LRB-

v

-RRB-

x

s

t

where

s

and

t

are

basic

terms

Since

s

t

y

BPA?drt?ID

y

we

also

have

s

y

BPA?drt?ID

y

and

t

y

BPA?drt?ID

y.

By

the

induction

hypothesis

we

then

have

BPA?drt?ID

s

y

y

t

y

y.

So

BPA?drt?ID

x

y

s

t

y

s

y

y.

-LRB-

vi

-RRB-

x

oerel?s

where

s

is

a

basic

term

From

the

deduction

rules

we

now

have

that

T?BPA?drt?ID

oerel?s

oe

s

and

since

x

y

BPA?drt?ID

y

we

also

have

T?BPA?drt?ID

y

oe

t

x

y

oe

s

t

for

some

t

such

that

s

t

BPA?drt?ID

t.

By

Lemma

-LRB-

v

-RRB-

we

have

BPA?drt?ID

y

oerel?t

rel?y

By

the

induction

hypothesis

we

have

BPA?drt?ID

s

t

t.

So

BPA?drt?ID

x

y

oerel?s

y

oerel?s

oerel?t

rel?y

oerel?s

t

rel?y

oerel?t

rel?y

y.

Remark

-LRB-

Completeness

of

BPA?drt?ID

-RRB-

Completeness

of

BPA?drt?ID

is

also

claimed

-LRB-

without

proof

-RRB-

in

Section

of

-LSB-

-RSB-

-LRB-

where

BPA?drt?ID

is

called

BPAffidt

-RRB-

and

in

Theorem

of

-LSB-

-RSB-

Soundness

and

Completeness

of

BPA?drt?ID

Definition

-LRB-

Signature

of

BPAdrt?ID

-RRB-

The

signature

of

BPAdrt?ID

consists

of

the

undelayable

atomic

actions

faja

Ag

the

delayable

atomic

actions

faja

Ag

the

undelayable

deadlock

constant

ffi

the

delayable

deadlock

constant

ffi

the

alternative

composition

operator

the

sequential

composition

operator

the

time

unit

delay

operator

oerel

the

now

operator

rel

and

the

unbounded

start

delay

operator

b

c

Definition

-LRB-

Axioms

of

BPAdrt?ID

-RRB-

The

process

algebra

BPAdrt?ID

is

axiomatized

by

the

axioms

of

BPA?drt?ID

given

in

Definition

on

page

and

Axioms

ATS

and

USD

shown

in

Table

BPAdrt?ID

ATS

USD

a

bac

ATS

bxc

rel?x

oerel?bxc

USD

Table

Axioms

for

delayable

actions

Definition

-LRB-

Recursion

Principle

for

BPAdrt?ID

-RRB-

Next

to

the

axioms

mentioned

in

Definition

the

system

BPA?drt?ID

also

contains

the

recursion

principle

RSP

-LRB-

USD

-RRB-

shown

in

Table

on

the

following

page

For

more

information

on

recursion

principles

and

their

status

with

respect

to

axioms

see

-LSB-

-RSB-

Remark

-LRB-

Notation

BPA?drt?ID

-RRB-

By

superscripting

a

theory

with

a

-LRB-

e.g.

BPA?drt?ID

-RRB-

we

indicate

the

presence

of

the

recursion

principle

RSP

-LRB-

USD

-RRB-

Note

that

in

-LSB-

-RSB-

the

notation

BPAdrt?ID

RSP

-LRB-

USD

-RRB-

is

used

instead

of

BPA?drt?ID

However

we

find

that

notation

cumbersome

as

it

clutters

up

the

formulae

y

rel?x

oerel?y

-RRB-

y

bxc

RSP

-LRB-

USD

-RRB-

Table

Recursive

Specification

Principle

for

the

Unbounded

Start

Delay

Definition

-LRB-

Semantics

of

BPAdrt?ID

-RRB-

The

semantics

of

BPAdrt?ID

are

given

by

the

term

deduction

system

T?BPAdrt?ID

induced

by

the

deduction

rules

for

BPA?drt?ID

given

in

Definition

on

page

and

the

deduction

rules

for

delayable

actions

given

in

Table

a

a

p

a

oe

a

ffi

oe

ffi

x

a

bxc

a

x

a

p

bxc

a

p

bxc

oe

bxc

Table

Deduction

rules

for

delayable

actions

Definition

-LRB-

Bisimulation

and

Bisimulation

Model

for

BPAdrt?ID

-RRB-

Bisimulation

for

BPAdrt?ID

and

the

corresponding

bisimulation

model

are

defined

in

the

same

way

as

for

BPA?drt?ffi

and

BPA

respectively

Replace

BPA?drt?ffi

by

BPAdrt?ID

in

Definition

on

page

and

BPA

by

BPAdrt?ID

in

Definition

on

page

Definition

-LRB-

Basic

Terms

of

BPAdrt?ID

-RRB-

We

define

oe

ffi

ffi

basic

terms

inductively

as

follows

-LRB-

i

-RRB-

If

a

Affi

then

a

and

a

are

oe

ffi

ffi

basic

terms

-LRB-

ii

-RRB-

if

a

Affi

and

t

is

a

oe

ffi

ffi

basic

term

then

a

t

and

a

t

are

oe

ffi

ffi

basic

terms

-LRB-

iii

-RRB-

if

t

and

s

are

oe

ffi

ffi

basic

terms

then

t

s

is

a

oe

ffi

ffi

basic

term

-LRB-

iv

-RRB-

if

t

is

a

oe

ffi

ffi

basic

term

then

oerel?t

is

a

oe

ffi

ffi

basic

term

From

now

on

if

we

speak

of

basic

terms

in

the

context

of

BPAdrt?ID

we

mean

oe

ffi

ffi

basic

terms

Definition

-LRB-

Number

of

Symbols

of

a

BPAdrt?ID

term

-RRB-

We

define

n?x

the

number

of

symbols

of

x

is

inductively

as

follows

-LRB-

i

-RRB-

For

a

Affi

we

define

n?a

n?a

-LRB-

ii

-RRB-

for

closed

BPAdrt?ID

terms

x

and

y

we

define

n?x?y

n?x

y

n?x

n?y

-LRB-

iii

-RRB-

for

a

closed

BPAdrt?ID

term

x

we

define

n?oerel?x

n

rel?x

n?bxc

n?x

Proposition

-LRB-

Properties

of

BPA?drt?ID

Part

I

-RRB-

For

BPAdrt?ID

terms

x

and

y

and

any

a

Affi

we

have

the

following

equalities

-LRB-

i

-RRB-

BPA?drt?ID

bac

a

-LRB-

ii

-RRB-

BPA?drt?ID

bx

yc

bxc

y

-LRB-

iii

-RRB-

BPA?drt?ID

bx

yc

bxc

byc

-LRB-

iv

-RRB-

BPA?drt?ID

boerel?x?c

ffi

-LRB-

v

-RRB-

BPAdrt?ID

rel?a

a

Proof

-LRB-

i

-RRB-

Consider

the

following

computation

BPAdrt?ID

a

bac

rel?a

oerel

bac

a

oerel?a

a

ffi

oerel?a

rel?a

rel

oerel

bac

oerel?a

rel

a

oerel

bac

oerel?a

rel

rel?a

oerel

bac

oerel?a

rel

bac

oerel?a

rel?a

oerel?a

Using

RSP

-LRB-

USD

-RRB-

we

obtain

BPA?drt?ID

a

bac

-LRB-

ii

-RRB-

Consider

the

following

computation

BPAdrt?ID

bxc

y

rel?x

oerel?bxc

y

rel?x

y

oerel?bxc

y

rel?x

y

oerel?bxc

y

Using

RSP

-LRB-

USD

-RRB-

-LRB-

with

x

instantiated

by

x

y

and

y

by

bxc

y

-RRB-

we

obtain

BPA?drt?ID

bxc

y

bx

yc

-LRB-

iii

-RRB-

Consider

the

following

computation

BPAdrt?ID

bxc

byc

rel?x

oerel?bxc

rel?y

oerel?byc

rel?x

y

oerel?bxc

byc

Using

RSP

-LRB-

USD

-RRB-

we

obtain

BPA?drt?ID

bxc

byc

bx

yc

-LRB-

iv

-RRB-

Consider

the

following

computation

BPAdrt?ID

ffi

bffic

rel?ffi

oerel

bffic

ffi

oerel

bffic

oerel?ffi

rel?oerel?x

oerel?ffi

Using

RSP

-LRB-

USD

-RRB-

we

obtain

BPA?drt?ID

ffi

boerel?x?c

-LRB-

v

-RRB-

Consider

the

following

computation

BPAdrt?ID

rel?a

rel?bac

rel

rel?a

oerel

bac

rel?a

oerel?a

rel?a

rel?oerel?a

a

ffi

a

So

we

obtain

BPAdrt?ID

rel?a

a

Remark

-LRB-

Properties

of

BPA?drt?ID

Part

I

-RRB-

The

equalities

of

Proposition

on

the

page

before

are

not

new

but

have

been

described

before

See

for

example

-LSB-

-RSB-

Proposition

-LRB-

Properties

of

BPA?drt?ID

Part

II

-RRB-

For

any

BPAdrt?ID

term

x

we

have

the

following

equality

BPA?drt?ID

ffi

x

ffi

Proof

Using

Proposition

-LRB-

ii

-RRB-

we

derive

BPA?drt?ID

ffi

x

bffic

x

bffi

xc

bffic

ffi

Theorem

-LRB-

Elimination

for

BPA?drt?ID

-RRB-

Let

t

be

a

closed

BPAdrt?ID

term

Then

there

is

a

basic

term

s

such

that

BPA?drt?ID

s

t.

Proof

First

a

term

rewriting

system

is

given

Then

it

is

shown

that

this

term

rewriting

system

is

strongly

normalizing

and

that

the

normal

forms

of

the

closed

BPAdrt?ID

terms

are

basic

terms

The

term

rewriting

system

is

given

in

Table

on

the

following

page

The

rewriting

rules

RATS

and

are

obtained

directly

from

the

axioms

x

y

z

x

z

y

z

x

y

z

x

y

z

oerel?x

y

oerel?x

y

bac

a

RATS

bac

a

bx

yc

bxc

y

bx

yc

bxc

byc

boerel?x?c

ffi

rel?a

a

rel?x

y

rel?x

rel?y

rel?x

y

rel?x

y

rel?oerel?x

ffi

rel?a

a

Table

Term

rewriting

system

for

BPAdrt?ID

The

rewriting

rules

and

are

added

to

deal

properly

with

the

recursive

definition

of

ultimate

start

delay

the

corresponding

equalities

are

derivable

from

the

axioms

as

is

shown

in

Proposition

With

the

method

of

the

lexicographical

path

ordering

it

is

shown

that

the

term

rewriting

system

is

strongly

normalizing

Thereto

the

operator

is

assigned

the

lexicographical

status

for

the

first

argument

and

the

following

well-founded

partial

ordering

on

the

signature

of

BPAdrt?ID

is

defined

b

c

a

rel

a

oerel

Now

we

show

that

the

left-hand

side

of

every

rewriting

rule

is

bigger

than

the

right-hand

side

with

respect

to

the

ordering

Olpo

This

is

done

by

the

following

reductions

bac!Olpo

bac

Olpo

a

bac!Olpo

bac

Olpo

a

bx

yc!Olpo

bx

yc

Olpo

bx

yc

bx

yc

Olpo

x

y

x

y

Olpo

bxc

x

y

Olpo

bxc

x

y

Olpo

bxc

y

bx

yc!Olpo

bx

yc

Olpo

bx

yc

bx

yc

Olpo

x

y

x

y

Olpo

bxc

byc

boerel?x?c!Olpo

boerel?x?c

Olpo

ffi

rel?a

Olpo

rel

a

Olpo

a

Note

that

we

do

not

give

reductions

for

and

as

these

already

have

been

given

in

the

proofs

of

previous

elimination

theorems

and

since

the

new

ordering

is

a

proper

extension

of

the

old

ones

these

proofs

remain

valid

It

remains

to

prove

that

every

normal

form

of

a

closed

BPAdrt?ID

term

is

a

basic

term

Suppose

that

s

is

the

normal

form

of

a

closed

BPAdrt?ID

term

Furthermore

suppose

that

s

is

not

a

basic

term

and

that

is

the

smallest

subterm

of

s

which

is

not

a

basic

term

We

distinguish

all

possible

cases

-LRB-

i

-RRB-

is

of

the

form

a

or

a

for

some

a

Affi

Then

is

clearly

a

basic

term

so

this

case

does

not

occur

-LRB-

ii

-RRB-

is

of

the

form

for

basic

terms

and

With

respect

to

basic

term

the

following

cases

can

be

distinguished

-LRB-

a

-RRB-

a

for

some

a

Affi

Then

is

a

basic

term

This

contradicts

the

assumption

that

is

not

a

basic

term

-LRB-

b

-RRB-

a

for

some

a

Affi

Then

is

a

basic

term

and

we

have

again

a

contradiction

-LRB-

c

-RRB-

a

for

some

a

Affi

and

basic

term

Then

rewriting

rule

is

applicable

so

is

not

a

normal

form

-LRB-

d

-RRB-

a

for

some

a

Affi

and

basic

term

Then

rewriting

rule

is

applicable

so

is

not

a

normal

form

-LRB-

e

-RRB-

for

some

basic

terms

and

Then

rewriting

rule

is

applicable

so

is

not

a

normal

form

-LRB-

f

-RRB-

for

some

basic

term

Then

rewriting

rule

is

applicable

so

is

not

a

normal

form

-LRB-

iii

-RRB-

is

of

the

form

for

basic

terms

and

Then

is

a

basic

term

itself

so

this

case

can

not

happen

-LRB-

iv

-RRB-

is

of

the

form

for

some

basic

term

Then

again

is

a

basic

term

itself

so

this

case

can

not

happen

either

-LRB-

v

-RRB-

is

of

the

form

where

is

a

basic

term

Then

one

of

can

be

applied

so

is

not

a

normal

form

-LRB-

vi

-RRB-

is

of

the

form

for

some

basic

term

Then

one

of

RATS

or

can

be

applied

so

is

not

a

normal

form

In

every

case

is

a

basic

term

or

a

rewriting

rule

is

applicable

If

is

a

basic

term

this

contradicts

the

assumption

that

it

is

not

If

a

rewriting

rule

is

applicable

then

and

s

are

not

a

normal

form

This

contradicts

the

assumption

that

s

is

a

normal

form

From

this

contradiction

we

conclude

that

s

is

a

basic

term

Remark

-LRB-

Elimination

for

BPA?drt?ID

-RRB-

Elimination

for

BPA?drt?ID

is

also

claimed

-LRB-

without

proof

-RRB-

in

Theorem

of

-LSB-

-RSB-

-LRB-

where

BPA?drt?ID

is

called

BPAdrt?ID

-RRB-

Theorem

-LRB-

Soundness

of

BPA?drt?ID

-RRB-

The

set

of

closed

BPAdrt?ID

terms

modulo

bisimulation

equivalence

is

a

model

of

BPA?drt?ID

Proof

Note

that

the

soundness

proofs

for

Axioms

and

given

in

the

previous

sections

also

remain

valid

in

the

setting

with

delayable

actions

This

is

due

to

the

fact

that

the

underlying

model

-LRB-

namely

finite

transition

systems

with

oe

's

-RRB-

has

not

changed

Or

stated

more

concretely

all

oe-transitions

were

considered

without

regard

for

whether

they

resulted

from

a

oerel

operator

or

otherwise

so

the

extra

oe-transitions

introduced

by

the

delayable

actions

do

not

matter

Axiom

ATS

Take

the

relation

R

a

bac

There

are

only

two

transitions

possible

on

the

left-hand

side

of

the

axiom

a

a

p

and

a

oe

a

The

right-hand

side

can

also

perform

two

transitions

bac

a

p

and

bac

oe

bac

and

note

that

a

bac

R.

Axiom

USD

Take

the

relation

R

s

s

bsc

rel?s

oerel?bsc

js

C?BPAdrt?ID

First

we

look

at

the

transitions

of

the

left-hand

side

-LRB-

i

-RRB-

Suppose

bsc

a

p

Then

this

must

be

due

to

s

a

p

But

then

also

rel?s

a

p

and

rel?s

oerel?bsc

a

p

and

note

that

p

p

R.

-LRB-

ii

-RRB-

Suppose

bsc

a

p

Then

this

must

be

due

to

s

a

p

But

then

also

rel?s

a

p

and

rel?s

oerel?bsc

a

p.

-LRB-

iii

-RRB-

Suppose

bsc

oe

p

Then

necessarily

p

bsc

We

also

have

oerel?bsc

oe

bsc

hence

rel?s

oerel?bsc

oe

bsc

and

note

that

bsc

bsc

R.

Secondly

we

look

at

the

transitions

of

the

right-hand

side

-LRB-

i

-RRB-

Suppose

rel?s

oerel?bsc

a

p

Then

this

must

be

due

to

s

a

p

But

then

also

bsc

a

p

and

note

that

p

p

R.

-LRB-

ii

-RRB-

Suppose

rel?s

oerel?bsc

a

p

Then

this

must

be

due

to

s

a

p

But

then

also

bsc

a

p.

-LRB-

iii

-RRB-

Suppose

rel?s

oerel?bsc

oe

p

Then

this

must

be

due

to

oerel?bsc

oe

p

with

p

bsc

Clearly

then

also

bsc

oe

bsc

and

note

that

bsc

p

R.

RSP

-LRB-

USD

-RRB-

Suppose

that

is

a

bisimulation

relation

between

y

and

rel?x

oerel?y

We

must

prove

that

y

BPA?drt?ID

bxc

Then

take

the

symmetric

transitive

closure

of

-LRB-

which

is

again

a

bisimulation

see

-LSB-

-RSB-

-RRB-

denoted

and

extend

it

to

a

bisimulation

relation

R

between

y

and

bxc

as

follows

R

-LSB-

s

bxc

js

C?BPAdrt?ID

y

oe

-RRB-

s

Now

let

s

be

any

closed

term

such

that

y

oe

-RRB-

s.

Then

using

induction

on

the

number

of

oe-transitions

we

have

that

y

s

First

we

look

at

the

transitions

of

the

left-hand

side

i.e.

the

transitions

of

s

-LRB-

i

-RRB-

Suppose

s

a

p.

Then

as

y

s

y

a

q

such

that

p

q

As

y

rel?x

oerel?y

we

have

rel?x

oerel?y

a

r

for

some

r

such

that

q

r

Hence

rel?x

a

r

hence

x

a

r

hence

bxc

a

r

and

note

that

by

transitivity

p

r

so

p

r

R.

-LRB-

ii

-RRB-

Suppose

s

a

p.

Then

as

y

s

y

a

p

As

y

rel?x

oerel?y

we

have

rel?x

oerel?y

a

p.

Hence

rel?x

a

p

hence

x

a

p

hence

bxc

a

p.

-LRB-

iii

-RRB-

Suppose

s

oe

p

Since

y

oe

-RRB-

s

we

have

y

oe

-RRB-

p

We

also

have

bxc

oe

bxc

and

note

that

p

bxc

R.

Secondly

we

look

at

the

transitions

of

the

right-hand

side

-LRB-

i

-RRB-

Suppose

bxc

a

q

Then

x

a

q

hence

rel?x

a

q

hence

rel?x

oerel?y

a

q

As

y

rel?x

oerel?y

we

know

that

y

a

p

for

some

p

such

that

p

q

Since

y

s

we

also

have

s

a

r

for

some

r

such

that

p

r

and

note

that

by

symmetry

and

transitivity

r

q

so

r

q

R.

-LRB-

ii

-RRB-

Suppose

bxc

a

p

Then

x

a

p

hence

rel?x

a

p

hence

rel?x

oerel?y

a

p

As

y

rel?x

oerel?y

we

know

that

y

a

p

Since

y

s

we

also

have

s

a

p.

-LRB-

iii

-RRB-

Suppose

bxc

oe

q.

Then

it

must

be

the

case

that

q

bxc

We

also

have

oerel?y

oe

y

hence

rel?x

oerel?y

oe

y

and

as

y

rel?x

oerel?y

we

know

that

y

oe

p

such

that

p

y

As

y

s

necessarily

s

oe

r

for

some

r

such

that

p

r

Since

y

oe

-RRB-

s

we

have

y

oe

-RRB-

r

and

note

that

r

bxc

R.

Remark

-LRB-

Soundness

of

BPA?drt?ID

-RRB-

Soundness

of

BPA?drt?ID

is

also

claimed

-LRB-

without

proof

-RRB-

in

Theorem

of

-LSB-

-RSB-

-LRB-

where

BPA?drt?ID

is

called

BPAdrt?ID

-RRB-

Lemma

-LRB-

Towards

Completeness

of

BPA?drt?ID

-RRB-

Let

x

and

y

be

closed

BPAdrt?ID

terms

and

let

a

A

Then

we

have

-LRB-

i

-RRB-

T?BPAdrt?ID

x

a

p

-RRB-

BPA?drt?ID

x

a

x

-LRB-

ii

-RRB-

T?BPAdrt?ID

x

a

y

-RRB-

BPA?drt?ID

x

a

y

x

-LRB-

iii

-RRB-

T?BPAdrt?ID

x

-RRB-

BPA?drt?ID

x

rel?x

-LRB-

iv

-RRB-

BPA?drt?ID

x

ffi

x

-LRB-

v

-RRB-

T?BPAdrt?ID

x

oe

y

-RRB-

BPA?drt?ID

x

oerel?y

rel?x

-LRB-

vi

-RRB-

T?BPAdrt?ID

x

oe

x

-RRB-

BPA?drt?ID

x

bxc

-LRB-

vii

-RRB-

T?BPAdrt?ID

x

a

y

-RRB-

n?x

n?y

-LRB-

viii

-RRB-

T?BPAdrt?ID

x

oe

y

-RRB-

x

y

n?x

n?y

Proof

For

part

-LRB-

i

-RRB-

-LRB-

vi

-RRB-

we

assume

by

Theorem

and

Theorem

without

loss

of

generality

that

x

is

a

basic

term

and

then

apply

induction

on

the

structure

of

basic

terms

For

part

-LRB-

vii

-RRB-

and

-LRB-

viii

-RRB-

we

again

have

to

use

induction

on

the

general

structure

of

terms

-LRB-

i

-RRB-

Suppose

that

T?BPAdrt?ID

x

a

p.

Case

x

b

where

b

Affi

Because

T?BPAdrt?ID

x

a

p

it

must

be

the

case

that

b

a

So

we

have

BPA?drt?ID

x

b

b

b

a

b

a

x.

Case

x

b

where

b

Affi

Because

T?BPAdrt?ID

x

a

p

it

must

be

the

case

that

b

a

So

we

have

BPA?drt?ID

x

b

bbc

rel?b

oerel

bbc

rel?b

rel?b

oerel

bbc

b

bbc

a

b

a

x.

Case

x

b

where

b

Affi

and

is

a

basic

term

This

is

in

contradiction

with

T?BPAdrt?ID

x

a

p

so

this

case

does

not

occur

Case

x

b

where

b

Affi

and

is

a

basic

term

This

is

in

contradiction

with

T?BPAdrt?ID

x

a

p

so

this

case

does

not

occur

Case

x

where

and

are

basic

terms

As

T?BPAdrt?ID

x

a

p

necessarily

T?BPAdrt?ID

a

p

or

T?BPAdrt?ID

a

p.

Therefore

by

the

induction

hypothesis

BPA?drt?ID

a

or

BPA?drt?ID

a

But

then

in

both

cases

BPA?drt?ID

x

a

a

x.

Case

x

where

is

a

basic

term

This

is

in

contradiction

with

T?BPAdrt?ID

x

a

p

so

this

case

does

not

occur

-LRB-

ii

-RRB-

Suppose

that

T?BPAdrt?ID

x

a

y.

Case

x

b

where

b

Affi

This

is

in

contradiction

with

T?BPAdrt?ID

x

a

y

so

this

case

does

not

occur

Case

x

b

where

b

Affi

This

is

in

contradiction

with

T?BPAdrt?ID

x

a

y

so

this

case

does

not

occur

Case

x

b

where

b

Affi

and

is

a

basic

term

Then

because

T?BPAdrt?ID

x

a

y

it

must

be

that

b

a

and

y.

So

BPA?drt?ID

x

x

x

b

x

a

y

x.

Case

x

b

where

b

Affi

and

is

a

basic

term

Then

because

T?BPAdrt?ID

x

a

y

it

must

be

that

b

a

and

y.

So

BPA?drt?ID

x

a?y

a?a

y

a?y?a?y

a?y?x

Case

x

where

and

are

basic

terms

As

T?BPAdrt?ID

x

a

y

necessarily

T?BPAdrt?ID

a

y

or

T?BPAdrt?ID

a

y.

Therefore

by

the

induction

hypothesis

BPA?drt?ID

a

y

or

BPA?drt?ID

a

y

But

then

in

both

cases

BPA?drt?ID

x

a

a

y?x

Case

x

where

is

a

basic

term

This

is

in

contradiction

with

T?BPAdrt?ID

x

a

y

so

this

case

does

not

occur

-LRB-

iii

-RRB-

Suppose

that

T?BPAdrt?ID

x

Case

x

a

where

a

Affi

We

have

BPA?drt?ID

x

a

rel?a

rel?x

Case

x

a

where

a

Affi

This

is

in

contradiction

with

T?BPAdrt?ID

x

so

this

case

does

not

occur

Case

x

a

where

a

Affi

and

is

a

basic

term

We

have

BPA?drt?ID

x

a

rel?a

rel?a

rel?x

Case

x

a

where

a

Affi

and

is

a

basic

term

This

is

in

contradiction

with

T?BPAdrt?ID

x

so

this

case

does

not

occur

Case

x

where

and

are

basic

terms

As

T?BPAdrt?ID

x

necessarily

T?BPAdrt?ID

and

T?BPAdrt?ID

Therefore

by

the

induction

hypothesis

BPA?drt?ID

and

BPA?drt?ID

But

then

also

BPA?drt?ID

x

rel?x

Case

x

where

is

a

basic

term

This

is

in

contradiction

with

T?BPAdrt?ID

x

so

this

case

does

not

occur

-LRB-

iv

-RRB-

Case

x

a

where

a

Affi

This

is

exactly

Axiom

Case

x

a

where

a

Affi

Then

we

have

BPA?drt?ID

x

ffi

a

ffi

bac

ffi

rel?a

oerel

bac

ffi

a

ffi

oerel

bac

a

oerel

bac

rel?a

oerel

bac

bac

a

x.

Case

x

a

where

a

Affi

and

is

a

basic

term

Then

we

have

BPA?drt?ID

x

ffi

a

ffi

a

ffi

a

ffi

a

x.

Case

x

a

where

a

Affi

and

is

a

basic

term

Then

by

case

BPA?drt?ID

x

ffi

a

ffi

a

ffi

a

ffi

a

x.

Case

x

where

and

are

basic

terms

By

the

induction

hypothesis

we

have

BPA?drt?ID

ffi

ffi

So

BPA?drt?ID

x

ffi

ffi

x.

Case

x

where

is

a

basic

term

This

is

exactly

Axiom

-LRB-

v

-RRB-

Suppose

that

T?BPAdrt?ID

x

oe

y.

Case

x

a

where

a

Affi

This

is

in

contradiction

with

T?BPAdrt?ID

x

oe

y

so

this

case

does

not

occur

Case

x

a

where

a

Affi

Because

T?BPAdrt?ID

x

oe

y

it

must

be

the

case

that

y

a

So

we

have

BPA?drt?ID

x

a

bac

rel?a

oerel

bac

oerel?a

rel?a

ffi

oerel?a

rel?a

rel

oerel

bac

oerel?a

rel

rel?a

oerel

bac

oerel?a

rel

bac

oerel?a

rel?a

oerel?y

rel?x

Case

x

a

where

a

Affi

and

is

a

basic

term

This

is

in

contradiction

with

T?BPAdrt?ID

x

oe

y

so

this

case

does

not

occur

Case

x

where

a

Affi

and

is

a

basic

term

Because

T?BPAdrt?ID

x

oe

y

it

must

be

the

case

that

y

a

So

we

have

BPA?drt?ID

x

a

bac

rel?a

oerel

bac

rel?a

oerel

bac

oerel

bac

rel?a

rel?a

oerel?y

rel?a

ffi

oerel?y

rel?a

rel

oerel

bac

oerel?y

rel

a

oerel

bac

oerel?y

rel

rel?a

oerel

bac

oerel?y

rel

bac

oerel?y

rel?a

oerel?y

oerel?y

rel?x

Case

x

where

and

are

basic

terms

As

T?BPAdrt?ID

x

oe

y

necessarily

-LRB-

-RRB-

T?BPAdrt?ID

oe

y

or

-LRB-

-RRB-

T?BPAdrt?ID

oe

y

or

-LRB-

-RRB-

T?BPAdrt?ID

oe

oe

where

y

In

the

first

case

by

the

induction

hypothesis

we

have

BPA?drt?ID

oerel?y

and

by

-LRB-

iii

-RRB-

BPA?drt?ID

Therefore

BPA?drt?ID

x

oerel?y

oerel?y

oerel?y

rel?x

The

second

case

is

treated

analogously

In

the

third

case

we

have

by

the

induction

hypothesis

BPA?drt?ID

Therefore

we

have

BPA?drt?ID

x

oerel?y

rel?x

Case

x

where

is

a

basic

term

Because

T?BPAdrt?ID

x

oe

y

it

must

be

the

case

that

y

So

we

have

BPA?drt?ID

x

oerel?y

oerel?y

ffi

oerel?y

oerel?y

rel?x

-LRB-

vi

-RRB-

Suppose

that

T?BPAdrt?ID

x

oe

x.

Case

x

a

where

a

Affi

This

is

in

contradiction

with

T?BPAdrt?ID

x

oe

x

so

this

case

does

not

occur

Case

x

a

where

a

Affi

Then

we

have

using

Proposition

-LRB-

i

-RRB-

BPA?drt?ID

x

a

bac

bxc

Case

x

a

where

a

Affi

and

is

a

basic

term

This

is

in

contradiction

with

T?BPAdrt?ID

x

oe

x

so

this

case

does

not

occur

Case

x

a

where

a

Affi

and

is

a

basic

term

Then

we

can

derive

using

Proposition

-LRB-

i

-RRB-

and

-LRB-

ii

-RRB-

BPA?drt?ID

x

bac

ba

bxc

Case

x

where

and

are

basic

terms

Then

we

can

derive

using

Proposition

-LRB-

iii

-RRB-

and

the

induction

hypothesis

BPA?drt?ID

x

bxc

bxc

Case

x

where

is

a

basic

term

This

is

in

contradiction

with

T?BPAdrt?ID

x

oe

x

so

this

case

does

not

occur

-LRB-

vii

-RRB-

Suppose

that

T?BPAdrt?ID

x

a

y.

Case

x

b

where

b

Affi

This

is

in

contradiction

with

T?BPAdrt?ID

x

a

y

so

this

case

does

not

occur

Case

x

b

where

b

Affi

This

is

in

contradiction

with

T?BPAdrt?ID

x

a

y

so

this

case

does

not

occur

Case

x

is

of

the

form

for

certain

terms

and

Then

by

T?BPAdrt?ID

x

a

y

we

either

have

T?BPAdrt?ID

a

p

and

y

or

we

have

T?BPAdrt?ID

a

and

y

for

a

certain

term

In

the

first

case

we

have

n?x

n?y

and

in

the

second

we

can

apply

the

induction

hypothesis

to

arrive

at

n?x

so

we

get

n?x

n?y

Case

x

for

certain

terms

and

Since

T?BPAdrt?ID

x

a

y

necessarily

T?BPAdrt?ID

a

y

or

T?BPAdrt?ID

a

y.

Therefore

by

the

induction

hypothesis

n?y

or

n?y

In

both

cases

n?x

n?y

Case

x

for

a

certain

term

This

is

in

contradiction

with

T?BPAdrt?ID

x

a

y

so

this

case

does

not

occur

Case

x

for

a

certain

term

Then

by

T?BPAdrt?ID

x

a

y

we

also

have

T?BPAdrt?ID

a

y

and

using

the

induction

hypothesis

n?x

n

n?y

Case

x

for

a

certain

term

Then

by

T?BPAdrt?ID

x

a

y

we

also

have

T?BPAdrt?ID

a

y

and

using

the

induction

hypothesis

n?x

n?y

-LRB-

viii

-RRB-

Suppose

that

T?BPAdrt?ID

x

oe

y.

Case

x

a

where

a

Affi

This

is

in

contradiction

with

T?BPAdrt?ID

x

oe

y

so

this

case

does

not

occur

Case

x

a

where

a

Affi

As

T?BPAdrt?ID

x

oe

y

necessarily

y

x.

Case

x

for

certain

terms

and

Then

because

T?BPAdrt?ID

x

oe

y

either

x

y

and

we

are

done

or

T?BPAdrt?ID

oe

and

y

for

some

term

In

that

case

we

can

apply

the

induction

hypothesis

to

arrive

at

so

we

get

n?x

n?y

Case

x

for

certain

terms

and

As

T?BPAdrt?ID

x

oe

y

necessarily

-LRB-

-RRB-

T?BPAdrt?ID

oe

y

or

-LRB-

-RRB-

T?BPAdrt?ID

oe

y

or

-LRB-

-RRB-

T?BPAdrt?ID

oe

oe

where

y

In

the

first

case

by

the

induction

hypothesis

y

or

n?y

If

y

we

have

n?x

n?y

otherwise

we

have

n?x

n?y

The

second

case

is

treated

analogously

In

the

third

case

by

the

induction

hypothesis

or

and

or

If

and

we

have

x

y

and

we

are

done

Otherwise

we

have

so

n?x

n?y

Case

x

for

a

certain

term

Because

T?BPAdrt?ID

x

oe

y

it

must

be

the

case

that

y

Then

we

have

n?x

n?y

n?y

Case

x

for

a

certain

term

This

is

in

contradiction

with

T?BPAdrt?ID

x

oe

y

so

this

case

does

not

occur

Case

x

for

a

certain

term

Because

T?BPAdrt?ID

x

oe

y

it

must

be

the

case

that

x

y

and

we

are

done

Theorem

-LRB-

Completeness

of

BPA?drt?ID

-RRB-

The

axiom

system

BPA?drt?ID

is

a

complete

axiomatization

of

the

set

of

closed

BPAdrt?ID

terms

modulo

bisimulation

equivalence

Proof

Suppose

that

s

t

BPAdrt?ID

t

We

will

prove

that

BPA?drt?ID

s

t

t.

By

Theorem

we

can

restrict

ourselves

to

basic

terms

s

and

t

The

proof

is

done

with

induction

on

n?s

using

Lemma

-LRB-

vii

-RRB-

-LRB-

viii

-RRB-

and

case

distinction

on

the

form

of

basic

term

s.

-LRB-

i

-RRB-

s

ffi

The

equality

that

we

must

prove

i.e.

BPA?drt?ID

ffi

t

t

is

Lemma

-LRB-

iv

-RRB-

-LRB-

ii

-RRB-

s

a

where

a

A

Then

s

a

p

Then

also

s

t

a

p

Since

s

t

and

t

are

bisimilar

also

t

a

p.

By

Lemma

-LRB-

i

-RRB-

we

have

BPA?drt?ID

t

a?t

Consider

the

following

computation

BPA?drt?ID

s

t

a

t

t.

-LRB-

iii

-RRB-

s

ffi

Then

ffi

oe

ffi

Therefore

s

t

oe

s

and

t

oe

with

s

BPAdrt?ID

With

Lemma

-LRB-

v

-RRB-

we

have

BPA?drt?ID

t

rel?t

Two

cases

need

to

be

considered

-LRB-

a

-RRB-

t

Now

s

t

oe

s

t

and

t

oe

t

so

by

Lemma

-LRB-

vi

-RRB-

we

have

BPA?drt?ID

s

t

bs

tc

and

BPA?drt?ID

t

btc

So

we

can

derive

using

Proposition

-LRB-

iii

-RRB-

and

Lemma

-LRB-

iv

-RRB-

BPA?drt?ID

s

t

bs

tc

bsc

btc

bffic

btc

ffi

btc

bffic

btc

bffi

tc

btc

t.

-LRB-

b

-RRB-

t

Now

by

Lemma

-LRB-

viii

-RRB-

n?t

Therefore

the

induction

hypothesis

is

applicable

BPA?drt?ID

ffi

Consider

the

following

computation

BPA?drt?ID

s

t

ffi

t

bffic

t

rel?ffi

oerel

bffic

t

ffi

oerel?ffi

t

oerel?ffi

t

oerel?ffi

rel?t

rel?t

rel?t

t.

-LRB-

iv

-RRB-

s

a

where

a

A

Then

s

a

p.

Therefore

s

t

a

p

and

since

s

t

BPAdrt?ID

t

t

a

p.

Using

Lemma

-LRB-

i

-RRB-

we

obtain

BPA?drt?ID

t

a

t

We

also

have

s

oe

s.

Therefore

s?t

oe

and

t

oe

From

Lemma

-LRB-

v

-RRB-

we

obtain

BPA?drt?ID

t

rel?t

Two

cases

can

be

distinguished

-LRB-

a

-RRB-

t

Now

s

t

oe

s

t

and

t

oe

t

so

by

Lemma

-LRB-

vi

-RRB-

we

have

BPA?drt?ID

s

t

bs

tc

and

BPA?drt?ID

t

btc

So

we

can

derive

using

Proposition

-LRB-

iii

-RRB-

BPA?drt?ID

s

t

bs

tc

bsc

btc

bac

btc

a

btc

bac

btc

ba

tc

btc

t.

-LRB-

b

-RRB-

t

Now

by

Lemma

-LRB-

viii

-RRB-

n?t

Therefore

the

induction

hypothesis

is

applicable

BPA?drt?ID

a

Consider

the

following

computation

BPA?drt?ID

s

t

a

t

bac

t

rel?a

oerel

bac

t

a?oerel?a

t

oerel?a

t

oerel?a

rel?t

rel?t

rel?t

t.

-LRB-

v

-RRB-

s

ffi

where

is

a

basic

term

Then

we

have

BPA?drt?ID

s

ffi

ffi

and

using

-LRB-

i

-RRB-

BPA?drt?ID

s

t

t.

-LRB-

vi

-RRB-

s

a

where

a

A

and

is

a

basic

term

Then

s

a

and

s

t

a

By

Lemma

-LRB-

ii

-RRB-

we

have

BPA?drt?ID

t

Since

s?t

BPAdrt?ID

t

we

also

have

t

a

for

some

such

that

BPAdrt?ID

By

induction

we

have

BPA?drt?ID

Consider

the

following

computation

BPA?drt?ID

s

t

a

t

a

t

t.

-LRB-

vii

-RRB-

s

ffi

where

is

a

basic

term

Then

we

have

using

Proposition

BPA?drt?ID

s

ffi

ffi

and

using

-LRB-

iii

-RRB-

BPA?drt?ID

s

t

t.

-LRB-

viii

-RRB-

s

a

where

a

A

and

is

a

basic

term

Then

s

a

and

s

t

a

Since

s

t

BPAdrt?ID

t

we

also

have

t

a

for

some

such

that

BPAdrt?ID

By

induction

we

therefore

have

BPA?drt?ID

We

also

have

s

oe

s

and

s

t

oe

s

and

t

oe

By

Lemma

-LRB-

ii

-RRB-

we

have

BPA?drt?ID

t

a

t

and

BPA?drt?ID

t

rel?t

Two

cases

can

be

distinguished

-LRB-

a

-RRB-

t

Now

s

t

oe

s

t

and

t

oe

t

so

by

Lemma

-LRB-

vi

-RRB-

we

have

BPA?drt?ID

s

t

bs

tc

and

BPA?drt?ID

t

btc

So

we

can

derive

using

Proposition

-LRB-

i

-RRB-

-LRB-

iii

-RRB-

BPA?drt?ID

s

t

bs

tc

bsc

btc

ba

btc

bac

bac

ba

btc

ba

tc

ba

tc

btc

t.

-LRB-

b

-RRB-

t

Now

by

Lemma

-LRB-

viii

-RRB-

n?t

By

the

induction

hypothesis

we

then

have

BPA?drt?ID

s

Consider

the

following

computation

BPA?drt?ID

s

t

a

t

bac

t

rel?a

oerel

bac

t

a

oerel?a

t

a

oerel?a

t

oerel?a

t

oerel?a

t

oerel?s

t

oerel?s

rel?t

oerel?s

rel?t

rel?t

t.

-LRB-

ix

-RRB-

s

where

and

are

basic

terms

Since

t

BPAdrt?ID

t

we

also

have

t

BPAdrt?ID

t

and

t

BPAdrt?ID

t.

By

induction

we

have

BPA?drt?ID

t

t

and

BPA?drt?ID

t

t

Then

BPA?drt?ID

s

t

t

t

t.

-LRB-

x

-RRB-

s

where

is

a

basic

term

Then

s

oe

and

s?t

oe

and

t

oe

for

some

such

that

BPAdrt?ID

By

induction

we

have

BPA?drt?ID

and

by

Lemma

-LRB-

v

-RRB-

we

have

BPA?drt?ID

t

rel?t

Then

BPA?drt?ID

s?t

t

rel?t

rel?t

rel?t

t.

Remark

-LRB-

Completeness

of

BPA?drt?ID

-RRB-

Completeness

of

BPA?drt?ID

is

also

claimed

-LRB-

without

proof

-RRB-

in

Theorem

of

-LSB-

-RSB-

-LRB-

where

BPA?drt?ID

is

called

BPAdrt?ID

-RRB-

Definition

-LRB-

Axioms

for

the

Ultimate

Start

Delay

-RRB-

We

define

Axioms

for

the

ultimate

start

delay

as

given

in

Table

on

the

next

page

Note

that

they

precisely

correspond

to

the

equalities

of

Proposition

-LRB-

i

-RRB-

-LRB-

iv

-RRB-

Remark

-LRB-

Proving

Soundness

and

Completeness

Indirectly

-RRB-

Next

to

proving

soundness

and

completeness

directly

-LRB-

outlined

in

Remarks

on

page

and

on

page

-RRB-

we

can

also

take

a

sound

and

complete

process

theory

P

and

replace

some

of

its

axioms

by

some

new

axioms

that

exactly

correspond

to

the

equalities

of

P

that

are

used

to

prove

the

completeness

of

P

The

resulting

new

theory

will

then

also

be

sound

and

complete

bac

a

bx

yc

bxc

y

bx

yc

bxc

byc

boerel?x?c

ffi

Table

Additional

axioms

for

the

ultimate

start

delay

For

an

example

of

this

method

see

Corollaries

and

below

Furthermore

this

method

is

used

in

the

proofs

of

Corollaries

and

Corollary

-LRB-

Soundness

of

BPAdrt?ID

-RRB-

The

set

of

closed

BPAdrt?ID

terms

modulo

bisimulation

equivalence

is

a

model

of

BPAdrt?ID

Proof

This

follows

directly

from

the

soundness

of

BPA?drt?ID

-LRB-

see

Theorem

on

page

-RRB-

and

the

fact

that

Axioms

are

derivable

in

BPA?drt?ID

-LRB-

see

Proposition

on

page

-RRB-

Corollary

-LRB-

Completeness

of

BPAdrt?ID

-RRB-

If

we

add

Axioms

of

Table

to

BPAdrt?ID

we

again

have

a

complete

axiomatization

of

the

set

of

closed

BPAdrt?ID

terms

modulo

bisimulation

equivalence

Proof

Careful

inspection

of

the

dependencies

between

the

proofs

in

this

section

reveals

that

the

proof

of

Theorem

only

relies

upon

RSP

-LRB-

USD

-RRB-

to

ensure

Proposition

-LRB-

i

-RRB-

-LRB-

iv

-RRB-

So

we

obviously

do

not

need

RSP

-LRB-

USD

-RRB-

anymore

if

we

add

the

corresponding

Axioms

Note

that

in

this

way

we

get

a

purely

equational

axiomatization

-LRB-

i.e.

without

conditional

axioms

or

principles

-RRB-

Soundness

and

Completeness

of

BPA?drt

Definition

-LRB-

Signature

of

BPA?drt

-RRB-

The

signature

of

BPA?drt

consists

of

the

undelayable

atomic

actions

faja

Ag

the

undelayable

deadlock

constant

ffi

the

immediate

deadlock

constant

ffi

the

alternative

composition

operator

the

sequential

composition

operator

the

time

unit

delay

operator

oerel

and

the

now

operator

rel

Definition

-LRB-

Axioms

of

BPA?drt

-RRB-

The

process

algebra

BPA?drt

is

axiomatized

by

the

axioms

of

BPA?drt?ID

given

in

Definition

on

page

and

Axioms

DRTSID

and

DCSID

shown

in

Table

on

the

following

page

BPA?drt

DRTSID

DCSID

oerel

ffi

ffi

DRTSID

x

ffi

x

ffi

x

ffi

rel

ffi

ffi

DCSID

Table

Axioms

for

ffi

Remark

-LRB-

versus

-RRB-

Note

that

in

a

setting

with

immediate

deadlock

we

do

not

have

anymore

that

x

ffi

x

as

ffi

ffi

ffi

We

do

however

still

have

a

ffi

a.

See

also

Remark

on

page

and

Remark

on

page

Remark

-LRB-

Derivability

of

and

-RRB-

Note

that

from

Axiom

and

DRTSID

we

can

derive

Axiom

and

as

we

have

ffi

x

oerel

ffi

x

oerel

ffi

x

oerel

ffi

ffi

and

similarly

oerel?x

ffi

oerel?x

oerel

ffi

oerel?x

ffi

oerel?x

However

we

still

choose

to

include

and

even

for

theories

that

contain

ffi

as

our

goal

is

not

to

find

a

minimal

axiomatization

but

instead

to

find

a

convenient

one

-LRB-

with

regard

to

ease

of

proofs

and

calculations

-RRB-

Definition

-LRB-

Summation

Convention

with

Respect

to

Immediate

Deadlock

-RRB-

In

a

setting

with

immediate

deadlock

we

will

use

the

convention

that

a

summation

over

the

empty

set

yields

the

immediate

deadlock

X

i

ti

ffi

Definition

-LRB-

Semantics

of

BPA?drt

-RRB-

The

semantics

of

BPA?drt

are

given

by

the

term

deduction

system

T?BPA?drt

induced

by

the

deduction

rules

for

BPA?drt?ID

given

in

Definition

on

page

minus

the

deduction

rule

oerel?x

oe

x

plus

the

deduction

rules

given

in

Table

on

the

following

page

Definition

-LRB-

Bisimulation

for

BPA?drt

-RRB-

Bisimulation

for

BPA?drt

is

defined

as

follows

a

binary

relation

R

on

closed

BPA?drt

terms

is

a

bisimulation

if

the

following

transfer

conditions

hold

for

all

closed

BPA?drt

terms

p

and

q

-LRB-

i

-RRB-

If

RS?p

q

and

T?BPA?drt

p

a

where

a

A

then

there

exists

a

closed

term

such

that

T?BPA?drt

q

a

and

ID

ffi

ID?x

ID?x

y

ID?x

ID?y

ID?x

y

ID?x

oerel?x

oe

x

Table

Deduction

rules

for

ID

-LRB-

ii

-RRB-

If

RS?p

q

and

T?BPA?drt

p

oe

then

there

exists

a

closed

term

such

that

T?BPA?drt

q

oe

and

-LRB-

iii

-RRB-

If

RS?p

q

and

T?BPA?drt

p

a

p

where

a

A

then

T?BPA?drt

q

a

p

-LRB-

iv

-RRB-

If

RS?p

q

and

T?BPA?drt

ID?p

then

T?BPA?drt

ID?q

Two

BPA?drt

terms

p

and

q

are

bisimilar

notation

p

BPA?drt

q

if

there

exists

a

bisimulation

relation

R

such

that

R?p

q

Definition

-LRB-

Bisimulation

Model

for

BPA?drt

-RRB-

The

bisimulation

model

for

BPA?drt

is

defined

in

the

same

way

as

for

BPA

Replace

BPA

by

BPA?drt

in

Definition

on

page

Definition

-LRB-

Basic

Terms

of

BPA?drt

-RRB-

We

define

oe

ffi

ffi

basic

terms

inductively

as

follows

-LRB-

i

-RRB-

Immediate

deadlock

ffi

is

a

oe

ffi

ffi

basic

term

-LRB-

ii

-RRB-

if

a

Affi

then

a

is

a

oe

ffi

ffi

basic

term

-LRB-

iii

-RRB-

if

a

Affi

and

t

is

a

oe

ffi

ffi

basic

term

then

a

t

is

a

oe

ffi

ffi

basic

term

-LRB-

iv

-RRB-

if

t

and

s

are

oe

ffi

ffi

basic

terms

then

t

s

is

a

oe

ffi

ffi

basic

term

-LRB-

v

-RRB-

if

t

is

a

oe

ffi

ffi

basic

term

then

oerel?t

is

a

oe

ffi

ffi

basic

term

From

now

on

if

we

speak

of

basic

terms

in

the

context

of

BPA?drt

we

mean

oe

ffi

ffi

basic

terms

Definition

-LRB-

Number

of

Symbols

of

a

BPA?drt

term

-RRB-

We

define

n?x

the

number

of

symbols

of

x

inductively

as

follows

-LRB-

i

-RRB-

We

define

n

ffi

-LRB-

ii

-RRB-

for

a

Affi

we

define

n?a

-LRB-

iii

-RRB-

for

closed

BPAdrt

terms

x

and

y

we

define

n?x

y

n?x

y

n?x

n?y

-LRB-

iv

-RRB-

for

a

closed

BPAdrt

term

x

we

define

n?oerel?x

n

rel?x

n?x

Theorem

-LRB-

Elimination

for

BPA?drt

-RRB-

Let

t

be

a

closed

BPA?drt

term

Then

there

is

a

basic

term

s

such

that

BPA?drt

t

s.

x

y

z

x

z

y

z

x

y

z

x

y

z

oerel?x

y

oerel?x

y

ffi

x

ffi

rel?a

a

rel?x

y

rel?x

rel?y

rel?x

y

rel?x

y

rel?oerel?x

ffi

rel

ffi

ffi

RDCSID

Table

Term

rewriting

system

for

BPA?drt

Proof

The

term

rewriting

system

of

Table

is

associated

to

BPA?drt

by

assigning

a

direction

to

the

axioms

With

the

method

of

the

lexicographical

path

ordering

it

is

easily

proven

that

this

term

rewriting

system

is

strongly

normalizing

Give

the

lexicographical

status

for

the

first

argument

and

define

the

following

well-founded

partial

ordering

on

constant

and

function

symbols

rel

ffi

oerel

We

give

the

following

reductions

for

the

rewriting

rules

and

RDCSID

ffi

x

Olpo

ffi

x

Olpo

ffi

rel

ffi

Olpo

rel

ffi

Olpo

ffi

Note

that

the

reductions

for

the

other

rewriting

rules

have

already

been

given

in

the

proofs

of

previous

elimination

theorems

Next

we

will

prove

that

the

normal

forms

of

the

closed

BPA?drt

terms

are

basic

terms

Thereto

suppose

that

s

is

a

normal

form

of

some

closed

BPA?drt

term

Furthermore

suppose

that

s

is

not

a

basic

term

Let

denote

the

smallest

subterm

of

s

which

is

not

a

basic

term

Then

we

can

prove

that

is

not

a

normal

form

by

case

analysis

We

distinguish

all

possible

cases

-LRB-

i

-RRB-

is

an

atomic

action

ffi

or

ffi

But

then

is

a

basic

term

This

is

in

contradiction

with

the

assumption

that

is

not

a

basic

term

so

this

case

does

not

occur

-LRB-

ii

-RRB-

is

of

the

form

for

basic

terms

and

With

case

analysis

on

the

structure

of

basic

term

-LRB-

a

-RRB-

If

is

ffi

then

rewriting

rule

can

be

applied

and

hence

is

not

a

normal

form

-LRB-

b

-RRB-

If

is

of

the

form

a

for

some

a

Affi

then

is

a

basic

term

and

so

is

a

basic

term

which

again

contradicts

the

assumption

that

is

not

a

basic

term

This

case

can

therefore

not

occur

-LRB-

c

-RRB-

If

is

of

the

form

a

t

for

some

a

Affi

and

some

basic

term

t

then

rewriting

rule

can

be

applied

So

is

not

a

normal

form

-LRB-

d

-RRB-

If

is

of

the

form

for

and

basic

terms

Then

rewriting

rule

is

applicable

Therefore

is

not

a

normal

form

-LRB-

e

-RRB-

If

is

of

the

form

oerel?t

for

some

basic

term

t

Then

rewriting

rule

is

applicable

So

is

not

a

normal

form

-LRB-

iii

-RRB-

is

of

the

form

for

basic

terms

and

In

this

case

would

be

a

basic

term

which

contradicts

the

assumption

that

is

not

a

basic

term

Therefore

this

case

can

not

occur

-LRB-

iv

-RRB-

is

of

the

form

oerel?t

for

some

basic

term

t

But

then

is

basic

term

too

so

the

case

does

not

occur

-LRB-

v

-RRB-

is

of

the

form

rel?t

for

some

basic

term

t

But

then

one

of

or

RDCSID

is

applicable

so

is

not

a

normal

form

In

any

case

that

can

occur

it

follows

that

is

not

a

normal

form

Since

is

a

subterm

of

s

we

conclude

that

s

is

not

a

normal

form

This

contradicts

the

assumption

that

s

is

a

normal

form

From

this

contradiction

we

conclude

that

s

is

a

basic

term

which

completes

the

proof

Remark

-LRB-

Elimination

for

BPA?drt

-RRB-

Elimination

for

a

slightly

different

version

of

BPA?drt

is

also

claimed

-LRB-

without

proof

-RRB-

in

Section

of

-LSB-

-RSB-

Theorem

-LRB-

Soundness

of

BPA?drt

-RRB-

The

set

of

closed

BPA?drt

terms

modulo

bisimulation

equivalence

is

a

model

of

BPA?drt

Proof

For

soundness

of

Axioms

and

we

refer

to

the

proof

of

soundness

of

BPA?drt?ID

To

extend

these

proofs

from

BPA?drt?ID

to

BPA?drt

we

have

to

check

that

the

bisimulations

given

in

previous

soundness

proofs

respect

the

ID

predicate

-LRB-

as

required

by

transfer

condition

-LRB-

iv

-RRB-

in

on

page

-RRB-

However

as

the

fact

that

they

do

can

be

easily

checked

we

will

not

give

details

It

remains

to

prove

soundness

of

the

axioms

from

Table

on

page

For

all

axioms

we

look

at

the

transitions

of

both

sides

at

the

same

time

Axiom

DRTSID

Take

the

relation

R

f?oerel

ffi

ffi?g

We

look

at

the

transitions

of

both

sides

at

the

same

time

We

have

oerel

ffi

and

ffi

Also

ID?oerel

ffi

and

ID?ffi

Axiom

Take

the

relation

R

s

s

s

ffi

s

s

C?BPA?drt

We

look

at

the

transitions

of

both

sides

at

the

same

time

We

have

s

ffi

p

iff

s

p

and

note

that

p

p

R.

Also

ID?s

ffi

iff

ID?s

ID

ffi

iff

ID?s

Axiom

Take

the

relation

R

ffi

s

ffi

s

C?BPA?drt

We

look

at

the

transitions

of

both

sides

at

the

same

time

We

have

ffi

s

and

ffi

Also

ID

ffi

s

and

ID

ffi

Axiom

DCSID

Take

the

relation

R

f

rel

ffi

ffi?g

We

look

at

the

transitions

of

both

sides

at

the

same

time

We

have

rel

ffi

and

ffi

Also

ID

rel

ffi

and

ID

ffi

Remark

-LRB-

Soundness

of

BPA?drt

-RRB-

Soundness

of

a

slightly

different

version

of

BPA?drt

is

also

claimed

-LRB-

without

proof

-RRB-

in

Section

of

-LSB-

-RSB-

Lemma

-LRB-

Towards

Completeness

of

BPA?drt

-RRB-

Let

x

be

a

closed

BPA?drt

term

and

let

a

A

Then

we

have

-LRB-

i

-RRB-

T?BPA?drt

x

a

p

-RRB-

BPA?drt

x

a

x

-LRB-

ii

-RRB-

T?BPA?drt

x

a

y

-RRB-

BPA?drt

x

a

y

x

-LRB-

iii

-RRB-

T?BPA?drt

ID?x

-RRB-

BPA?drt

x

ffi

-LRB-

iv

-RRB-

T?BPA?drt

ID?x

-RRB-

BPA?drt

x

ffi

x

-LRB-

v

-RRB-

T?BPA?drt

x

-RRB-

BPA?drt

x

rel?x

-LRB-

vi

-RRB-

T?BPA?drt

x

oe

y

-RRB-

BPA?drt

x

oerel?y

rel?x

-LRB-

vii

-RRB-

T?BPA?drt

x

a

y

-RRB-

n?x

n?y

-LRB-

viii

-RRB-

T?BPA?drt

x

oe

y

-RRB-

n?x

n?y

Proof

For

part

-LRB-

i

-RRB-

-LRB-

vi

-RRB-

we

assume

by

Theorem

and

Theorem

without

loss

of

generality

that

x

is

a

basic

term

and

apply

induction

on

the

structure

of

basic

terms

For

part

-LRB-

vii

-RRB-

and

-LRB-

viii

-RRB-

we

again

have

to

use

induction

on

the

general

structure

of

terms

-LRB-

i

-RRB-

Suppose

that

T?BPA?drt

x

a

p.

Case

x

ffi

This

is

in

contradiction

with

T?BPA?drt

x

a

p

so

this

case

does

not

occur

Case

x

b

where

b

Affi

Because

T?BPA?drt

x

a

p

it

must

be

the

case

that

b

a

So

we

have

BPA?drt

x

b

b

b

a

b

a

x.

Case

x

b

where

b

Affi

and

is

a

basic

term

This

is

in

contradiction

with

T?BPA?drt

x

a

p

so

this

case

does

not

occur

Case

x

where

and

are

basic

terms

As

T?BPA?drt

x

a

p

necessarily

T?BPA?drt

a

p

or

T?BPA?drt

a

p.

Therefore

by

the

induction

hypothesis

BPA?drt

a

or

BPA?drt

a

But

then

in

both

cases

BPA?drt

x

a

a

x.

Case

x

where

is

a

basic

term

This

is

in

contradiction

with

T?BPA?drt

x

a

p

so

this

case

does

not

occur

-LRB-

ii

-RRB-

Suppose

that

T?BPA?drt

x

a

y.

Case

x

ffi

This

is

in

contradiction

with

T?BPA?drt

x

a

y

so

this

case

does

not

occur

Case

x

b

where

b

Affi

This

is

in

contradiction

with

T?BPA?drt

x

a

y

so

this

case

does

not

occur

Case

x

b

where

b

Affi

and

is

a

basic

term

Then

because

T?BPA?drt

x

a

y

it

must

be

that

b

a

and

y.

So

BPA?drt

x

x

x

b

x

a

y

x.

Case

x

where

and

are

basic

terms

As

T?BPA?drt

x

a

y

necessarily

T?BPA?drt

a

y

or

T?BPA?drt

a

y.

Therefore

by

the

induction

hypothesis

BPA?drt

a

y

or

BPA?drt

a

y

But

then

in

both

cases

BPA?drt

x

a

y

a

y

x.

Case

x

where

is

a

basic

term

This

is

in

contradiction

with

T?BPA?drt

x

a

y

so

this

case

does

not

occur

-LRB-

iii

-RRB-

Suppose

that

T?BPA?drt

ID?x

Case

x

ffi

Then

we

have

BPA?drt

x

ffi

is

trivially

fulfilled

Case

x

a

where

a

Affi

This

is

in

contradiction

with

T?BPA?drt

ID?x

so

this

case

does

not

occur

Case

x

a

where

a

Affi

and

is

a

basic

term

This

is

in

contradiction

with

T?BPA?drt

ID?x

so

this

case

does

not

occur

Case

x

where

and

are

basic

terms

Then

because

T?BPA?drt

ID?x

it

must

be

the

case

that

T?BPA?drt

So

by

the

induction

hypothesis

we

have

that

BPA?drt

ffi

ffi

But

then

also

BPA?drt

x

ffi

ffi

ffi

Case

x

where

is

a

basic

term

This

is

in

contradiction

with

T?BPA?drt

ID?x

so

this

case

does

not

occur

-LRB-

iv

-RRB-

Suppose

that

T?BPA?drt

ID?x

Case

x

ffi

This

is

in

contradiction

with

T?BPA?drt

ID?x

so

this

case

does

not

occur

Case

x

a

where

a

Affi

Then

we

have

BPA?drt

x

ffi

a

ffi

a

x.

Case

x

a

where

a

Affi

and

is

a

basic

term

Then

BPA?drt

x

ffi

a

ffi

a

ffi

a

ffi

a

x.

Case

x

where

and

are

basic

terms

As

T?BPA?drt

ID?x

necessarily

T?BPA?drt

or

T?BPA?drt

Therefore

by

the

induction

hypothesis

BPA?drt

ffi

or

BPA?drt

ffi

So

in

both

cases

BPA?drt

x

ffi

ffi

x.

Case

x

where

is

a

basic

term

Then

BPA?drt

x

ffi

ffi

x.

-LRB-

v

-RRB-

Suppose

that

T?BPA?drt

x

Case

x

ffi

By

Axiom

DCSID

we

have

BPA?drt

x

ffi

rel

ffi

rel?x

Case

x

a

where

a

Affi

We

have

BPA?drt

x

a

rel?a

rel?x

Case

x

a

where

a

Affi

and

is

a

basic

term

We

have

BPA?drt

x

a

rel?a

rel?a

rel?x

Case

x

where

and

are

basic

terms

As

T?BPA?drt

x

necessarily

T?BPA?drt

and

T?BPA?drt

Therefore

by

the

induction

hypothesis

BPA?drt

and

BPA?drt

But

then

also

BPA?drt

x

rel?x

Case

x

where

is

a

basic

term

with

ID?x

This

is

in

contradiction

with

T?BPA?drt

x

so

this

case

does

not

occur

Case

x

where

is

a

basic

term

Then

by

T?BPA?drt

x

it

must

be

the

case

that

So

by

-LRB-

iii

-RRB-

we

have

that

BPA?drt

ffi

Therefore

BPA?drt

x

oerel

ffi

ffi

rel?x

-LRB-

vi

-RRB-

Suppose

that

T?BPA?drt

x

oe

y.

Case

x

ffi

This

is

in

contradiction

with

T?BPA?drt

x

oe

y

so

this

case

does

not

occur

Case

x

a

where

a

Affi

This

is

in

contradiction

with

T?BPA?drt

x

oe

y

so

this

case

does

not

occur

Case

x

a

where

a

Affi

and

is

a

basic

term

This

is

in

contradiction

with

T?BPA?drt

x

oe

y

so

this

case

does

not

occur

Case

x

where

and

are

basic

terms

As

T?BPA?drt

x

oe

y

necessarily

-LRB-

-RRB-

T?BPA?drt

oe

y

or

-LRB-

-RRB-

T?BPA?drt

oe

y

or

-LRB-

-RRB-

T?BPA?drt

oe

oe

where

y

In

the

first

case

by

the

induction

hypothesis

we

have

BPA?drt

oerel?y

and

by

-LRB-

v

-RRB-

BPA?drt

Therefore

BPA?drt

x

oerel?y

oerel?y

oerel?y

rel?x

The

second

case

is

treated

analogously

In

the

third

case

we

have

by

the

induction

hypothesis

BPA?drt

Therefore

we

have

BPA?drt

x

oerel?y

rel?x

Case

x

where

is

a

basic

term

Because

T?BPA?drt

x

oe

y

it

must

be

the

case

that

y

So

we

have

BPA?drt

x

oerel?y

oerel?y

ffi

oerel?y

oerel?y

rel?x

-LRB-

vii

-RRB-

Suppose

that

T?BPA?drt

x

a

y.

Case

x

ffi

This

is

in

contradiction

with

T?BPA?drt

x

oe

y

so

this

case

does

not

occur

Case

x

b

where

b

Affi

This

is

in

contradiction

with

T?BPA?drt

x

a

y

so

this

case

does

not

occur

Case

x

for

certain

terms

and

Then

because

T?BPA?drt

x

a

y

we

either

have

T?BPA?drt

a

p

and

y

or

we

have

T?BPA?drt

a

and

y

for

some

term

In

the

first

case

we

have

n?x

n?y

and

in

the

second

we

can

apply

the

induction

hypothesis

to

arrive

at

so

we

get

n?x

n?y

Case

x

for

certain

terms

and

Since

T?BPA?drt

x

a

y

necessarily

T?BPA?drt

a

y

or

T?BPA?drt

a

y.

Therefore

by

the

induction

hypothesis

n?y

or

n?y

In

both

cases

n?x

n?y

Case

x

for

a

certain

term

This

is

in

contradiction

with

T?BPA?drt

x

a

y

so

this

case

does

not

occur

Case

x

for

a

certain

term

Since

T?BPA?drt

x

a

y

necessarily

T?BPA?drt

a

y.

Therefore

by

the

induction

hypothesis

n?y

So

n?x

n

n?y

-LRB-

viii

-RRB-

Suppose

that

T?BPA?drt

x

oe

y.

Case

x

ffi

This

is

in

contradiction

with

T?BPA?drt

x

oe

y

so

this

case

does

not

occur

Case

x

a

where

a

Affi

This

is

in

contradiction

with

T?BPA?drt

x

oe

y

so

this

case

does

not

occur

Case

x

for

certain

terms

and

Then

necessarily

oe

and

y

for

some

term

We

now

can

apply

the

induction

hypothesis

to

arrive

at

so

we

get

n?x

n?y

Case

x

for

certain

terms

and

As

T?BPA?drt

x

oe

y

necessarily

-LRB-

-RRB-

T?BPA?drt

oe

y

or

-LRB-

-RRB-

T?BPA?drt

oe

y

or

-LRB-

-RRB-

T?BPA?drt

oe

oe

where

y

In

the

first

case

by

the

induction

hypothesis

n?y

So

n?x

n?y

The

second

case

is

treated

analogously

In

the

third

case

by

the

induction

hypothesis

and

So

n?x

n?y

Case

x

for

a

certain

term

Because

T?BPA?drt

x

oe

y

it

must

be

the

case

that

y

Then

we

have

n?x

n?y

n?y

Case

x

for

a

certain

term

This

is

in

contradiction

with

T?BPA?drt

x

oe

y

so

this

case

does

not

occur

Theorem

-LRB-

Completeness

of

BPA?drt

-RRB-

The

axiom

system

BPA?drt

is

a

complete

axiomatization

of

the

set

of

closed

BPA?drt

terms

modulo

-LRB-

strong

-RRB-

bisimulation

equivalence

Proof

Suppose

s

t

BPA?drt

t

We

will

prove

with

induction

on

the

structure

of

basic

term

s

that

BPA?drt

s

t

t.

By

Theorem

we

can

restrict

ourselves

to

basic

terms

without

loss

of

generality

The

proof

is

done

with

induction

on

n?s

using

Lemma

-LRB-

vii

-RRB-

-LRB-

viii

-RRB-

and

case

distinction

on

the

form

of

basic

term

s.

-LRB-

i

-RRB-

s

ffi

Using

Axiom

we

have

BPA?drt

s

t

ffi

t

t

ffi

t.

-LRB-

ii

-RRB-

s

ffi

Then

since

for

u

Aoe

we

have

T?BPA?drt

s

also

T?BPA?drt

s

t

u

p

iff

T?BPA?drt

t

u

p

and

for

a

A

T?BPA?drt

s

t

a

p

iff

T?BPA?drt

t

a

p.

Furthermore

we

have

ID?s

t

since

T?BPA?drt

ID?s

Since

s

t

BPA?drt

t

we

also

have

T?BPA?drt

ID?t

Using

Lemma

-LRB-

iv

-RRB-

we

have

BPA?drt

s

t

ffi

t

t

ffi

t.

-LRB-

iii

-RRB-

s

a

where

a

A.

From

the

deduction

rules

we

have

T?BPA?drt

s

a

p

and

T?BPA?drt

s?t

a

p

Since

s?t

BPA?drt

t

we

also

have

T?BPA?drt

t

a

p.

By

Lemma

-LRB-

i

-RRB-

we

obtain

BPA?drt

t

a

t.

So

BPA?drt

s

t

a

t

t.

-LRB-

iv

-RRB-

s

ffi

where

is

a

basic

term

Then

we

have

BPA?drt

s

ffi

ffi

and

using

-LRB-

ii

-RRB-

BPA?drt

s

t

t.

-LRB-

v

-RRB-

s

a

where

a

A

and

is

a

basic

term

From

the

deduction

rules

we

obtain

T?BPA?drt

s

a

and

T?BPA?drt

s

t

a

Since

s

t

BPA?drt

t

we

then

also

have

T?BPA?drt

t

a

for

some

such

that

BPA?drt

By

the

induction

hypothesis

we

have

BPA?drt

From

Lemma

-LRB-

ii

-RRB-

we

have

BPA?drt

t

a

t.

So

BPA?drt

s

t

a

t

a

t

t.

-LRB-

vi

-RRB-

s

where

and

are

basic

terms

Since

t

BPA?drt

t

we

also

have

t

BPA?drt

t

and

t

BPA?drt

t.

By

the

induction

hypothesis

we

then

have

BPA?drt

t

t

t

t.

So

BPA?drt

s

t

t

t

t.

-LRB-

vii

-RRB-

s

where

is

a

basic

term

From

the

deduction

rules

we

have

T?BPA?drt

oe

and

since

s

t

BPA?drt

t

we

also

have

T?BPA?drt

t

oe

s

t

oe

for

some

such

that

BPA?drt

By

Lemma

-LRB-

vi

-RRB-

we

have

BPA?drt

t

rel?t

By

the

induction

hypothesis

we

have

BPA?drt

So

BPA?drt

s

t

t

rel?t

rel?t

rel?t

t.

Remark

-LRB-

Completeness

of

BPA?drt

-RRB-

Completeness

of

a

slightly

different

version

of

BPA?drt

is

also

claimed

-LRB-

without

proof

-RRB-

in

Section

of

-LSB-

-RSB-

Soundness

and

Completeness

of

BPA?drt

Definition

-LRB-

Signature

of

BPAdrt

-RRB-

The

signature

of

BPAdrt

consists

of

the

undelayable

atomic

actions

faja

Ag

the

delayable

deadlock

actions

faja

Ag

the

undelayable

deadlock

constant

ffi

the

delayable

deadlock

constant

ffi

the

immediate

deadlock

constant

ffi

the

alternative

composition

operator

the

sequential

composition

operator

the

time

unit

delay

operator

oerel

the

now

operator

rel

and

the

unbounded

start

delay

operator

b

c

Definition

-LRB-

Axioms

of

BPAdrt

-RRB-

The

process

algebra

BPAdrt

is

axiomatized

by

the

axioms

of

BPA?drt

given

in

Definition

on

page

and

Axioms

ATS

and

USD

shown

in

Table

on

page

BPAdrt

DRTSID

DCSID

ATS

USD

Definition

-LRB-

Recursion

Principle

for

BPAdrt

-RRB-

Next

to

the

axioms

mentioned

in

Definition

the

system

BPA?drt

also

contains

the

recursion

principle

RSP

-LRB-

USD

-RRB-

shown

in

Table

on

page

For

more

information

on

recursion

principles

and

their

status

with

respect

to

axioms

see

-LSB-

-RSB-

Definition

-LRB-

Semantics

of

BPAdrt

-RRB-

The

semantics

of

BPAdrt

are

given

by

the

term

deduction

system

T?BPAdrt

induced

by

the

deduction

rules

for

BPA?drt

given

in

Definition

on

page

and

the

deduction

rules

given

in

Table

on

page

Remark

-LRB-

Semantics

of

BPAdrt

-RRB-

Note

that

for

any

closed

BPAdrt

term

x

we

have

that

ID?bxc

Hence

there

is

no

deduction

rule

for

ID?bxc

Definition

-LRB-

Bisimulation

and

Bisimulation

Model

for

BPAdrt

-RRB-

Bisimulation

for

BPAdrt

and

the

corresponding

bisimulation

model

are

defined

in

the

same

way

as

for

BPA?drt

and

BPA

respectively

Replace

BPA?drt

by

BPAdrt

in

Definition

on

page

and

BPA

by

BPAdrt

in

Definition

on

page

Definition

-LRB-

Basic

Terms

of

BPAdrt

-RRB-

We

define

oe

ffi

ffi

ffi

basic

terms

inductively

as

follows

-LRB-

i

-RRB-

Immediate

deadlock

ffi

is

a

oe

ffi

ffi

ffi

basic

term

-LRB-

ii

-RRB-

If

a

Affi

then

a

and

a

are

oe

ffi

ffi

ffi

basic

terms

-LRB-

iii

-RRB-

if

a

Affi

and

t

is

a

oe

ffi

ffi

ffi

basic

term

then

a

t

and

a

t

are

oe

ffi

ffi

ffi

basic

terms

-LRB-

iv

-RRB-

if

t

and

s

are

oe

ffi

ffi

ffi

basic

terms

then

t

s

is

a

oe

ffi

ffi

ffi

basic

term

-LRB-

v

-RRB-

if

t

is

a

oe

ffi

ffi

ffi

basic

term

then

oerel?t

is

a

oe

ffi

ffi

ffi

basic

term

From

now

on

if

we

speak

of

basic

terms

in

the

context

of

BPAdrt

we

mean

oe

ffi

ffi

ffi

basic

terms

Theorem

-LRB-

General

Form

of

Basic

Terms

of

BPAdrt

-RRB-

Modulo

the

commutativity

and

associativity

of

the

and

modulo

superfluous

ffi

summands

all

basic

terms

t

of

BPAdrt

are

of

the

form

t

X

i

m

ai

si

X

j

n

bj

X

k

p

ck

uk

X

l

q

dl

X

o

r

oerel?vo

for

m

n

p

q

r

N

ai

bj

ck

dl

Affi

and

basic

terms

si

ul

and

vo

Proof

Trivial

by

inspection

of

the

definition

of

basic

terms

Definition

Observe

that

the

general

form

of

basic

terms

is

closed

under

the

formation

rules

gives

in

Definition

Remark

-LRB-

General

Form

of

Basic

Terms

of

BPAdrt

-RRB-

Note

that

the

case

t

ffi

is

generated

when

we

take

m

n

p

q

r

See

also

Definition

Definition

-LRB-

Number

of

Symbols

of

a

BPAdrt

term

-RRB-

We

define

n?x

the

number

of

symbols

of

x

inductively

as

follows

-LRB-

i

-RRB-

We

define

n

ffi

-LRB-

ii

-RRB-

for

a

Affi

we

define

n?a

n?a

-LRB-

iii

-RRB-

for

closed

BPAdrt

terms

x

and

y

we

define

n?x

y

n?x

y

n?x

n?y

-LRB-

iv

-RRB-

for

a

closed

BPAdrt

term

x

we

define

n?oerel?x

n

rel?x

n?bxc

n?x

Proposition

-LRB-

Properties

of

BPA?drt

Part

I

-RRB-

For

BPAdrt

terms

x

and

y

and

any

a

Affi

we

have

the

following

equalities

-LRB-

i

-RRB-

BPA?drt

bac

a

-LRB-

ii

-RRB-

BPA?drt

bx

yc

bxc

y

-LRB-

iii

-RRB-

BPA?drt

bx

yc

bxc

byc

-LRB-

iv

-RRB-

BPA?drt

boerel?x?c

ffi

-LRB-

v

-RRB-

BPA?drt

b?ffic

ffi

-LRB-

vi

-RRB-

BPAdrt

rel?a

a

-LRB-

vii

-RRB-

BPAdrt

bxc

ffi

bxc

Proof

The

proofs

for

equality

-LRB-

i

-RRB-

-LRB-

iv

-RRB-

and

-LRB-

vi

-RRB-

given

in

Proposition

on

page

with

respect

to

BPAdrt?ID

remain

valid

in

the

setting

of

BPAdrt

as

can

be

easily

checked

Equality

-LRB-

v

-RRB-

and

-LRB-

vii

-RRB-

do

not

appear

in

Proposition

Consider

the

following

computation

for

equality

-LRB-

v

-RRB-

BPAdrt

ffi

bffic

rel?ffi

oerel?bffic

ffi

oerel?ffi

oerel?ffi

ffi

oerel?ffi

rel

ffi

oerel?ffi

Using

RSP

-LRB-

USD

-RRB-

we

obtain

BPA?drt

ffi

b?ffic

Finally

consider

the

following

computation

for

equality

-LRB-

vii

-RRB-

BPAdrt

bxc

ffi

rel?x

oerel?bxc

ffi

rel?x

oerel?bxc

bxc

Note

the

use

of

Axiom

in

the

second

step

Proposition

-LRB-

Properties

of

BPA?drt

Part

II

-RRB-

For

any

BPAdrt

term

x

we

have

the

following

equality

BPA?drt

ffi

x

ffi

Proof

Using

Proposition

-LRB-

ii

-RRB-

we

derive

BPA?drt

ffi

x

bffic

x

bffi

xc

bffic

ffi

Lemma

-LRB-

Representation

of

BPA?drt

Terms

-RRB-

Let

t

be

a

basic

term

Then

either

-LRB-

i

-RRB-

BPA?drt

t

ffi

or

-LRB-

ii

-RRB-

BPA?drt

t

rel?t

ffi

or

-LRB-

iii

-RRB-

BPA?drt

t

btc

or

-LRB-

iv

-RRB-

there

exists

a

basic

term

s

such

that

BPA?drt

t

rel?t

oerel?s

and

n?s

n?t

Proof

Let

t

be

a

basic

term

By

Theorem

we

may

now

proceed

by

case

analysis

on

the

form

of

basic

terms

Suppose

by

Theorem

that

t

has

the

following

general

form

t

X

i

m

ai

si

X

j

n

bj

X

k

p

ck

uk

X

l

q

dl

X

o

r

oerel?vo

for

m

n

p

q

r

N

ai

bj

ck

dl

Affi

and

basic

terms

si

ul

and

vo

We

distinguish

four

cases

-LRB-

i

-RRB-

There

are

no

summands

p

q

m

n

r

-LRB-

ii

-RRB-

Every

and

at

least

one

summand

starts

with

an

undelayable

action

m?n

and

p

q

r

-LRB-

iii

-RRB-

Every

and

at

least

one

summand

starts

with

a

delayable

action

p

q

and

m

n

r

-LRB-

iv

-RRB-

Neither

of

the

above

there

are

both

summands

that

start

with

delayable

action

and

ones

the

start

with

undelayable

actions

or

there

are

summands

that

start

with

the

time

unit

delay

operator

p

q

r

and

m

n

r

As

can

be

easily

seen

this

covers

all

cases

We

now

prove

that

the

four

cases

we

distinguish

exactly

correspond

to

the

four

cases

in

the

formulation

of

the

theorem

-LRB-

i

-RRB-

We

have

p

q

m

n

r

So

by

Definition

t

ffi

and

BPA?drt

t

ffi

-LRB-

ii

-RRB-

We

have

m

n

and

p

q

r

So

t

X

i

m

ai

si

X

j

n

bj

for

m

n

N

ai

bj

Affi

and

basic

terms

si

Then

we

have

the

following

computation

BPA?drt

t

X

i

m

ai

si

X

j

n

bj

X

i

m

rel?ai

si

X

j

n

rel?bj

X

i

m

rel?ai

si

X

j

n

rel?bj

rel

@X

i

m

ai

si

X

j

n

bj

A

rel

@X

i

m

ai

si

X

j

n

bj

ffi

A

rel

@X

i

m

ai

si

X

j

n

bj

ffi

A

rel?t

ffi

rel?t

rel?ffi

rel?t

ffi

-LRB-

iii

-RRB-

We

have

p

q

and

m

n

r

So

t

X

k

p

ck

uk

X

l

q

dl

for

p

q

N

ck

dl

Affi

and

basic

terms

uk

Using

Proposition

-LRB-

i

-RRB-

-LRB-

iii

-RRB-

we

then

have

the

following

computation

BPA?drt

t

X

k

p

ck

uk

X

l

q

dl

X

k

p

bckc

uk

X

l

q

bdlc

X

k

p

bck

ukc

X

l

q

bdlc

X

k

p

ck

uk

X

l

q

dl

btc

-LRB-

iv

-RRB-

We

have

p

q

r

and

m

n

r

So

t

X

i

m

ai

si

X

j

n

bj

X

k

p

ck

uk

X

l

q

dl

X

o

r

oerel?vo

for

m

n

p

q

r

N

ai

bj

ck

dl

Affi

and

basic

terms

si

ul

and

vo

Then

we

have

the

following

computation

BPA?drt

t

X

i

m

ai

si

X

j

n

bj

X

k

p

ck

uk

X

l

q

dl

X

o

r

oerel?vo

X

i

m

rel?ai

si

X

j

n

rel?bj

X

k

p

bckc

uk

X

l

q

bdlc

X

o

r

oerel?vo

X

i

m

rel?ai

si

X

j

n

rel?bj

X

k

p

rel?ck

oerel?bckc

uk

X

l

q

rel?dl

oerel?bdlc

X

o

r

oerel?vo

X

i

m

rel?ai

si

X

j

n

rel?bj

X

k

p

rel?ck

oerel?ck

uk

X

l

q

rel?dl

oerel?dl

X

o

r

oerel?vo

X

i

m

rel?ai

si

X

j

n

rel?bj

X

k

p

rel?ck

uk

oerel?ck

uk

X

l

q

rel?dl

oerel?dl

X

o

r

oerel?vo

X

i

m

rel?ai

si

X

j

n

rel?bj

X

k

p

rel?ck

uk

oerel?ck

uk

X

l

q

rel?dl

oerel?dl

X

o

r

oerel?vo

rel

@X

i

m

ai

si

X

j

n

bj

X

k

p

ck

uk

X

l

q

dl

A

oerel

@X

k

p

ck

uk

X

l

q

dl

X

o

r

vo

A

rel

@X

i

m

ai

si

X

j

n

bj

X

k

p

ck

uk

oerel?bckc

uk

X

l

q

dl

oerel?bdlc

X

o

r

oerel?vo

oerel

@X

k

p

ck

uk

X

l

q

dl

X

o

r

vo

A

rel

@X

i

m

ai

si

X

j

n

bj

X

k

p

ck

oerel?bckc

uk

X

l

q

dl

oerel?bdlc

X

o

r

oerel?vo

oerel

@X

k

p

ck

uk

X

l

q

dl

X

o

r

vo

A

rel

@X

i

m

ai

si

X

j

n

bj

X

k

p

bckc

uk

X

l

q

bdlc

X

o

r

oerel?vo

A

oerel

@X

k

p

ck

uk

X

l

q

dl

X

o

r

vo

A

rel

@X

i

m

ai

si

X

j

n

bj

X

k

p

ck

uk

X

l

q

dl

X

o

r

oerel?vo

A

oerel

@X

k

p

ck

uk

X

l

q

dl

X

o

r

vo

A

rel?t

oerel?s

Where

we

define

s

X

k

p

ck

uk

X

l

q

dl

X

o

r

vo

Note

that

n?s

n?t

is

now

trivially

satisfied

every

summand

of

s

also

appears

as

a

subterm

of

t

and

by

m?n

r

t

must

contain

summands

that

do

not

appear

in

s.

Therefore

t

must

contain

at

least

more

symbols

than

s.

Lemma

-LRB-

Simplified

Representation

of

BPA?drt

Terms

-RRB-

Let

t

be

a

basic

term

Then

either

-LRB-

i

-RRB-

BPA?drt

t

ffi

or

-LRB-

ii

-RRB-

BPA?drt

t

t

ffi

Proof

This

lemma

follows

almost

immediately

from

Lemma

case

-LRB-

i

-RRB-

mentioned

there

corresponds

to

case

-LRB-

i

-RRB-

here

and

cases

-LRB-

ii

-RRB-

-LRB-

iv

-RRB-

mentioned

there

correspond

to

case

-LRB-

ii

-RRB-

here

We

distinguish

the

four

cases

from

Lemma

-LRB-

i

-RRB-

BPA?drt

t

ffi

-LRB-

ii

-RRB-

BPA?drt

t

rel?t

ffi

Then

we

have

using

Axiom

BPA?drt

t

rel?t

ffi

rel?t

ffi

ffi

t

ffi

-LRB-

iii

-RRB-

BPA?drt

t

btc

Then

we

have

using

Proposition

-LRB-

vii

-RRB-

BPA?drt

t

btc

btc

ffi

t

ffi

-LRB-

iv

-RRB-

BPA?drt

t

rel?t

oerel?s

Then

we

have

using

Axiom

BPA?drt

t

rel?t

oerel?s

rel?t

oerel?s

ffi

t

ffi

Remark

-LRB-

Representation

of

BPA?drt

Terms

-RRB-

The

main

use

of

Lemmata

and

will

be

in

induction

proofs

regarding

regarding

the

-LRB-

yet

to

be

treated

-RRB-

theories

PA?drt

and

ACP?drt

-LRB-

see

Sections

and

-RRB-

Theorem

-LRB-

Elimination

for

BPA?drt

-RRB-

Let

t

be

a

closed

BPAdrt

term

Then

there

is

a

basic

term

s

such

that

BPA?drt

t

s.

Proof

First

a

term

rewriting

system

is

given

Then

it

is

shown

that

this

term

rewriting

system

is

strongly

normalizing

and

that

the

normal

forms

of

the

closed

BPAdrt

terms

are

basic

terms

The

term

rewriting

system

is

given

in

Table

The

rewriting

rules

x

y

z

x

z

y

z

x

y

z

x

y

z

oerel?x

y

oerel?x

y

bac

a

RATS

bac

a

bx

yc

bxc

y

bx

yc

bxc

byc

boerel?x?c

ffi

b?ffic

ffi

ffi

x

ffi

rel?a

a

rel?x

y

rel?x

rel?y

rel?x

y

rel?x

y

rel?oerel?x

ffi

rel?a

a

rel

ffi

ffi

RDCSID

Table

Term

rewriting

system

for

BPAdrt

RATS

and

RDCSID

are

obtained

directly

from

the

axioms

The

rewriting

rules

and

are

added

to

deal

properly

with

the

recursive

definition

of

ultimate

start

delay

The

corresponding

equalities

are

derivable

from

the

axioms

as

is

shown

in

Proposition

With

the

method

of

the

lexicographical

path

ordering

it

is

shown

that

the

term

rewriting

system

is

strongly

normalizing

Thereto

the

operator

is

assigned

the

lexicographical

status

for

the

first

argument

and

the

following

well-founded

partial

ordering

on

the

signature

of

BPAdrt

is

defined

b

c

a

rel

a

oerel

We

give

the

following

reduction

for

the

rewriting

rule

b?ffic

Olpo

b?ffic

Olpo

ffi

Note

that

the

reductions

for

the

other

rewriting

rules

have

already

been

given

in

the

proofs

of

previous

elimination

theorems

It

remains

to

prove

that

every

normal

form

of

a

closed

BPAdrt

term

is

a

basic

term

Suppose

that

s

is

the

normal

form

of

a

closed

BPAdrt

term

Furthermore

suppose

that

s

is

not

a

basic

term

and

that

is

the

smallest

subterm

of

s

which

is

not

a

basic

term

We

distinguish

all

possible

cases

-LRB-

i

-RRB-

is

of

the

form

a

or

a

for

some

a

Affi

or

of

the

form

ffi

Then

is

clearly

a

basic

term

so

this

case

does

not

occur

-LRB-

ii

-RRB-

is

of

the

form

for

basic

terms

and

With

respect

to

basic

term

the

following

cases

can

be

distinguished

-LRB-

a

-RRB-

ffi

then

is

applicable

so

is

not

a

normal

form

-LRB-

b

-RRB-

a

for

some

a

Affi

Then

is

a

basic

term

This

contradicts

the

assumption

that

is

not

a

basic

term

-LRB-

c

-RRB-

a

for

some

a

Affi

Then

is

a

basic

term

and

we

have

again

a

contradiction

-LRB-

d

-RRB-

a

for

some

a

Affi

and

basic

term

Then

rewriting

rule

is

applicable

so

is

not

a

normal

form

-LRB-

e

-RRB-

a

for

some

a

Affi

and

some

basic

term

Then

rewriting

rule

is

applicable

so

is

not

a

normal

form

-LRB-

f

-RRB-

for

some

basic

terms

and

Then

rewriting

rule

is

applicable

so

is

not

a

normal

form

-LRB-

g

-RRB-

for

some

basic

term

Then

rewriting

rule

is

applicable

so

is

not

a

normal

form

-LRB-

iii

-RRB-

is

of

the

form

for

basic

terms

and

Then

is

a

basic

term

itself

so

this

case

can

not

happen

-LRB-

iv

-RRB-

is

of

the

form

for

some

basic

term

Then

again

is

a

basic

term

itself

so

this

case

can

not

happen

either

-LRB-

v

-RRB-

is

of

the

form

where

is

a

basic

term

Then

one

of

or

RDCSID

can

be

applied

so

is

not

a

normal

form

-LRB-

vi

-RRB-

is

of

the

form

for

some

basic

term

Then

one

of

RATS

or

can

be

applied

so

is

not

a

normal

form

In

every

case

is

a

basic

term

or

a

rewriting

rule

is

applicable

If

is

a

basic

term

this

contradicts

the

assumption

that

it

is

not

If

a

rewriting

rule

is

applicable

then

and

s

are

not

a

normal

form

This

contradicts

the

assumption

that

s

is

a

normal

form

From

this

contradiction

we

conclude

that

s

is

a

basic

term

Remark

-LRB-

Elimination

for

BPAdrt

-RRB-

Elimination

for

a

somewhat

different

version

of

BPAdrt

is

also

claimed

-LRB-

without

proof

-RRB-

in

Section

of

-LSB-

-RSB-

Theorem

-LRB-

Soundness

of

BPA?drt

-RRB-

The

set

of

closed

BPAdrt

terms

modulo

bisimulation

equivalence

is

a

model

of

BPA?drt

Proof

The

soundness

of

each

of

the

axioms

of

BPAdrt

has

already

been

proven

in

the

previous

soundness

theorems

so

we

do

not

repeat

those

proofs

here

Remark

-LRB-

Soundness

of

BPAdrt

-RRB-

Soundness

of

a

somewhat

different

version

of

BPAdrt

is

also

claimed

-LRB-

without

proof

-RRB-

in

Section

of

-LSB-

-RSB-

Lemma

-LRB-

Towards

Completeness

of

BPA?drt

-RRB-

Let

x

be

a

closed

BPAdrt

term

and

let

a

A

Then

we

have

-LRB-

i

-RRB-

T?BPAdrt

x

a

p

-RRB-

BPA?drt

x

a

x

-LRB-

ii

-RRB-

T?BPAdrt

x

a

y

-RRB-

BPA?drt

x

a

y

x

-LRB-

iii

-RRB-

T?BPAdrt

ID?x

-RRB-

BPA?drt

x

ffi

-LRB-

iv

-RRB-

T?BPAdrt

ID?x

-RRB-

BPA?drt

x

ffi

x

-LRB-

v

-RRB-

T?BPAdrt

x

-RRB-

BPA?drt

x

rel?x

-LRB-

vi

-RRB-

T?BPAdrt

x

oe

y

-RRB-

BPA?drt

x

oerel?y

rel?x

-LRB-

vii

-RRB-

T?BPAdrt

x

oe

x

-RRB-

BPA?drt

x

bxc

-LRB-

viii

-RRB-

T?BPAdrt

x

a

y

-RRB-

n?x

n?y

-LRB-

ix

-RRB-

T?BPAdrt

x

oe

y

-RRB-

x

y

n?x

n?y

Proof

For

part

-LRB-

i

-RRB-

-LRB-

vii

-RRB-

we

assume

by

Theorem

and

Theorem

without

loss

of

generality

that

x

is

a

basic

term

and

then

apply

induction

on

the

structure

of

basic

terms

For

part

-LRB-

viii

-RRB-

and

-LRB-

ix

-RRB-

we

again

have

to

use

induction

on

the

general

structure

of

terms

-LRB-

i

-RRB-

Suppose

that

T?BPAdrt

x

a

p.

Case

x

ffi

This

is

in

contradiction

with

T?BPAdrt

x

a

p

so

this

case

does

not

occur

Case

x

b

where

b

Affi

Because

T?BPAdrt

x

a

p

it

must

be

the

case

that

b

a

So

we

have

BPA?drt

x

b

b

b

a

b

a

x.

Case

x

b

where

b

Affi

Because

T?BPAdrt

x

a

p

it

must

be

the

case

that

b

a

So

we

have

BPA?drt

x

b

bbc

rel?b

oerel

bbc

rel?b

rel?b

oerel

bbc

b?bbc

a?b

a?x

Case

x

b

where

b

Affi

and

is

a

basic

term

This

is

in

contradiction

with

T?BPAdrt

x

a

p

so

this

case

does

not

occur

Case

x

b

where

b

Affi

and

is

a

basic

term

This

is

in

contradiction

with

T?BPAdrt

x

a

p

so

this

case

does

not

occur

Case

x

where

and

are

basic

terms

As

T?BPAdrt

x

a

p

necessarily

T?BPAdrt

a

p

or

T?BPAdrt

a

p.

Therefore

by

the

induction

hypothesis

BPA?drt

a

or

BPA?drt

a

But

then

in

both

cases

BPA?drt

x

a

a

x.

Case

x

where

is

a

basic

term

This

is

in

contradiction

with

T?BPAdrt

x

a

p

so

this

case

does

not

occur

-LRB-

ii

-RRB-

Suppose

that

T?BPAdrt

x

a

y.

Case

x

ffi

This

is

in

contradiction

with

T?BPAdrt

x

a

y

so

this

case

does

not

occur

Case

x

b

where

b

Affi

This

is

in

contradiction

with

T?BPAdrt

x

a

y

so

this

case

does

not

occur

Case

x

b

where

b

Affi

This

is

in

contradiction

with

T?BPAdrt

x

a

y

so

this

case

does

not

occur

Case

x

b

where

b

Affi

and

is

a

basic

term

Then

because

T?BPAdrt

x

a

y

it

must

be

that

b

a

and

y.

So

BPA?drt

x

x

x

b

x

a

y

x.

Case

x

b

where

b

Affi

and

is

a

basic

term

Then

because

T?BPAdrt

x

a

y

it

must

be

that

b

a

and

y.

So

BPA?drt

x

b

a

y

a

a

y

a

y

a

y

a

y

x.

Case

x

where

and

are

basic

terms

As

T?BPAdrt

x

a

y

necessarily

T?BPAdrt

a

y

or

T?BPAdrt

a

y.

Therefore

by

the

induction

hypothesis

BPA?drt

a

y

or

BPA?drt

a

y

But

then

in

both

cases

BPA?drt

x

a

y

a

y

x.

Case

x

where

is

a

basic

term

This

is

in

contradiction

with

T?BPAdrt

x

a

y

so

this

case

does

not

occur

-LRB-

iii

-RRB-

Suppose

that

T?BPAdrt

ID?x

Case

x

ffi

Then

BPA?drt

x

ffi

is

trivially

fulfilled

Case

x

a

where

a

Affi

This

is

in

contradiction

with

T?BPAdrt

ID?x

so

this

case

does

not

occur

Case

x

a

where

a

Affi

This

is

in

contradiction

with

T?BPAdrt

ID?x

so

this

case

does

not

occur

Case

x

a

where

a

Affi

and

is

a

basic

term

This

is

in

contradiction

with

T?BPAdrt

ID?x

so

this

case

does

not

occur

Case

x

a

where

a

Affi

and

is

a

basic

term

This

is

in

contradiction

with

T?BPAdrt

ID?x

so

this

case

does

not

occur

Case

x

where

and

are

basic

terms

Then

because

T?BPAdrt

ID?x

it

must

be

the

case

that

T?BPAdrt

So

by

the

induction

hypothesis

we

have

that

BPA?drt

ffi

ffi

But

then

also

BPA?drt

x

ffi

ffi

ffi

Case

x

where

is

a

basic

term

This

is

in

contradiction

with

T?BPAdrt

ID?x

so

this

case

does

not

occur

-LRB-

iv

-RRB-

Suppose

that

T?BPAdrt

ID?x

Case

x

ffi

This

is

in

contradiction

with

T?BPAdrt

ID?x

so

this

case

does

not

occur

Case

x

a

where

a

Affi

BPA?drt

x

ffi

a

ffi

a

x.

Case

x

a

where

a

Affi

Then

we

have

BPA?drt

x

ffi

a

ffi

bac

ffi

rel?a

oerel

bac

ffi

a

ffi

oerel

bac

a?oerel

bac

rel?a

oerel

bac

bac

a

x.

Case

x

where

a

Affi

and

is

a

basic

term

BPA?drt

x?ffi

a?ffi

x.

Case

x

a

where

a

Affi

and

is

a

basic

term

Then

using

Case

BPA?drt

x

ffi

ffi

ffi

a

ffi

x.

Case

x

where

and

are

basic

terms

As

T?BPAdrt

ID?x

necessarily

T?BPAdrt

or

T?BPAdrt

Therefore

by

the

induction

hypothesis

BPA?drt

ffi

or

BPA?drt

ffi

So

in

both

cases

BPA?drt

x

ffi

ffi

x.

Case

x

where

is

a

basic

term

Then

we

have

BPA?drt

x

ffi

ffi

x.

-LRB-

v

-RRB-

Suppose

that

T?BPAdrt

x

Case

x

ffi

By

Axiom

DCSID

we

have

BPA?drt

x

ffi

rel

ffi

rel?x

Case

x

a

where

a

Affi

We

have

BPA?drt

x

a

rel?a

rel?x

Case

x

a

where

a

Affi

This

is

in

contradiction

with

T?BPAdrt

x

so

this

case

does

not

occur

Case

x

a

where

a

Affi

and

is

a

basic

term

We

have

BPA?drt

x

a

rel?a

rel?a

rel?x

Case

x

a

where

a

Affi

and

is

a

basic

term

This

is

in

contradiction

with

T?BPAdrt

x

so

this

case

does

not

occur

Case

x

where

and

are

basic

terms

As

T?BPAdrt

x

necessarily

T?BPAdrt

and

T?BPAdrt

Therefore

by

the

induction

hypothesis

BPAdrt

and

BPA?drt

But

then

also

BPA?drt

x

rel?x

Case

x

where

is

a

basic

term

Then

by

T?BPAdrt

x

it

must

be

the

case

that

So

by

-LRB-

iii

-RRB-

we

have

that

BPA?drt

ffi

Therefore

BPA?drt

x

oerel

ffi

ffi

rel?x

-LRB-

vi

-RRB-

Suppose

that

T?BPAdrt

x

oe

y.

Case

x

ffi

This

is

in

contradiction

with

T?BPAdrt

x

oe

y

so

this

case

does

not

occur

Case

x

a

where

a

Affi

This

is

in

contradiction

with

T?BPAdrt

x

oe

y

so

this

case

does

not

occur

Case

x

a

where

a

Affi

Because

T?BPAdrt

x

oe

y

it

must

be

the

case

that

y

a

So

we

have

BPA?drt

x

a

bac

rel?a

oerel

bac

oerel?a

rel?a

ffi

oerel?a

rel?a

rel

oerel

bac

oerel?a

rel

rel?a

oerel

bac

oerel?a

rel

bac

oerel?a

rel?a

oerel?y

rel?x

Case

x

a

where

a

Affi

and

is

a

basic

term

This

is

in

contradiction

with

T?BPAdrt

x

oe

y

so

this

case

does

not

occur

Case

x

a

where

a

Affi

and

is

a

basic

term

Because

T?BPAdrt

x

oe

y

it

must

be

the

case

that

y

a

So

we

have

BPA?drt

x

a

bac

rel?a

oerel

bac

rel?a

oerel

bac

oerel

bac

rel?a

rel?a

oerel?y

rel?a

ffi

oerel?y

rel?a

rel

oerel

bac

oerel?y

rel

a

oerel

bac

oerel?y

rel

rel?a

oerel

bac

oerel?y

rel

bac

oerel?y

rel?a

oerel?y

rel?a

oerel?y

rel?x

Case

x

where

and

are

basic

terms

As

T?BPAdrt

x

oe

y

necessarily

-LRB-

-RRB-

T?BPAdrt

oe

y

or

-LRB-

-RRB-

T?BPAdrt

oe

y

or

-LRB-

-RRB-

T?BPAdrt

oe

oe

where

y

In

the

first

case

by

the

induction

hypothesis

we

have

BPA?drt

oerel?y

and

by

-LRB-

v

-RRB-

BPA?drt

Therefore

BPA?drt

x

oerel?y

oerel?y

oerel?y

rel?x

The

second

case

is

treated

analogously

In

the

third

case

we

have

by

the

induction

hypothesis

BPAdrt

Therefore

we

have

BPA?drt

x

oerel?y

rel?x

Case

x

where

is

a

basic

term

Because

T?BPAdrt

x

oe

y

it

must

be

the

case

that

y

So

we

have

BPA?drt

x

oerel?y

oerel?y

ffi

oerel?y

oerel?y

rel?x

-LRB-

vii

-RRB-

Suppose

that

T?BPAdrt

x

oe

x.

Case

x

ffi

This

is

in

contradiction

with

T?BPAdrt

x

oe

x

so

this

case

does

not

occur

Case

x

a

where

a

Affi

This

is

in

contradiction

with

T?BPAdrt

x

oe

x

so

this

case

does

not

occur

Case

x

a

where

a

Affi

Then

we

have

using

Proposition

-LRB-

i

-RRB-

BPA?drt

x

a

bac

bxc

Case

x

a

where

a

Affi

and

is

a

basic

term

This

is

in

contradiction

with

T?BPAdrt

x

oe

x

so

this

case

does

not

occur

Case

x

a

where

a

Affi

and

is

a

basic

term

Then

we

can

derive

using

Proposition

-LRB-

i

-RRB-

and

-LRB-

ii

-RRB-

BPA?drt

x

a

bac

ba

bxc

Case

x

where

and

are

basic

terms

Then

we

can

derive

using

Proposition

-LRB-

iii

-RRB-

and

the

induction

hypothesis

BPA?drt

x

bxc

bxc

Case

x

where

is

a

basic

term

This

is

in

contradiction

with

T?BPAdrt

x

oe

x

so

this

case

does

not

occur

-LRB-

viii

-RRB-

Suppose

that

T?BPAdrt

x

a

y.

Case

x

ffi

This

is

in

contradiction

with

T?BPAdrt

x

oe

y

so

this

case

does

not

occur

Case

x

b

where

b

Affi

This

is

in

contradiction

with

T?BPAdrt

x

a

y

so

this

case

does

not

occur

Case

x

b

where

b

Affi

This

is

in

contradiction

with

T?BPAdrt

x

a

y

so

this

case

does

not

occur

Case

x

for

certain

terms

and

Then

because

T?BPAdrt

x

a

y

we

either

have

T?BPAdrt

a

p

and

y

or

we

have

T?BPAdrt

a

and

y

for

some

term

In

the

first

case

we

have

n?x

n?y

and

in

the

second

we

can

apply

the

induction

hypothesis

to

arrive

at

so

we

get

n?x

n?y

Case

x

for

certain

terms

and

Since

T?BPAdrt

x

a

y

necessarily

T?BPAdrt

a

y

or

T?BPAdrt

a

y.

Therefore

by

the

induction

hypothesis

n?y

or

n?y

In

both

cases

n?x

n?y

Case

x

for

a

certain

term

This

is

in

contradiction

with

T?BPAdrt

x

a

y

so

this

case

does

not

occur

Case

x

for

a

certain

term

Since

T?BPAdrt

x

a

y

necessarily

T?BPAdrt

a

y.

Therefore

by

the

induction

hypothesis

n?y

So

n?x

n

n?y

Case

x

for

a

certain

term

Since

T?BPAdrt

x

a

y

necessarily

T?BPAdrt

a

y.

Therefore

by

the

induction

hypothesis

n?y

So

n?x

n?y

-LRB-

ix

-RRB-

Suppose

that

T?BPAdrt

x

oe

y.

Case

x

ffi

This

is

in

contradiction

with

T?BPAdrt

x

oe

y

so

this

case

does

not

occur

Case

x

a

where

a

Affi

This

is

in

contradiction

with

T?BPAdrt

x

oe

y

so

this

case

does

not

occur

Case

x

a

where

a

Affi

Because

T?BPAdrt

x

oe

y

it

must

be

the

case

that

x

y

and

we

are

done

Case

x

for

certain

terms

and

Then

necessarily

oe

and

y

for

some

term

We

now

can

apply

the

induction

hypothesis

to

arrive

at

so

we

get

n?x

n?y

Case

x

for

certain

terms

and

As

T?BPAdrt

x

oe

y

necessarily

-LRB-

-RRB-

T?BPAdrt

oe

y

or

-LRB-

-RRB-

T?BPAdrt

oe

y

or

-LRB-

-RRB-

T?BPAdrt

oe

oe

where

y

In

the

first

case

by

the

induction

hypothesis

n?y

So

n?x

n?y

The

second

case

is

treated

analogously

In

the

third

case

by

the

induction

hypothesis

and

So

n?x

n?y

Case

x

for

a

certain

term

Because

T?BPAdrt

x

oe

y

it

must

be

the

case

that

y

Then

we

have

n?x

n?y

n?y

Case

x

for

a

certain

term

This

is

in

contradiction

with

T?BPAdrt

x

oe

y

so

this

case

does

not

occur

Case

x

for

a

certain

term

Because

T?BPAdrt

x

oe

y

it

must

be

the

case

that

x

y

and

we

are

done

Remark

-LRB-

Towards

Completeness

of

BPA?drt

-RRB-

Note

that

Lemma

on

page

now

also

follows

as

a

corollary

from

Lemma

-LRB-

iii

-RRB-

and

-LRB-

iv

-RRB-

by

the

law

of

the

excluded

middle

Theorem

-LRB-

Completeness

of

BPA?drt

-RRB-

The

axiom

system

BPA?drt

is

a

complete

axiomatization

of

the

set

of

closed

BPAdrt

terms

modulo

bisimulation

equivalence

Proof

Suppose

that

s

t

BPAdrt

t

We

will

prove

that

BPA?drt

s

t

t.

By

Theorem

we

can

restrict

ourselves

to

basic

terms

s

and

t

The

proof

is

done

with

induction

on

n?s

using

Lemma

-LRB-

viii

-RRB-

-LRB-

ix

-RRB-

and

case

distinction

on

the

form

of

basic

term

s.

-LRB-

i

-RRB-

s

ffi

Using

Axiom

we

have

BPA?drt

s

t

ffi

t

t

ffi

t.

-LRB-

ii

-RRB-

s

ffi

Then

since

for

u

Aoe

we

have

T?BPAdrt

s

also

T?BPAdrt

s

t

u

p

iff

T?BPAdrt

t

u

p

and

for

a

A

T?BPAdrt

s

t

a

p

iff

T?BPAdrt

t

a

p.

Furthermore

we

have

ID?s

t

since

T?BPAdrt

ID?s

Since

s

t

BPAdrt

t

we

also

have

T?BPAdrt

ID?t

Using

Lemma

-LRB-

iv

-RRB-

we

have

BPA?drt

s

t

ffi

t

t

ffi

t.

-LRB-

iii

-RRB-

s

a

where

a

A.

From

the

deduction

rules

we

have

T?BPAdrt

s

a

p

and

T?BPAdrt

s?t

a

p

Since

s?t

BPAdrt

t

we

also

have

T?BPAdrt

t

a

p.

By

Lemma

-LRB-

i

-RRB-

we

obtain

BPA?drt

t

a

t.

So

BPA?drt

s

t

a

t

t.

-LRB-

iv

-RRB-

s

ffi

Then

ffi

oe

ffi

Therefore

s?t

oe

and

t

oe

with

BPAdrt

With

Lemma

-LRB-

vi

-RRB-

we

have

BPA?drt

t

rel?t

Two

cases

need

to

be

considered

-LRB-

a

-RRB-

t

Now

s

t

oe

s

t

and

t

oe

t

so

by

Lemma

-LRB-

vii

-RRB-

we

have

BPA?drt

s

t

bs

tc

and

BPA?drt

t

btc

So

we

can

derive

using

Proposition

-LRB-

i

-RRB-

and

-LRB-

iii

-RRB-

and

Lemma

-LRB-

iv

-RRB-

BPA?drt

s?t

bs

tc

bsc

btc

bffic

btc

ffi

btc

bffic

btc

bffi

tc

btc

t.

-LRB-

b

-RRB-

t

Now

by

Lemma

-LRB-

ix

-RRB-

n?t

Therefore

the

induction

hypothesis

is

applicable

BPA?drt

ffi

Consider

the

following

computation

BPA?drt

s

t

ffi

t

bffic

t

rel?ffi

oerel

bffic

t

ffi

oerel?ffi

t

oerel?ffi

t

oerel?ffi

rel?t

rel?t

rel?t

t.

-LRB-

v

-RRB-

s

a

where

a

A

Then

s

a

p.

Therefore

s

t

a

p

and

since

s

t

BPAdrt

t

t

a

p.

Using

Lemma

-LRB-

i

-RRB-

we

obtain

BPA?drt

t

a

t

We

also

have

s

oe

s.

Therefore

s

t

oe

s

and

t

oe

From

Lemma

-LRB-

vi

-RRB-

we

obtain

BPA?drt

t

rel?t

Two

cases

can

be

distinguished

-LRB-

a

-RRB-

t

Now

s

t

oe

s

t

and

t

oe

t

so

by

Lemma

-LRB-

vii

-RRB-

we

have

BPA?drt

s

t

bs

tc

and

BPA?drt

t

btc

So

we

can

derive

using

Proposition

-LRB-

i

-RRB-

and

-LRB-

iii

-RRB-

BPA?drt

s

t

bs

tc

bsc

btc

bac

btc

a

btc

bac

btc

ba

tc

btc

t.

-LRB-

b

-RRB-

t

Now

by

Lemma

-LRB-

ix

-RRB-

n?t

Therefore

the

induction

hypothesis

is

applicable

BPA?drt

a

Consider

the

following

computation

BPA?drt

s

t

a

t

bac

t

rel?a

oerel

bac

t

a?oerel?a

t

oerel?a

t

oerel?a

rel?t

rel?t

rel?t

t.

-LRB-

vi

-RRB-

s

ffi

where

is

a

basic

term

Then

we

have

BPA?drt

s

ffi

ffi

and

using

-LRB-

ii

-RRB-

BPA?drt

s

t

t.

-LRB-

vii

-RRB-

s

a

where

a

A

and

is

a

basic

term

From

the

deduction

rules

we

obtain

T?BPAdrt

s

a

and

T?BPAdrt

s

t

a

Since

s

t

BPAdrt

t

we

then

also

have

T?BPAdrt

t

a

for

some

such

that

BPAdrt

By

the

induction

hypothesis

we

have

BPA?drt

From

Lemma

-LRB-

ii

-RRB-

we

have

BPA?drt

t

a

t.

So

BPA?drt

s

t

a

t

a

t

t.

-LRB-

viii

-RRB-

s

ffi

where

is

a

basic

term

Then

we

have

using

Proposition

BPA?drt

s

ffi

ffi

and

using

-LRB-

iv

-RRB-

BPA?drt

s

t

t.

-LRB-

ix

-RRB-

s

a

where

a

A

and

is

a

basic

term

Then

s

a

and

s

t

a

Since

s

t

BPAdrt

t

we

also

have

t

a

for

some

such

that

BPAdrt

By

induction

we

therefore

have

BPA?drt

We

also

have

s

oe

s

and

s

t

oe

s

and

t

oe

By

Lemma

-LRB-

ii

-RRB-

we

have

BPA?drt

t

a

t

and

BPA?drt

t

rel?t

Two

cases

can

be

distinguished

-LRB-

a

-RRB-

t

Now

s

t

oe

s

t

and

t

oe

t

so

by

Lemma

-LRB-

vii

-RRB-

we

have

BPA?drt

s

t

bs

tc

and

BPA?drt

t

btc

So

we

can

derive

using

Proposition

-LRB-

i

-RRB-

-LRB-

iii

-RRB-

BPA?drt

s

t

bs

tc

bsc

btc

ba

btc

bac

bac

ba

btc

ba

tc

ba

tc

btc

t.

-LRB-

b

-RRB-

t

Now

by

Lemma

-LRB-

ix

-RRB-

n?t

By

the

induction

hypothesis

we

then

have

BPA?drt

s

Consider

the

following

computation

BPA?drt

s

t

a

t

bac

t

rel?a

oerel

bac

t

a

oerel?a

oerel?a

t

oerel?s

t

oerel?s

rel?t

oerel?s

rel?t

rel?t

t.

-LRB-

x

-RRB-

s

where

and

are

basic

terms

Since

t

BPAdrt

t

we

also

have

t

BPAdrt

t

and

t

BPAdrt

t.

By

the

induction

hypothesis

we

then

have

BPA?drt

t

t

t

t.

So

BPA?drt

s

t

t

t

t.

-LRB-

xi

-RRB-

s

where

is

a

basic

term

From

the

deduction

rules

we

have

T?BPAdrt

oe

and

since

s

t

BPAdrt

t

we

also

have

T?BPAdrt

t

oe

s

t

oe

for

some

such

that

BPAdrt

By

Lemma

-LRB-

vi

-RRB-

we

have

BPA?drt

t

rel?t

By

the

induction

hypothesis

we

have

BPA?drt

So

BPA?drt

s

t

t

rel?t

rel?t

rel?t

t.

Remark

-LRB-

Completeness

of

BPAdrt

-RRB-

Completeness

of

a

somewhat

different

version

of

BPAdrt

is

also

claimed

-LRB-

without

proof

-RRB-

in

Section

of

-LSB-

-RSB-

Definition

-LRB-

Axiom

for

the

Ultimate

Start

Delay

and

Immediate

Deadlock

-RRB-

We

define

Axiom

for

the

ultimate

start

delay

of

immediate

deadlock

as

shown

in

Table

Note

that

it

precisely

corresponds

to

the

equality

of

Proposition

-LRB-

v

-RRB-

b?ffic

ffi

Table

Axiom

for

b?ffic

Corollary

-LRB-

Soundness

of

BPAdrt

-RRB-

The

set

of

closed

BPAdrt

terms

modulo

bisimulation

equivalence

is

a

model

of

BPAdrt

Proof

This

follows

directly

from

the

soundness

of

BPA?drt

-LRB-

see

Theorem

on

page

-RRB-

combined

with

the

fact

that

Axioms

are

derivable

in

BPA?drt

-LRB-

see

Proposition

on

page

-RRB-

Corollary

-LRB-

Completeness

of

BPAdrt

-RRB-

If

we

add

Axioms

of

Table

on

page

and

Axiom

of

Table

to

BPAdrt

we

again

have

a

complete

axiomatization

of

the

set

of

closed

BPAdrt

terms

modulo

bisimulation

equivalence

Proof

Careful

inspection

of

the

dependencies

between

the

proofs

in

this

section

reveals

that

the

proof

of

Theorem

only

relies

upon

RSP

-LRB-

USD

-RRB-

to

ensure

Proposition

-LRB-

i

-RRB-

-LRB-

v

-RRB-

So

we

obviously

do

not

need

RSP

-LRB-

USD

-RRB-

anymore

if

we

add

the

corresponding

Axioms

Note

that

in

this

way

we

get

a

purely

equational

axiomatization

-LRB-

i.e.

without

conditional

axioms

or

principles

-RRB-

Concurrent

Process

Algebras

Introduction

In

this

section

we

prove

soundness

and

completeness

for

some

discrete-time

concurrent

process

algebras

i.e.

process

algebras

containing

a

merge

operator

We

will

not

explore

the

field

of

concurrent

process

algebras

as

thoroughly

as

we

did

for

basic

process

algebra

However

we

feel

that

we

have

shown

which

paths

to

take

in

proving

soundness

and

completeness

for

concurrent

process

algebras

The

first

algebra

we

examine

is

PA?drt?ID

which

is

basically

BPA?drt?ID

with

a

-LRB-

free

-RRB-

merge

operator

added

We

also

look

at

which

intuitively

very

similar

to

PA?drt?ID

but

only

defined

in

a

slightly

different

way

We

then

proceed

by

replacing

the

-LRB-

free

-RRB-

merge

operator

with

a

merge

operator

capable

of

communication

this

yields

ACP?drt?ID

We

again

examine

a

slightly

different

version

Finally

we

take

a

look

at

PAdrt

and

ACPdrt

These

theories

also

contain

the

ultimate

start

delay

and

the

immediate

deadlock

As

ACPdrt

combines

all

features

described

here

in

one

theory

the

completeness

result

for

ACPdrt

is

in

a

sense

the

mother

of

all

completeness

results

when

it

comes

to

discrete-time

process

algebra

's

Soundness

and

Completeness

of

PA?drt?ID

Definition

-LRB-

Signature

of

PA?drt?ID

-RRB-

The

signature

of

PA?drt?ID

consists

of

the

undelayable

atomic

actions

faja

Ag

the

undelayable

deadlock

constant

ffi

the

alternative

composition

operator

the

sequential

composition

operator

the

time

unit

delay

operator

oerel

the

now

operator

rel

the

-LRB-

free

-RRB-

merge

operator

k

and

the

left

merge

operator

k

Definition

-LRB-

Axioms

of

PA?drt?ID

-RRB-

The

process

algebra

PA?drt?ID

is

axiomatized

by

the

axioms

of

BPA?drt?ID

given

in

Definition

on

page

Axioms

shown

in

Table

and

Axioms

shown

in

Table

on

the

following

page

PA?drt?ID

x

k

y

x

k

y

y

k

x

a

k

x

a

x

a

x

k

y

a

x

k

y

x

y

k

z

x

k

z

y

k

z

Table

Axioms

for

the

-LRB-

free

-RRB-

merge

Definition

-LRB-

Semantics

of

PA?drt?ID

-RRB-

The

semantics

of

PA?drt?ID

are

given

by

the

term

deduction

system

T?PA?drt?ID

induced

by

the

deduction

rules

for

BPA?drt?ID

given

in

Definition

on

page

and

the

deduction

rules

for

the

-LRB-

free

-RRB-

merge

given

in

Table

on

the

following

page

oerel?x

k

rel?y

ffi

oerel?x

k

rel?y

oerel?z

oerel?x

k

z

Table

Additional

axioms

for

PA?drt?ID

x

a

x

k

y

a

k

y

y

a

x

k

y

a

x

k

x

a

x

k

y

a

k

y

x

a

p

x

k

y

a

y

y

a

p

x

k

y

a

x

x

a

p

x

k

y

a

y

x

oe

y

oe

x

k

y

oe

k

x

oe

y

oe

x

k

y

oe

k

Table

Deduction

rules

for

the

-LRB-

free

-RRB-

merge

Definition

-LRB-

Bisimulation

and

Bisimulation

Model

for

PA?drt?ID

-RRB-

Bisimulation

for

PA?drt?ID

and

the

corresponding

bisimulation

model

are

defined

in

the

same

way

as

for

BPA?drt?ffi

and

BPA

respectively

Replace

BPA?drt?ffi

by

PA?drt?ID

in

Definition

on

page

and

BPA

by

PA?drt?ID

in

Definition

on

page

Definition

-LRB-

Basic

Terms

of

PA?drt?ID

-RRB-

If

we

speak

of

basic

terms

in

the

context

of

PA?drt?ID

we

mean

oe

ffi

basic

terms

as

defined

in

Definition

on

page

Definition

-LRB-

Number

of

Symbols

of

a

PA?drt?ID

Term

-RRB-

We

define

n?x

the

number

of

symbols

of

x

inductively

as

follows

-LRB-

i

-RRB-

For

a

Affi

we

define

n?a

-LRB-

ii

-RRB-

for

closed

PA?drt?ID

terms

x

and

y

we

define

n?x

y

n?x

y

n?x

k

y

n?x

k

y

n?x

n?y

-LRB-

iii

-RRB-

for

a

closed

PA?drt?ID

term

x

we

define

n?oerel?x

n

rel?x

n?x

Remark

-LRB-

Proving

Elimination

using

the

Direct

Method

-RRB-

In

some

settings

the

term

rewriting

analysis

method

for

proving

elimination

mentioned

in

Remark

on

page

does

not

work

as

the

term

rewriting

system

we

arrive

at

is

not

strongly

terminating

In

these

cases

we

apply

a

direct

method

we

simply

prove

that

for

all

closed

terms

elimination

can

be

achieved

by

examining

all

possible

cases

Although

conceptually

simple

this

method

often

gives

rise

to

very

lengthy

proofs

exponentially

so

for

theories

that

contain

many

features

See

Theorem

on

the

next

page

for

an

example

of

this

method

Theorem

-LRB-

Elimination

for

PA?drt?ID

-RRB-

Let

t

be

a

closed

PA?drt?ID

term

Then

there

is

a

closed

BPA?drt?ID

term

s

such

that

PA?drt?ID

s

t.

Proof

Let

t

be

a

closed

PA?drt?ID

term

The

theorem

is

proven

by

induction

on

n?t

and

case

distinction

on

the

general

structure

of

t.

-LRB-

i

-RRB-

t

a

for

some

a

Affi

Then

t

is

a

closed

BPA?drt?ID

term

-LRB-

ii

-RRB-

t

for

closed

PA?drt?ID

terms

and

By

induction

there

are

closed

BPA?drt?ID

terms

and

such

that

PA?drt?ID

and

PA?drt?ID

But

then

also

PA?drt?ID

and

is

a

closed

BPA?drt?ID

term

-LRB-

iii

-RRB-

t

for

closed

PA?drt?ID

terms

and

This

case

is

treated

analogously

to

case

-LRB-

ii

-RRB-

-LRB-

iv

-RRB-

t

for

a

closed

PA?drt?ID

term

This

case

is

treated

analogously

to

case

-LRB-

ii

-RRB-

-LRB-

v

-RRB-

t

for

a

closed

PA?drt?ID

term

This

case

is

treated

analogously

to

case

-LRB-

ii

-RRB-

-LRB-

vi

-RRB-

t

k

for

closed

PA?drt?ID

terms

and

By

induction

there

are

closed

BPA?drt?ID

terms

and

such

that

PA?drt?ID

and

PA?drt?ID

By

Theorem

the

elimination

theorem

for

BPA?drt?ID

there

are

basic

terms

and

such

that

BPA?drt?ID

and

BPA?drt?ID

But

then

also

PA?drt?ID

PA?drt?ID

and

PA?drt?ID

k

k

We

prove

this

case

by

induction

on

the

structure

of

basic

term

-LRB-

a

-RRB-

a

for

some

a

Affi

Then

PA?drt?ID

k

k

a

k

a

and

a

is

a

closed

BPA?drt?ID

term

-LRB-

b

-RRB-

a

for

some

a

Affi

and

basic

term

Then

PA?drt?ID

k

k

a

k

a

k

By

the

induction

hypothesis

there

exists

a

closed

BPA?drt?ID

term

p

such

that

PA?drt?ID

k

p.

Then

PA?drt?ID

k

a

k

a

p

and

a

p

is

a

closed

BPA?drt?ID

term

-LRB-

c

-RRB-

for

basic

terms

and

Then

PA?drt?ID

k

k

k

k

k

By

induction

there

exist

closed

BPA?drt?ID

terms

and

such

that

PA?drt?ID

k

and

PA?drt?ID

k

Then

also

PA?drt?ID

k

k

k

k

and

is

a

closed

BPA?drt?ID

term

-LRB-

d

-RRB-

for

a

basic

term

By

Lemma

there

is

a

basic

term

such

that

either

PA?drt?ID

or

PA?drt?ID

with

With

case

analysis

we

obtain

i.

Then

PA?drt?ID

k

k

k

k

ffi

and

ffi

is

a

closed

BPA?drt?ID

term

ii

for

a

basic

term

Then

PA?drt?ID

k

k

k

k

k

By

the

induction

hypothesis

there

is

a

closed

BPA?drt?ID

term

p

such

that

PA?drt?ID

k

p

But

then

also

PA?drt?ID

k

k

oerel?p

and

oerel?p

is

a

closed

BPA?drt?ID

term

-LRB-

vii

-RRB-

t

k

for

closed

PA?drt?ID

terms

and

Then

PA?drt?ID

k

k

k

By

-LRB-

vi

-RRB-

there

are

closed

BPA?drt?ID

terms

and

such

that

PA?drt?ID

k

and

PA?drt?ID

k

But

then

also

PA?drt?ID

k

k

k

and

is

a

closed

BPA?drt?ID

term

Corollary

-LRB-

Elimination

for

PA?drt?ID

-RRB-

Let

t

be

a

closed

PA?drt?ID

term

Then

there

is

a

basic

term

s

such

that

PA?drt?ID

s

t.

Proof

This

follows

immediately

from

-LRB-

i

-RRB-

The

elimination

theorem

for

PA?drt?ID

-LRB-

see

Theorem

-RRB-

-LRB-

ii

-RRB-

the

elimination

theorem

for

BPA?drt?ID

-LRB-

see

Theorem

-RRB-

-LRB-

iii

-RRB-

the

fact

that

all

axioms

of

BPA?drt?ID

are

also

contained

in

PA?drt?ID

Remark

-LRB-

Elimination

for

PA?drt?ID

-RRB-

Elimination

for

PA?drt?ID

is

also

claimed

-LRB-

without

proof

-RRB-

in

Theorem

of

-LSB-

-RSB-

Theorem

-LRB-

Soundness

of

PA?drt?ID

-RRB-

The

set

of

closed

PA?drt?ID

terms

modulo

bisimulation

equivalence

is

a

model

of

PA?drt?ID

Proof

Soundness

is

proven

following

the

same

lines

as

in

the

previous

soundness

theorems

Axiom

Take

the

relation

R

s

s

s

k

t

t

k

s

s

k

t

s

k

t

t

k

s

s

t

C?PA?drt?ID

First

we

look

at

the

transitions

of

the

left-hand

side

-LRB-

i

-RRB-

Suppose

that

s

k

t

a

p.

First

we

look

at

the

s

k

t

t

k

s

pairs

By

inspection

of

the

deduction

rules

we

can

conclude

that

either

s

a

and

p

k

t

or

t

a

and

p

s

k

or

s

a

p

and

p

t

or

t

a

p

and

p

s.

Therefore

either

t

k

s

a

t

k

or

t

k

s

a

k

s

or

t

k

s

a

t

or

t

k

s

a

s

respectively

and

note

that

k

t

t

k

R

s

k

k

s

R

t

t

R

and

s

s

R.

Continuing

with

the

s

k

t

s

k

t

t

k

s

pairs

we

also

have

either

s

k

t

a

k

t

or

t

k

s

a

k

s

or

s

k

t

a

t

or

t

k

s

a

s.

Therefore

either

s

k

t

t

k

s

a

k

t

or

s

k

t

t

k

s

a

k

s

or

s

k

t

t

k

s

a

t

or

s

k

t

t

k

s

a

s

respectively

and

again

note

that

k

t

k

t

R

s

k

k

s

R

t

t

R

and

s

s

R.

-LRB-

ii

-RRB-

Suppose

that

s

k

t

a

p

This

case

can

not

occur

-LRB-

iii

-RRB-

Suppose

that

s

k

t

oe

p.

First

we

look

at

the

s

k

t

t

k

s

pairs

By

inspection

of

the

deduction

rules

we

can

conclude

that

s

oe

t

oe

and

p

k

Therefore

t

k

s

oe

k

and

note

that

k

k

R.

Continuing

with

the

s

k

t

s

k

t

t

k

s

pairs

we

also

have

s

k

t

oe

k

and

t

k

s

oe

k

Therefore

s

k

t

t

k

s

oe

k

k

and

note

that

k

k

k

R.

Secondly

we

look

at

the

transitions

of

the

right-hand

side

-LRB-

i

-RRB-

Suppose

that

t

k

s

a

p

This

case

is

handled

in

the

same

way

as

the

corresponding

-LRB-

sub

-RRB-

case

for

the

left-hand

side

shown

above

-LRB-

ii

-RRB-

Suppose

that

t

k

s

a

p

This

case

can

not

occur

-LRB-

iii

-RRB-

Suppose

that

t

k

s

oe

p

This

case

is

handled

in

the

same

way

as

the

corresponding

-LRB-

sub

-RRB-

case

for

the

left-hand

side

shown

above

-LRB-

iv

-RRB-

Suppose

that

s

k

t

t

k

s

a

p.

By

inspection

of

the

deduction

rules

we

can

conclude

that

either

s

a

and

p

k

t

or

t

a

and

p

k

s

or

s

a

p

and

p

t

or

t

a

p

and

p

s.

Therefore

either

s

k

t

a

k

t

or

s

k

t

a

s

k

or

s

k

t

a

t

or

s

k

t

a

s

respectively

and

note

that

k

t

k

t

R

s

k

k

s

R

t

t

R

and

s

s

R.

-LRB-

v

-RRB-

Suppose

that

s

k

t

t

k

s

a

p

This

case

can

not

occur

-LRB-

vi

-RRB-

Suppose

that

s

k

t

t

k

s

oe

p.

By

inspection

of

the

deduction

rules

we

can

conclude

that

s

oe

t

oe

and

p

k

k

Since

both

s

and

t

can

perform

a

oe

transition

we

obtain

s

k

t

oe

k

and

note

that

k

k

k

R.

Axiom

Take

the

relation

R

s

s

a

k

s

a

s

s

C?PA?drt?ID

We

look

at

the

transitions

of

both

sides

at

the

same

time

The

only

possible

transition

of

the

left-hand

side

is

a

k

s

a

s

the

only

possible

transition

of

the

right-hand

side

is

a

s

a

s

and

note

that

s

s

R.

Axiom

Take

the

relation

R

s

s

a

s

k

t

a

s

k

t

s

C?PA?drt?ID

We

look

at

the

transitions

of

both

sides

at

the

same

time

The

only

possible

transition

of

the

left-hand

side

is

a

s

k

t

a

s

k

t

the

only

possible

transition

of

the

right-hand

side

is

a

s

k

t

a

s

k

t

and

note

that

s

k

t

s

k

t

R.

Axiom

Take

the

relation

R

s

s

s

t

k

u

s

k

u

t

k

u

s

t

u

C?PA?drt?ID

First

we

look

at

the

transitions

of

the

left-hand

side

-LRB-

i

-RRB-

Suppose

s?t

k

u

a

p.

By

inspection

of

the

deduction

rules

we

can

conclude

that

either

s

a

and

p

k

u

or

t

a

and

p

k

u

or

s

a

p

and

p

u

or

t

a

p

and

p

u.

So

either

s

k

u

a

k

u

or

t

k

u

a

k

u

or

s

k

u

a

u

or

t

k

u

a

u.

respectively

Therefore

either

s

k

u

t

k

u

a

k

u

or

s

k

u?t

k

u

a

k

u

or

s

k

u?t

k

u

a

u

or

s

k

u?t

k

u

a

u

respectively

and

note

that

k

u

k

u

R

k

u

k

u

R

u

u

R

and

u

u

R.

-LRB-

ii

-RRB-

Suppose

s

t

k

u

a

p

This

case

can

not

occur

-LRB-

iii

-RRB-

Suppose

s

t

k

u

oe

p

Then

s

t

oe

u

oe

and

p

k

Then

one

of

the

following

situations

has

occurred

-LRB-

a

-RRB-

s

oe

and

t

then

s

k

u

oe

k

and

t

k

u

Therefore

s

k

u

t

k

u

oe

k

and

note

that

k

k

R.

-LRB-

b

-RRB-

s

and

t

oe

this

case

is

handled

in

the

same

way

as

the

previous

one

-LRB-

c

-RRB-

s

oe

t

oe

and

then

s

k

u

oe

k

and

t

k

u

oe

k

Therefore

s

k

u?t

k

u

oe

k

k

and

note

that

k

k

k

R.

Secondly

we

look

at

the

transitions

of

the

right-hand

side

-LRB-

i

-RRB-

Suppose

s

k

u

t

k

u

a

p.

By

inspection

of

the

deduction

rules

we

can

conclude

that

either

s

a

and

p

k

u

or

t

a

and

p

k

u

or

s

a

p

and

p

u

or

t

a

p

and

p

u.

Therefore

either

s

t

k

u

a

k

u

or

s

t

k

u

a

k

u

or

s

t

k

u

a

u

or

s

t

k

u

a

u

and

note

that

k

u

k

u

R

k

u

k

u

R

u

u

R

and

u

u

R.

-LRB-

ii

-RRB-

Suppose

s

k

u

t

k

u

a

p

This

case

can

not

occur

-LRB-

iii

-RRB-

Suppose

s

k

u

t

k

u

oe

p

Then

this

must

be

due

to

one

of

the

following

-LRB-

a

-RRB-

s

k

u

oe

p

and

t

k

u

then

s

oe

u

oe

and

p

k

Therefore

s

t

oe

and

s

t

k

u

oe

k

and

note

that

k

k

R.

-LRB-

b

-RRB-

s

k

u

and

t

k

u

oe

p

this

case

is

handled

in

the

same

way

as

the

previous

one

-LRB-

c

-RRB-

s

k

u

oe

t

k

u

oe

and

p

then

s

oe

t

oe

u

oe

k

and

k

Therefore

s?t

oe

and

s?t

k

u

oe

k

and

note

that

k

k

k

R.

Axiom

Take

the

relation

R

oerel?s

k

rel?t

ffi

s

t

C?PA?drt?ID

We

look

at

the

transitions

of

both

sides

at

the

same

time

Observe

that

there

are

no

transitions

possible

on

the

left-hand

side

oerel?s

k

rel?t

Also

for

the

righthand

side

there

are

no

transitions

possible

ffi

Axiom

Take

the

relation

R

s

s

oerel?s

k

rel?t

oerel?u

oerel?s

k

u

s

t

u

C?PA?drt?ID

We

look

at

the

transitions

of

both

sides

at

the

same

time

Observe

that

the

only

possible

transition

of

the

left-hand

side

is

oerel?s

k

rel?t

oerel?u

oe

s

k

u

and

that

the

only

possible

transition

of

the

right-hand

side

is

oerel?s

k

u

oe

s

k

u

and

note

that

s

k

u

s

k

u

R.

Remark

-LRB-

Soundness

of

PA?drt?ID

-RRB-

Soundness

of

PA?drt?ID

is

also

claimed

-LRB-

without

proof

-RRB-

in

Theorem

of

-LSB-

-RSB-

Remark

-LRB-

Conservativity

etc.

-RRB-

In

Theorem

and

below

we

use

the

concepts

conservative

extension

operationally

conservative

extension

path

format

and

sum

of

term

deduction

systems

For

a

formal

definition

of

these

concepts

see

Section

of

-LSB-

-RSB-

and

the

references

given

there

Remark

-LRB-

Proving

Completeness

using

Verhoef

's

Theorem

-RRB-

Instead

of

using

the

direct

proof

method

-LRB-

outlined

in

Remark

on

page

-RRB-

to

prove

completeness

we

can

also

use

Verhoef

's

General

Completeness

Theorem

-LSB-

-RSB-

to

derive

completeness

of

a

process

theory

P

given

the

completeness

of

a

subtheory

of

P

In

order

to

do

this

we

need

to

prove

that

P

has

the

elimination

property

for

and

that

P

is

a

conservative

extension

of

See

Theorems

and

below

for

a

example

of

this

proof

method

Furthermore

this

method

is

used

in

the

proofs

of

Theorems

and

Theorem

-LRB-

Conservativity

of

PA?drt?ID

with

respect

to

BPA?drt?ID

-RRB-

The

equational

specification

PA?drt?ID

is

a

conservative

extension

of

the

equational

specification

BPA?drt?ID

Proof

In

order

to

prove

conservativity

it

is

sufficient

to

verify

that

the

following

conditions

are

satisfied

-LRB-

i

-RRB-

Bisimulation

equivalence

is

definable

in

terms

of

predicate

and

relation

symbols

only

-LRB-

ii

-RRB-

BPA?drt?ID

is

a

complete

axiomatization

with

respect

to

the

bisimulation

equivalence

model

induced

by

T?BPA?drt?ID

-LRB-

see

Theorem

-RRB-

-LRB-

iii

-RRB-

PA?drt?ID

is

a

sound

axiomatization

with

respect

to

the

bisimulation

equivalence

model

induced

by

T?PA?drt?ID

-LRB-

see

Theorem

-RRB-

-LRB-

iv

-RRB-

T?PA?drt?ID

is

an

operationally

conservative

extension

of

T?BPA?drt?ID

And

in

order

for

T?PA?drt?ID

indeed

to

be

an

operationally

conservative

extension

of

T?BPA?drt?ID

we

must

verify

the

following

conditions

-LRB-

i

-RRB-

T?BPA?drt?ID

is

a

pure

well-founded

term

deduction

system

in

path

format

-LRB-

ii

-RRB-

T?PA?drt?ID

is

a

term

deduction

system

in

path

format

-LRB-

iii

-RRB-

T?BPA?drt?ID

T?PA?drt?ID

is

defined

That

the

above

properties

hold

can

be

trivially

checked

from

the

relevant

definitions

Theorem

-LRB-

Completeness

of

PA?drt?ID

-RRB-

The

equational

specification

PA?drt?ID

is

a

complete

axiomatization

of

the

set

of

closed

PA?drt?ID

terms

modulo

bisimulation

equivalence

Proof

By

Verhoef

's

General

Completeness

Theorem

-LRB-

see

-LSB-

-RSB-

or

Theorem

of

-LSB-

-RSB-

-RRB-

this

follows

immediately

from

-LRB-

i

-RRB-

PA?drt?ID

has

the

elimination

property

for

BPA?drt?ID

-LRB-

see

Theorem

-RRB-

-LRB-

ii

-RRB-

PA?drt?ID

is

a

conservative

extension

of

BPA?drt?ID

-LRB-

see

Theorem

-RRB-

Remark

-LRB-

Completeness

of

PA?drt?ID

-RRB-

Completeness

of

PA?drt?ID

is

also

claimed

-LRB-

without

proof

-RRB-

in

Theorem

of

-LSB-

-RSB-

Soundness

and

Completeness

of

Definition

-LRB-

Signature

of

-RRB-

The

signature

of

is

identical

to

the

signature

of

PA?drt?ID

as

given

in

Definition

it

consists

of

the

undelayable

atomic

actions

faja

Ag

the

undelayable

deadlock

constant

ffi

the

alternative

composition

operator

the

sequential

composition

operator

the

time

unit

delay

operator

oerel

the

now

operator

rel

the

-LRB-

free

-RRB-

merge

operator

k

and

the

left

merge

operator

k

Definition

-LRB-

Axioms

of

-RRB-

The

process

algebra

is

axiomatized

by

the

axioms

of

BPA?drt?ID

given

in

Definition

on

page

Axioms

shown

in

Table

on

page

and

Axioms

shown

in

Table

oerel?x

k

a

ffi

oerel?x

k

a

y

ffi

oerel?x

k

a

y

oerel?x

k

y

oerel?x

k

a

y

z

oerel?x

k

z

oerel?x

k

oerel?y

oerel?x

k

y

Table

Additional

axioms

for

Definition

-LRB-

Semantics

of

-RRB-

The

semantics

of

are

given

by

the

term

deduction

system

which

is

identical

to

the

term

deduction

system

T?PA?drt?ID

given

in

Definition

on

page

Definition

-LRB-

Bisimulation

and

Bisimulation

Model

for

-RRB-

Bisimulation

for

and

the

corresponding

bisimulation

model

are

defined

in

the

same

way

as

for

BPA?drt?ffi

and

BPA

respectively

Replace

BPA?drt?ffi

by

in

Definition

on

page

and

BPA

by

in

Definition

on

page

Definition

-LRB-

Basic

Terms

of

-RRB-

If

we

speak

of

basic

terms

in

the

context

of

we

mean

oe

ffi

basic

terms

as

defined

in

Definition

on

page

Definition

-LRB-

Number

of

Symbols

of

a

Term

-RRB-

We

define

n?x

the

number

of

symbols

of

x

inductively

as

follows

-LRB-

i

-RRB-

For

a

Affi

we

define

n?a

-LRB-

ii

-RRB-

for

closed

terms

x

and

y

we

define

n?x

y

n?x

y

n?x

k

y

n?x

k

y

n?x

n?y

-LRB-

iii

-RRB-

for

a

closed

term

x

we

define

n?oerel?x

n

rel?x

n?x

Theorem

-LRB-

Elimination

for

-RRB-

Let

t

be

a

closed

term

Then

there

is

a

closed

BPA?drt?ID

term

s

such

that

s

t.

Proof

First

a

term

rewriting

system

is

given

which

is

proven

to

be

strongly

normalizing

Thereafter

it

is

shown

that

every

normal

form

of

a

closed

term

is

a

closed

BPA?drt?ID

term

The

term

rewriting

system

used

is

given

in

Table

Note

that

we

have

added

natural

number

subscripts

n

to

the

merge

operators

in

order

to

deal

with

the

mutually

recursive

nature

of

definition

of

these

operators

For

a

description

of

this

technique

and

a

rigorous

formal

justification

of

its

use

see

Theorem

of

-LSB-

-RSB-

and

the

references

given

there

x

kn

y

x

k

n

y

y

k

n

x

a

k

n

x

a

x

a

x

k

n

y

a

x

kn

y

x

y

k

n

z

x

k

n

z

y

k

n

z

oerel?x

k

n

a

ffi

oerel?x

k

n

a

y

ffi

oerel?x

k

n

a

y

oerel?x

k

n

y

oerel?x

k

n

a

y

z

oerel?x

k

n

z

oerel?x

k

n

oerel?y

oerel?x

k

n

y

Table

Additional

rewriting

rules

for

Using

the

method

of

the

lexicographical

path

ordering

we

prove

that

the

term

rewriting

system

associated

with

is

strongly

normalizing

Thereto

the

operator

is

assigned

the

lexicographical

status

of

the

first

argument

and

the

following

well-founded

ordering

on

constant

and

function

symbols

is

used

a

oerel

k

k

k

n

kn

k

n

That

the

left-hand

side

of

every

rewriting

rule

is

bigger

than

the

right-hand

side

with

respect

to

the

ordering

Olpo

is

shown

by

the

following

reductions

x

kn

y

Olpo

x

k?n

y

Olpo

x

k?n

y

x

k?n

y

Olpo

x

k?n

y

k

n

x

k?n

y

x

k?n

y

k

n

x

k?n

y

Olpo

x

k

n

y

y

k

n

x

a

k

n

x

Olpo

a

k?n

x

Olpo

a

k?n

x

a

k?n

x

Olpo

a

x

a

x

k

n

y

Olpo

a

x

k?n

y

Olpo

a

x

k?n

y

a

x

k?n

y

Olpo

a

x

a

x

k?n

y

kn

a

x

k?n

y

Olpo

a

x

a

x

kn

y

Olpo

a

x

a

x

kn

y

Olpo

a

x

kn

y

x

y

k

n

z

Olpo

x

y

k?n

z

Olpo

x

y

k?n

z

x

y

k?n

z

Olpo

x

y

k

n

z

x

y

k

n

z

Olpo

x

k

n

z

y

k

n

z

oerel?x

k

n

a

Olpo

oerel?x

k?n

a

Olpo

ffi

oerel?x

k

n

a

y

Olpo

oerel?x

k?n

a

y

Olpo

ffi

oerel?x

k

n

a

y

Olpo

oerel?x

k?n

a

y

Olpo

oerel?x

k

n

a

y

Olpo

oerel?x

k

n

y

oerel?x

k

n

a

y

z

Olpo

oerel?x

k?n

a

y

z

Olpo

oerel?x

k

n

a

y

z

Olpo

oerel?x

k

n

z

oerel?x

k

n

oerel?y

Olpo

oerel?x

k?n

oerel?y

Olpo

oerel?oerel?x

k?n

oerel?y

Olpo

oerel?oe?rel?x

k

n

oe?rel?y

Olpo

oerel?x

k

n

y

It

remains

to

prove

that

every

normal

form

of

a

closed

term

is

a

closed

BPA?drt?ID

term

We

prove

this

as

follows

suppose

that

s

is

a

normal

form

of

a

closed

term

and

furthermore

suppose

that

s

is

not

a

closed

BPA?drt?ID

term

Now

consider

the

smallest

subterm

of

s

that

is

not

a

closed

BPA?drt?ID

term

Then

must

be

of

the

form

k

or

of

the

form

k

for

closed

BPA?drt?ID

terms

and

By

the

elimination

theorem

for

BPA?drt?ID

Theorem

we

may

assume

that

and

are

basic

terms

Now

in

the

first

case

k

clearly

rewriting

rule

is

applicable

This

contradicts

the

assumption

that

s

is

a

normal

form

so

this

case

can

not

occur

In

the

second

case

k

it

is

not

clear

at

first

sight

that

a

contradiction

can

be

derived

So

the

following

cases

have

to

be

considered

for

basic

term

-LRB-

i

-RRB-

a

for

some

a

Affi

Then

rewriting

rule

is

applicable

-LRB-

ii

-RRB-

a

for

some

a

Affi

and

closed

BPA?drt?ID

term

Clearly

rewriting

rule

is

applicable

-LRB-

iii

-RRB-

for

some

closed

BPA?drt?ID

terms

and

This

time

rewriting

rule

is

applicable

-LRB-

iv

-RRB-

for

some

closed

BPA?drt?ID

term

The

following

cases

can

be

considered

for

the

general

form

of

basic

term

-LRB-

a

-RRB-

a

for

some

a

Affi

In

this

case

rewriting

rule

is

applicable

-LRB-

b

-RRB-

for

some

a

A

and

basic

term

In

this

case

rewriting

rule

is

applicable

-LRB-

c

-RRB-

P

i

m

i

P

j

n

bj

P

k

p

k

for

m

n

p

N

ai

bj

Affi

and

i

and

k

basic

terms

In

this

case

at

least

one

of

the

rewriting

rules

and

is

applicable

In

every

case

a

rewriting

rule

is

applicable

Therefore

is

not

a

normal

form

We

see

that

in

both

cases

is

not

a

normal

form

But

this

contradicts

the

assumption

that

s

is

a

normal

form

From

this

contradiction

we

conclude

that

s

does

not

contain

a

merge

operator

Therefore

s

must

be

closed

BPA?drt?ID

term

which

had

to

be

proven

Corollary

-LRB-

Elimination

for

-RRB-

Let

t

be

a

closed

term

Then

there

is

a

basic

term

s

such

that

s

t.

Proof

This

follows

immediately

from

-LRB-

i

-RRB-

The

elimination

theorem

for

-LRB-

see

Theorem

-RRB-

-LRB-

ii

-RRB-

the

elimination

theorem

for

BPA?drt?ID

-LRB-

see

Theorem

-RRB-

-LRB-

iii

-RRB-

the

fact

that

all

axioms

of

BPA?drt?ID

are

also

contained

in

Remark

-LRB-

Elimination

for

-RRB-

Elimination

for

a

slightly

different

version

of

is

also

claimed

-LRB-

without

proof

-RRB-

in

Theorem

of

-LSB-

-RSB-

-LRB-

where

is

called

PAffidt

-RRB-

Definition

-LRB-

Ground

Equivalence

of

Equational

Specifications

-RRB-

Two

equational

specifications

and

are

called

ground

equivalent

if

they

have

identical

signatures

i.e.

and

the

same

equalities

over

closed

terms

hold

in

both

systems

i.e.

for

all

closed

terms

s

and

t

over

the

signature

-LRB-

or

-RRB-

we

have

s

t

iff

s

t.

Definition

-LRB-

Ground

Equivalence

of

Term

Deduction

Systems

-RRB-

Two

equational

specifications

and

are

called

ground

equivalent

if

they

have

identical

signatures

i.e.

and

the

same

equalities

over

closed

terms

hold

in

both

systems

i.e.

for

all

closed

terms

s

and

t

over

the

signature

-LRB-

or

-RRB-

we

have

s

t

iff

s

t.

Theorem

-LRB-

Ground

Equivalence

of

PA?drt?ID

and

-RRB-

For

all

closed

PA?drt?ID

terms

s

and

t

we

have

PA?drt?ID

s

t

if

and

only

if

s

t.

Proof

It

suffices

to

show

that

for

closed

terms

every

axiom

of

is

derivable

from

the

axioms

of

PA?drt?ID

and

vice

versa

that

for

closed

terms

every

axiom

of

PA?drt?ID

is

derivable

from

the

axioms

of

We

can

restrict

ourselves

to

the

axioms

that

are

not

shared

by

both

theories

Part

I

First

we

show

that

Axioms

of

are

derivable

in

PA?drt?ID

Axiom

PA?drt?ID

oerel?x

k

a

ffi

Consider

the

following

derivation

PA?drt?ID

oerel?x

k

a

oerel?x

k

rel?a

ffi

Axiom

PA?drt?ID

oerel?x

k

a

y

ffi

Consider

the

following

derivation

PA?drt?ID

oerel?x

k

a

y

oerel?x

k

rel?a

y

oerel?x

k

rel?a

y

ffi

Axiom

PA?drt?ID

oerel?x

k

a

y

oerel?x

k

y

By

the

Lemma

there

is

a

basic

terms

s

such

that

either

PA?drt?ID

y

rel?y

or

PA?drt?ID

y

rel?y

oerel?s

So

there

are

two

cases

to

be

distinguished

-LRB-

i

-RRB-

PA?drt?ID

y

rel?y

Then

we

have

PA?drt?ID

oerel?x

k

a

y

oerel?x

k

a

rel?y

oerel?x

k

rel?a

rel?y

oerel?x

k

rel?a

y

ffi

oerel?x

k

rel?y

oerel?x

k

y

-LRB-

ii

-RRB-

PA?drt?ID

y

rel?y

oerel?s

Then

we

have

PA?drt?ID

oerel?x

k

a

y

oerel?x

k

a

rel?y

oerel?s

oerel?x

k

rel?a

rel?y

oerel?s

oerel?x

k

rel?a

y

oerel?s

oerel?x

k

oerel?s

oerel?x

k

rel?y

oerel?s

oerel?x

k

y

Axiom

PA?drt?ID

oerel?x

k

a

y

z

oerel?x

k

z

Handled

in

the

same

way

as

the

previous

case

Axiom

PA?drt?ID

oerel?x

k

oerel?y

oerel?x

k

y

Consider

the

following

derivation

PA?drt?ID

oerel?x

k

oerel?y

oerel?x

k

ffi

oerel?y

oerel?x

k

rel?ffi

oerel?y

oerel?x

k

y

Part

II

Secondly

we

show

that

Axioms

of

PA?drt?ID

are

derivable

in

Axiom

oerel?x

k

rel?y

ffi

Use

the

general

form

of

basic

term

y.

Take

y

X

i

m

ai

ti

X

j

n

bj

X

k

p

oerel?uk

for

m

n

p

N

ai

bj

Affi

and

basic

terms

ti

and

uk

Then

we

have

oerel?x

k

rel?y

oerel?x

k

rel

@X

i

m

ai

ti

X

j

n

bj

X

k

p

oerel?uk

A

oerel?x

k

@X

i

m

rel?ai

ti

X

j

n

rel?bj

X

k

p

rel?oerel?uk

A

oerel?x

k

@X

i

m

rel?ai

ti

X

j

n

rel?bj

X

k

p

rel?oerel?uk

A

oerel?x

k

@X

i

m

ai

ti

X

j

n

bj

X

k

p

ffi

A

oerel?x

k

@X

i

m

ai

ti

X

j

n

bj

ffi

A

oerel?x

k

ffi

ffi

Axiom

oerel?x

k

rel?y

oerel?z

oerel?x

k

z

Use

the

general

form

of

basic

term

y.

Take

y

X

i

m

ai

ti

X

j

n

bj

X

k

p

oerel?uk

for

m

n

p

N

ai

bj

Affi

and

basic

terms

ti

and

uk

Then

we

have

oerel?x

k

rel?y

oerel?z

oerel?x

k

rel

@X

i

m

ai

ti

X

j

n

bj

X

k

p

oerel?uk

A

oerel?z

A

oerel?x

k

@X

i

m

rel?ai

ti

X

j

n

rel?bj

X

k

p

rel?oerel?uk

oerel?z

A

oerel?x

k

@X

i

m

rel?ai

ti

X

j

n

rel?bj

X

k

p

rel?oerel?uk

oerel?z

A

oerel?x

k

@X

i

m

ai

ti

X

j

n

bj

X

k

p

ffi

oerel?z

A

oerel?x

k

@X

i

m

ai

ti

X

j

n

bj

oerel?z

A

oerel?x

k

oerel?z

oerel?x

k

z

Theorem

-LRB-

Ground

Equivalence

of

T?PA?drt?ID

and

-RRB-

The

term

deduction

systems

T?PA?drt?ID

and

are

ground

equivalent

Proof

Both

term

deduction

systems

have

the

same

signature

and

the

same

set

of

deduction

rules

Then

it

is

trivial

that

the

same

equalities

hold

between

closed

terms

Remark

-LRB-

Proving

Soundness

and

Completeness

using

Ground

Equivalence

-RRB-

Next

to

the

direct

method

for

proving

completeness

-LRB-

outlined

in

Remark

on

page

-RRB-

the

indirect

method

-LRB-

outlined

in

Remark

on

page

or

Verhoef

's

method

-LRB-

outlined

in

Remark

on

page

-RRB-

we

can

also

derive

completeness

of

a

process

theory

P

from

the

given

completeness

of

a

sufficiently

similar

process

theory

As

it

turns

out

the

intuitive

notion

sufficiently

similar

can

be

formalized

by

the

notion

of

ground

equivalence

Using

this

method

we

can

also

derive

the

soundness

of

P.

See

Theorem

below

for

technical

details

and

the

proofs

of

Corollaries

and

for

applications

of

this

proof

method

Theorem

-LRB-

Soundness

and

Completeness

-RRB-

Let

and

be

ground

equivalent

equational

specifications

Let

and

be

ground

equivalent

term

deduction

systems

Then

is

a

sound

axiomatization

with

respect

to

the

bisimulation

equivalence

model

induced

by

if

and

only

if

is

a

sound

axiomatization

with

respect

to

the

bisimulation

equivalence

model

induced

by

Also

is

a

complete

axiomatization

with

respect

to

the

bisimulation

equivalence

model

induced

by

if

and

only

if

is

a

complete

axiomatization

with

respect

to

the

bisimulation

equivalence

model

induced

by

Proof

Let

and

be

ground

equivalent

equational

specifications

Let

and

be

ground

equivalent

term

deduction

systems

Suppose

that

is

a

sound

axiomatization

with

respect

to

the

bisimulation

equivalence

model

induced

by

Let

s

and

t

be

closed

terms

Suppose

that

s

t.

By

the

ground

equivalence

of

and

we

obtain

s

t.

Using

that

is

a

sound

axiomatization

with

respect

to

the

bisimulation

equivalence

model

induced

by

we

have

that

s

t.

By

the

ground

equivalence

of

and

it

follows

that

s

t

The

proof

in

the

other

direction

is

analogous

The

second

part

of

the

theorem

is

proven

analogously

Remark

-LRB-

Soundness

and

Completeness

versus

Elimination

-RRB-

Note

that

it

is

useless

to

extend

Theorem

to

include

an

elimination

result

as

in

general

elimination

is

used

to

prove

ground

equivalence

between

and

In

the

proof

of

Theorem

on

page

for

example

we

use

elimination

to

allow

ourselves

to

restrict

the

proof

to

basic

terms

Corollary

-LRB-

Soundness

of

-RRB-

The

set

of

closed

terms

modulo

bisimulation

equivalence

is

a

model

of

Proof

This

follows

immediately

from

the

following

observations

and

Theorem

-LRB-

i

-RRB-

PA?drt?ID

and

are

ground

equivalent

equational

specifications

-LRB-

see

Theorem

-RRB-

-LRB-

ii

-RRB-

T?PA?drt?ID

and

are

ground

equivalent

term

deduction

systems

-LRB-

see

Theorem

-RRB-

-LRB-

iii

-RRB-

Soundness

of

PA?drt?ID

-LRB-

see

Theorem

-RRB-

Remark

-LRB-

Soundness

of

-RRB-

Soundness

of

a

slightly

different

version

of

is

also

claimed

-LRB-

without

proof

-RRB-

in

Theorem

of

-LSB-

-RSB-

-LRB-

where

is

called

PAffidt

-RRB-

Corollary

-LRB-

Completeness

of

-RRB-

The

equational

specification

is

a

complete

axiomatization

of

the

set

of

closed

terms

modulo

bisimulation

equivalence

Proof

This

follows

immediately

from

the

following

observations

and

Theorem

-LRB-

i

-RRB-

PA?drt?ID

and

are

ground

equivalent

equational

specifications

-LRB-

see

Theorem

-RRB-

-LRB-

ii

-RRB-

T?PA?drt?ID

and

are

ground

equivalent

term

deduction

systems

-LRB-

see

Theorem

-RRB-

-LRB-

iii

-RRB-

Completeness

of

PA?drt?ID

-LRB-

see

Theorem

-RRB-

Remark

-LRB-

Completeness

of

-RRB-

Completeness

of

a

slightly

different

version

of

is

also

claimed

-LRB-

without

proof

-RRB-

in

Theorem

of

-LSB-

-RSB-

-LRB-

where

is

called

PAffidt

-RRB-

Soundness

and

Completeness

of

ACP?drt?ID

Definition

-LRB-

Communication

Function

-RRB-

For

this

section

and

all

sections

to

come

we

presume

the

existence

of

a

fixed

commutative

associative

complete

function

Affi

Affi

Affi

that

can

be

considered

a

parameter

of

the

respective

theories

The

function

has

to

be

strict

in

the

sense

that

it

should

always

evaluate

to

ffi

when

one

or

both

of

its

parameters

is

ffi

Definition

-LRB-

Signature

of

ACP?drt?ID

-RRB-

The

signature

of

ACP?drt?ID

consists

of

the

undelayable

atomic

actions

faja

Ag

the

undelayable

deadlock

constant

ffi

the

alternative

composition

operator

the

sequential

composition

operator

the

time

unit

delay

operator

oerel

the

now

operator

rel

the

-LRB-

communicating

-RRB-

merge

operator

k

the

left

merge

operator

k

and

the

communication

merge

operator

j

Definition

-LRB-

Axioms

of

ACP?drt?ID

-RRB-

The

process

algebra

ACP?drt?ID

is

axiomatized

by

the

axioms

of

PA?drt?ID

given

in

Definition

on

page

minus

Axiom

plus

Axioms

and

shown

in

Table

and

Axioms

shown

in

Table

on

the

next

page

ACP?drt?ID

x

k

y

x

k

y

y

k

x

x

j

y

a

j

b

x

a

j

b

x

a

x

j

b

a

j

b

x

a

x

j

b

y

a

j

b

x

k

y

oerel?x

j

oerel?y

oerel?x

j

y

x

y

j

z

x

j

z

y

j

z

x

j

y

z

x

j

y

x

j

z

a

j

b

c

if

fl?a

b

c

ffi

a

j

b

ffi

if

fl?a

b

ffi

Table

Axioms

for

the

-LRB-

communicating

-RRB-

merge

Remark

-LRB-

Axioms

-RRB-

Note

that

by

Definition

if

we

dropped

the

condition

c

ffi

in

then

would

be

a

special

case

of

and

so

would

not

be

needed

anymore

However

for

historical

reasons

we

retain

and

as

given

in

Table

Definition

-LRB-

Semantics

of

ACP?drt?ID

-RRB-

The

semantics

of

ACP?drt?ID

are

given

by

the

term

deduction

system

T?ACP?drt?ID

induced

by

the

deduction

rules

for

PA?drt?ID

given

in

Definition

on

page

and

the

deduction

rules

for

the

-LRB-

communicating

-RRB-

merge

given

in

Table

on

the

next

page

oerel?x

j

rel?y

ffi

rel?x

j

oerel?y

ffi

Table

Additional

axioms

for

ACP?drt?ID

x

a

y

b

fl?a

b

c

x

k

y

c

k

x

a

y

b

fl?a

b

c

x

j

y

c

k

x

a

y

b

p

fl?a

b

c

x

k

y

c

x

a

y

b

p

fl?a

b

c

x

j

y

c

x

a

p

y

b

fl?a

b

c

x

k

y

c

x

a

p

y

b

fl?a

b

c

x

j

y

c

x

a

p

y

b

p

fl?a

b

c

x

k

y

c

p

x

a

p

y

b

p

fl?a

b

c

x

j

y

c

p

x

oe

y

oe

x

j

y

oe

j

Table

Deduction

rules

for

the

-LRB-

communicating

-RRB-

merge

Definition

-LRB-

Bisimulation

and

Bisimulation

Model

for

ACP?drt?ID

-RRB-

Bisimulation

for

ACP?drt?ID

and

the

corresponding

bisimulation

model

are

defined

in

the

same

way

as

for

BPA?drt?ffi

and

BPA

respectively

Replace

BPA?drt?ffi

by

ACP?drt?ID

in

Definition

on

page

and

BPA

by

ACP?drt?ID

in

Definition

on

page

Definition

-LRB-

Basic

Terms

of

ACP?drt?ID

-RRB-

If

we

speak

of

basic

terms

in

the

context

of

ACP?drt?ID

we

mean

oe

ffi

basic

terms

as

defined

in

Definition

on

page

Definition

-LRB-

Number

of

Symbols

of

an

ACP?drt?ID

Term

-RRB-

We

define

n?x

the

number

of

symbols

of

x

inductively

as

follows

-LRB-

i

-RRB-

For

a

Affi

we

define

n?a

-LRB-

ii

-RRB-

for

closed

ACP?drt?ID

terms

x

and

y

we

define

n?x

y

n?x

y

n?x

k

y

n?x

k

y

n?x

j

y

n?x

n?y

-LRB-

iii

-RRB-

for

a

closed

ACP?drt?ID

term

x

we

define

n?oerel?x

n

rel?x

n?x

Theorem

-LRB-

Elimination

for

ACP?drt?ID

-RRB-

Let

t

be

a

closed

ACP?drt?ID

term

Then

there

is

a

closed

BPA?drt?ID

term

s

such

that

ACP?drt?ID

s

t.

Proof

Let

t

be

a

closed

ACP?drt?ID

term

The

theorem

is

proven

by

induction

on

n?t

and

case

distinction

on

the

general

structure

of

t.

-LRB-

i

-RRB-

t

a

for

some

a

Affi

Then

t

is

a

closed

BPA?drt?ID

term

-LRB-

ii

-RRB-

t

for

closed

ACP?drt?ID

terms

and

By

induction

there

are

closed

BPA?drt?ID

terms

and

such

that

ACP?drt?ID

and

ACP?drt?ID

But

then

also

ACP?drt?ID

and

is

a

closed

BPA?drt?ID

term

-LRB-

iii

-RRB-

t

for

closed

ACP?drt?ID

terms

and

This

case

is

treated

analogously

to

case

-LRB-

ii

-RRB-

-LRB-

iv

-RRB-

t

for

a

closed

ACP?drt?ID

term

This

case

is

treated

analogously

to

case

-LRB-

ii

-RRB-

-LRB-

v

-RRB-

t

for

a

closed

ACP?drt?ID

term

This

case

is

treated

analogously

to

case

-LRB-

ii

-RRB-

-LRB-

vi

-RRB-

t

k

for

closed

ACP?drt?ID

terms

and

By

induction

there

are

closed

BPA?drt?ID

terms

and

such

that

ACP?drt?ID

and

ACP?drt?ID

By

Theorem

the

elimination

theorem

for

BPA?drt?ID

there

are

basic

terms

and

such

that

BPA?drt?ID

and

BPA?drt?ID

But

then

also

ACP?drt?ID

ACP?drt?ID

and

ACP?drt?ID

k

k

We

prove

this

case

by

induction

on

the

structure

of

basic

term

-LRB-

a

-RRB-

a

for

some

a

Affi

Then

ACP?drt?ID

k

k

a

k

a

and

a

is

a

closed

BPA?drt?ID

term

-LRB-

b

-RRB-

a

for

some

a

Affi

and

basic

term

Then

ACP?drt?ID

k

k

a

k

a

k

By

the

induction

hypothesis

there

exists

a

closed

BPA?drt?ID

term

p

such

that

ACP?drt?ID

k

p.

Then

ACP?drt?ID

k

a

k

a

p

and

a

p

is

a

closed

BPA?drt?ID

term

-LRB-

c

-RRB-

for

basic

terms

and

Then

ACP?drt?ID

k

k

k

k

k

By

induction

there

exist

closed

BPA?drt?ID

terms

and

such

that

ACP?drt?ID

k

and

ACP?drt?ID

k

Then

also

ACP?drt?ID

k

k

k

and

is

a

closed

BPA?drt?ID

term

-LRB-

d

-RRB-

for

a

basic

term

By

Lemma

there

is

a

basic

term

such

that

either

ACP?drt?ID

or

ACP?drt?ID

with

With

case

analysis

we

obtain

i.

Then

ACP?drt?ID

k

k

k

k

ffi

and

ffi

is

a

closed

BPA?drt?ID

term

ii

for

a

basic

term

Then

ACP?drt?ID

k

k

k

k

k

k

By

the

induction

hypothesis

there

is

a

closed

BPA?drt?ID

term

p

such

that

ACP?drt?ID

k

p

But

then

also

ACP?drt?ID

k

k

oerel?p

and

oerel?p

is

a

closed

BPA?drt?ID

term

-LRB-

vii

-RRB-

t

j

for

closed

ACP?drt?ID

terms

and

By

induction

there

are

closed

BPA?drt?ID

terms

and

such

that

ACP?drt?ID

and

ACP?drt?ID

By

Theorem

the

elimination

theorem

for

BPA?drt?ID

there

are

basic

terms

and

such

that

BPA?drt?ID

and

BPA?drt?ID

But

then

also

ACP?drt?ID

ACP?drt?ID

and

ACP?drt?ID

j

j

We

prove

this

case

by

simultaneous

induction

on

the

structure

of

basic

terms

and

We

examine

all

possible

cases

-LRB-

a

-RRB-

a

and

b

for

some

a

b

Affi

Suppose

that

fl?a

b

c

Then

we

have

ACP?drt?ID

j

j

a

j

b

c

and

c

is

a

closed

BPA?drt?ID

term

-LRB-

b

-RRB-

a

and

for

some

a

b

Affi

and

some

basic

term

Suppose

that

fl?a

b

c

Then

we

have

ACP?drt?ID

j

j

a

j

b

c

and

c

is

a

closed

BPA?drt?ID

term

-LRB-

c

-RRB-

a

and

b

for

some

a

b

Affi

and

some

basic

term

This

case

is

treated

symmetrically

to

the

previous

case

-LRB-

d

-RRB-

a

and

b

for

some

a

b

Affi

and

some

basic

terms

and

Suppose

that

fl?a

b

c

Then

we

have

ACP?drt?ID

j

j

a

j

b

c

k

By

the

induction

hypothesis

there

exists

a

closed

BPA?drt?ID

term

such

that

ACP?drt?ID

k

So

ACP?drt?ID

j

c

k

c

and

c

is

a

closed

BPA?drt?ID

term

-LRB-

e

-RRB-

for

some

basic

terms

and

and

is

of

arbitrary

form

Then

ACP?drt?ID

j

j

j

j

j

By

the

induction

hypothesis

there

exist

closed

BPA?drt?ID

terms

and

such

that

ACP?drt?ID

j

and

ACP?drt?ID

j

So

we

have

ACP?drt?ID

j

j

j

and

is

a

closed

BPA?drt?ID

term

-LRB-

f

-RRB-

is

of

arbitrary

form

and

for

some

basic

terms

and

This

case

is

treated

symmetrically

to

the

previous

case

-LRB-

g

-RRB-

for

some

basic

term

and

is

of

arbitrary

form

By

Lemma

there

is

a

basic

term

such

that

either

ACP?drt?ID

or

ACP?drt?ID

with

With

case

analysis

we

obtain

i.

Then

ACP?drt?ID

j

j

j

ffi

and

ffi

is

a

closed

BPA?drt?ID

term

ii

for

a

basic

term

Then

ACP?drt?ID

j

j

j

j

j

By

the

induction

hypothesis

there

is

a

closed

BPA?drt?ID

term

p

such

that

ACP?drt?ID

j

p

But

then

also

ACP?drt?ID

j

j

oerel?p

and

oerel?p

is

a

closed

BPA?drt?ID

term

-LRB-

h

-RRB-

is

of

arbitrary

form

and

for

some

basic

term

This

case

is

treated

symmetrically

to

the

previous

case

-LRB-

viii

-RRB-

t

k

for

closed

ACP?drt?ID

terms

and

Then

ACP?drt?ID

k

k

k

j

By

-LRB-

vi

-RRB-

and

-LRB-

vii

-RRB-

there

are

closed

BPA?drt?ID

terms

and

such

that

ACP?drt?ID

k

ACP?drt?ID

k

and

ACP?drt?ID

j

But

then

also

ACP?drt?ID

k

k

k

j

and

is

a

closed

BPA?drt?ID

term

Corollary

-LRB-

Elimination

for

ACP?drt?ID

-RRB-

Let

t

be

a

closed

ACP?drt?ID

term

Then

there

is

a

basic

term

s

such

that

ACP?drt?ID

s

t.

Proof

This

follows

immediately

from

-LRB-

i

-RRB-

The

elimination

theorem

for

ACP?drt?ID

-LRB-

see

Theorem

-RRB-

-LRB-

ii

-RRB-

the

elimination

theorem

for

BPA?drt?ID

-LRB-

see

Theorem

-RRB-

-LRB-

iii

-RRB-

the

fact

that

all

axioms

of

BPA?drt?ID

are

also

contained

in

ACP?drt?ID

Remark

-LRB-

Elimination

for

ACP?drt?ID

-RRB-

Elimination

for

ACP?drt?ID

is

also

claimed

-LRB-

without

proof

-RRB-

in

Theorem

of

-LSB-

-RSB-

Theorem

-LRB-

Soundness

of

ACP?drt?ID

-RRB-

The

set

of

closed

ACP?drt?ID

terms

modulo

bisimulation

equivalence

is

a

model

of

ACP?drt?ID

Proof

We

only

prove

soundness

for

the

axioms

which

are

added

to

BPA?drt?ID

to

obtain

ACP?drt?ID

Axioms

The

proofs

for

the

soundness

of

these

axioms

with

respect

to

PA?drt?ID

that

are

given

in

Theorem

remain

valid

since

there

are

no

new

deduction

rules

dealing

with

k

Axiom

Take

the

relation

R

s

s

s

k

t

t

k

s

s

k

t

s

k

t

t

k

s

s

j

t

s

t

C?ACP?drt?ID

First

we

look

at

the

transitions

of

the

left-hand

side

-LRB-

i

-RRB-

Suppose

that

s

k

t

a

p.

First

we

look

at

the

s

k

t

t

k

s

pairs

By

inspection

of

the

deduction

rules

we

distinguish

the

following

cases

-LRB-

a

-RRB-

s

a

and

p

k

t

Then

t

k

s

a

t

k

and

k

t

t

k

R

-LRB-

b

-RRB-

t

a

and

p

s

k

Then

t

k

s

a

k

s

and

s

k

k

s

R.

-LRB-

c

-RRB-

s

a

p

and

p

t

Then

t

k

s

a

t

and

t

t

R.

-LRB-

d

-RRB-

t

a

p

and

p

s

Then

t

k

s

a

s

and

s

s

R.

-LRB-

e

-RRB-

s

b

t

c

fl?b

c

a

and

p

k

Then

t

k

s

a

k

and

k

k

R.

-LRB-

f

-RRB-

s

b

p

t

c

fl?b

c

a

and

p

Then

t

k

s

a

and

R.

-LRB-

g

-RRB-

s

b

t

c

p

fl?b

c

a

and

p

Then

t

k

s

a

and

R

We

continue

with

the

s

k

t

s

k

t

t

k

s

pairs

Distinguishing

the

same

cases

as

above

we

derive

-LRB-

a

-RRB-

s

k

t

a

k

t

Then

s

k

t

t

k

s

s

j

t

a

k

t

and

k

t

t

k

R.

-LRB-

b

-RRB-

t

k

s

a

k

s

Then

s

k

t

t

k

s

s

j

t

a

t

k

and

s

k

k

s

R.

-LRB-

c

-RRB-

s

k

t

a

t

Then

s

k

t

t

k

s

s

j

t

a

t

and

t

t

R.

-LRB-

d

-RRB-

t

k

s

a

s

Then

s

k

t

t

k

s

s

j

t

a

s

and

s

s

R.

-LRB-

e

-RRB-

sjt

a

k

Then

s

k

t?t

k

s?sjt

a

k

and

k

k

R.

-LRB-

f

-RRB-

s

j

t

a

Then

s

k

t

t

k

s

s

j

t

a

and

R.

-LRB-

g

-RRB-

s

j

t

a

Then

s

k

t

t

k

s

s

j

t

a

and

R.

-LRB-

ii

-RRB-

Suppose

that

s

k

t

a

p.

By

inspection

of

the

deduction

rules

we

can

conclude

that

s

b

p

t

c

p

and

fl?b

c

a.

Therefore

t

k

s

a

p

and

continuing

s

j

t

a

p

so

s

k

t

t

k

s

s

j

t

a

p.

-LRB-

iii

-RRB-

Suppose

that

s

k

t

oe

p.

By

inspection

of

the

deduction

rules

we

can

conclude

that

s

oe

and

t

oe

and

p

k

Therefore

t

k

s

oe

k

and

note

that

k

k

R.

Continuing

we

also

have

s

k

t

oe

k

t

k

s

oe

k

and

s

j

t

oe

j

Therefore

s

k

t

t

k

s

s

j

t

oe

k

k

j

and

note

that

k

k

k

j

R.

Secondly

we

look

at

the

transitions

of

the

right-hand

side

-LRB-

i

-RRB-

Suppose

that

t

k

s

a

p

This

case

is

handled

in

the

same

way

as

the

corresponding

-LRB-

sub

-RRB-

case

for

the

left-hand

side

shown

above

-LRB-

ii

-RRB-

Suppose

that

t

k

s

a

p

This

case

is

handled

in

the

same

way

as

the

corresponding

-LRB-

sub

-RRB-

case

for

the

left-hand

side

shown

above

-LRB-

iii

-RRB-

Suppose

that

t

k

s

oe

p

This

case

is

handled

in

the

same

way

as

the

corresponding

-LRB-

sub

-RRB-

case

for

the

left-hand

side

shown

above

-LRB-

iv

-RRB-

Suppose

that

s

k

t

t

k

s

s

j

t

a

p.

By

inspection

of

the

deduction

rules

we

distinguish

the

following

cases

-LRB-

a

-RRB-

s

a

and

p

k

t

Then

s

k

t

a

k

t

and

k

t

k

t

R.

-LRB-

b

-RRB-

t

a

and

p

k

s

Then

s

k

t

a

s

k

and

s

k

k

s

R.

-LRB-

c

-RRB-

s

a

p

and

p

t

Then

s

k

t

a

t

and

t

t

R.

-LRB-

d

-RRB-

t

a

p

and

p

s

Then

s

k

t

a

s

and

s

s

R.

-LRB-

e

-RRB-

s

b

t

c

fl?b

c

a

and

p

k

Then

s

k

t

a

k

and

k

k

R.

-LRB-

f

-RRB-

s

b

p

t

c

fl?b

c

a

and

p

Then

s

k

t

a

and

R.

-LRB-

g

-RRB-

s

b

t

c

p

fl?b

c

a

and

p

Then

s

k

t

a

and

R.

-LRB-

v

-RRB-

Suppose

that

s

k

t

t

k

s

s

j

t

a

p.

By

inspection

of

the

deduction

rules

we

can

conclude

that

s

b

p

t

c

p

and

fl?b

c

a.

Therefore

s

k

t

a

p.

-LRB-

vi

-RRB-

Suppose

that

s

k

t

t

k

s

s

j

t

oe

p.

By

inspection

of

the

deduction

rules

we

can

conclude

that

s

oe

t

oe

and

p

k

k

j

Since

both

s

and

t

can

perform

a

oe

transition

we

obtain

s

k

t

oe

k

and

note

that

k

k

k

j

R.

Axiom

Take

the

relation

R

s

s

a

j

b

s

a

j

b

s

s

C?ACP?drt?ID

We

look

at

the

transitions

of

both

sides

at

the

same

time

First

if

fl?a

b

ffi

there

are

no

transitions

possible

on

either

side

and

we

are

done

Otherwise

suppose

fl?a

b

c

Then

the

only

possible

transition

on

the

left-hand

side

is

a

s

j

b

c

s

and

the

only

possible

transition

on

the

right-hand

side

is

a

j

b

s

c

s

and

note

that

s

s

R.

Axiom

Take

the

relation

R

s

s

a

s

j

b

a

j

b

s

s

C?ACP?drt?ID

This

axiom

is

treated

symmetrically

to

the

previous

axiom

Axiom

Take

the

relation

R

s

s

a

s

j

b

t

a

j

b

s

k

t

s

t

C?ACP?drt?ID

We

look

at

the

transitions

of

both

sides

at

the

same

time

First

if

fl?a

b

ffi

there

are

no

transitions

possible

on

either

side

and

we

are

done

Otherwise

suppose

fl?a

b

c

Then

the

only

possible

transition

on

the

left-hand

side

is

a?sjb?t

c

s

k

t

and

the

only

possible

transition

on

the

right-hand

side

is

a

j

b

s

k

t

c

s

k

t

and

note

that

s

k

t

s

k

t

R.

Axiom

Take

the

relation

R

oerel?s

j

oerel?t

oerel?s

j

t

s

t

C?ACP?drt?ID

We

look

at

the

transitions

of

both

sides

at

the

same

time

The

only

possible

transition

of

the

left-hand

side

is

oerel?s

j

oerel?t

oe

s

j

t

and

the

only

possible

transition

of

the

right-hand

side

is

oerel?s

j

t

oe

s

j

t

and

note

that

s

j

t

s

j

t

R.

Axiom

Take

the

relation

R

oerel?s

j

rel?t

ffi

s

C?ACP?drt?ID

We

look

at

the

transitions

of

both

sides

at

the

same

time

Observe

that

there

are

no

transitions

possible

on

the

left-hand

side

oerel?s?j?rel?t

Also

for

the

right-hand

side

there

are

no

transitions

possible

ffi

Axiom

Take

the

relation

R

f

rel?x

j

oerel?y

ffi?g

This

axiom

is

treated

symmetrically

to

the

previous

axiom

Axiom

Take

the

relation

R

s

s

s

t

j

u

s

j

u

t

j

u

s

t

u

C?ACP?drt?ID

First

we

look

at

the

transitions

of

the

left-hand

side

-LRB-

i

-RRB-

Suppose

that

s?t?ju

a

p.

By

inspection

of

the

deduction

rules

we

distinguish

the

following

cases

-LRB-

a

-RRB-

s

b

u

c

fl?b

c

a

and

p

k

Then

s

j

u

a

k

so

also

s

j

u

t

j

u

a

k

and

note

that

k

k

R.

-LRB-

b

-RRB-

t

b

u

c

fl?b

c

a

and

p

k

This

case

is

treated

symmetrically

to

the

previous

case

-LRB-

c

-RRB-

s

b

p

u

c

fl?b

c

a

and

p

Then

s

j

u

a

so

also

s

j

u

t

j

u

a

and

note

that

R.

-LRB-

d

-RRB-

t

b

p

u

c

fl?b

c

a

and

p

This

case

is

treated

symmetrically

to

the

previous

case

-LRB-

e

-RRB-

s

b

u

c

p

fl?b

c

a

and

p

Then

s

j

u

a

so

also

s

j

u

t

j

u

a

and

note

that

R.

-LRB-

f

-RRB-

t

b

u

c

p

fl?b

c

a

and

p

This

case

is

treated

symmetrically

to

the

previous

case

-LRB-

ii

-RRB-

Suppose

that

s

t

j

u

a

p.

-LRB-

a

-RRB-

s

b

p

u

c

p

and

fl?b

c

a

Then

s

j

u

a

p

so

also

s

j

u

t

j

u

a

p.

-LRB-

b

-RRB-

t

b

p

u

c

p

and

fl?b

c

a

This

case

is

treated

symmetrically

to

the

previous

case

-LRB-

iii

-RRB-

Suppose

that

s?t?ju

oe

p.

By

inspection

of

the

deduction

rules

we

distinguish

the

following

cases

-LRB-

a

-RRB-

s

oe

t

u

oe

and

p

j

Then

s

j

u

oe

j

and

t

j

u

so

s

j

u

t

j

u

oe

j

and

note

that

j

j

R.

-LRB-

b

-RRB-

s

t

oe

u

oe

and

p

j

This

case

is

treated

symmetrically

to

the

previous

case

-LRB-

c

-RRB-

s

oe

t

oe

u

oe

and

p

j

Then

s

j

u

oe

j

and

tju

oe

so

sju?tju

oe

and

note

that

j

R.

Secondly

we

look

at

the

transitions

of

the

right-hand

side

-LRB-

i

-RRB-

Suppose

that

sju?tju

a

p.

By

inspection

of

the

deduction

rules

we

distinguish

the

following

cases

-LRB-

a

-RRB-

s

b

u

c

fl?b

c

a

and

p

k

Then

s

t

b

so

s

t

j

u

a

k

and

note

that

k

k

R.

-LRB-

b

-RRB-

t

b

u

c

fl?b

c

a

and

p

k

This

case

is

treated

symmetrically

to

the

previous

case

-LRB-

c

-RRB-

s

b

p

u

c

fl?b

c

a

and

p

Then

s

t

b

p

so

s

t

j

u

a

and

note

that

-LRB-

d

-RRB-

t

b

p

u

c

fl?b

c

a

and

p

This

case

is

treated

symmetrically

to

the

previous

case

-LRB-

e

-RRB-

s

b

u

c

p

fl?b

c

a

and

p

Then

s

t

b

so

s

t

j

u

a

and

note

that

-LRB-

f

-RRB-

t

b

u

c

p

fl?b

c

a

and

p

This

case

is

treated

symmetrically

to

the

previous

case

-LRB-

ii

-RRB-

Suppose

that

sju?tju

a

p.

By

inspection

of

the

deduction

rules

we

distinguish

the

following

cases

-LRB-

a

-RRB-

s

b

p

u

c

p

and

fl?b

c

a

Then

s

t

b

p

so

s

t

j

u

a

p.

-LRB-

b

-RRB-

t

b

p

u

c

p

and

fl?b

c

a

This

case

is

treated

symmetrically

to

the

previous

case

-LRB-

iii

-RRB-

Suppose

that

sju?tju

oe

p.

By

inspection

of

the

deduction

rules

we

distinguish

the

following

cases

-LRB-

a

-RRB-

s

oe

t

u

oe

and

p

j

Then

s

t

oe

so

s

t

j

u

oe

j

and

note

that

j

j

R.

-LRB-

b

-RRB-

s

t

oe

u

oe

and

p

j

This

case

is

treated

symmetrically

to

the

previous

case

-LRB-

c

-RRB-

s

oe

t

oe

u

oe

and

p

j

j

Then

s

t

oe

so

s?t

j

u

oe

j

and

note

that

j

j

j

R.

Axiom

Take

the

relation

R

s

s

s

j

t

u

s

j

t

s

j

u

s

t

u

C?ACP?drt?ID

This

axiom

is

treated

symmetrically

to

the

previous

axiom

Axiom

Take

the

relation

R

f?a

j

b

c?g

We

look

at

the

transitions

of

both

sides

at

the

same

time

By

the

definition

of

Axiom

we

have

that

fl?a

b

c

ffi

Then

the

only

possible

transition

of

the

lefthand

side

is

a

j

b

c

p

and

the

only

possible

transition

of

the

right-hand

side

is

c

c

p.

Axiom

Take

the

relation

R

f?a

j

b

ffi?g

We

look

at

the

transitions

of

both

sides

at

the

same

time

As

by

the

definition

of

Axiom

we

have

that

fl?a

b

ffi

there

are

no

transitions

possible

on

either

side

Remark

-LRB-

Soundness

of

ACP?drt?ID

-RRB-

Soundness

of

ACP?drt?ID

is

also

claimed

-LRB-

without

proof

-RRB-

in

Theorem

of

-LSB-

-RSB-

Theorem

-LRB-

Conservativity

of

ACP?drt?ID

with

respect

to

BPA?drt?ID

-RRB-

The

equational

specification

ACP?drt?ID

is

a

conservative

extension

of

the

equational

specification

BPA?drt?ID

Proof

In

order

to

prove

conservativity

it

is

sufficient

to

verify

that

the

following

conditions

are

satisfied

-LRB-

i

-RRB-

Bisimulation

equivalence

is

definable

in

terms

of

predicate

and

relation

symbols

only

-LRB-

ii

-RRB-

BPA?drt?ID

is

a

complete

axiomatization

with

respect

to

the

bisimulation

equivalence

model

induced

by

T?BPA?drt?ID

-LRB-

see

Theorem

-RRB-

-LRB-

iii

-RRB-

ACP?drt?ID

is

a

sound

axiomatization

with

respect

to

the

bisimulation

equivalence

model

induced

by

T?ACP?drt?ID

-LRB-

see

Theorem

-RRB-

-LRB-

iv

-RRB-

T?ACP?drt?ID

is

an

operationally

conservative

extension

of

T?BPA?drt?ID

And

in

order

for

T?ACP?drt?ID

indeed

to

be

an

operationally

conservative

extension

of

T?BPA?drt?ID

we

must

verify

the

following

conditions

-LRB-

i

-RRB-

T?BPA?drt?ID

is

a

pure

well-founded

term

deduction

system

in

path

format

-LRB-

ii

-RRB-

T?ACP?drt?ID

is

a

term

deduction

system

in

path

format

-LRB-

iii

-RRB-

T?BPA?drt?ID

T?ACP?drt?ID

is

defined

That

the

above

properties

hold

can

be

trivially

checked

from

the

relevant

definitions

Theorem

-LRB-

Completeness

of

ACP?drt?ID

-RRB-

The

equational

specification

ACP?drt?ID

is

a

complete

axiomatization

of

the

set

of

closed

ACP?drt?ID

terms

modulo

bisimulation

equivalence

Proof

By

Verhoef

's

General

Completeness

Theorem

-LRB-

see

-LSB-

-RSB-

or

Theorem

of

-LSB-

-RSB-

-RRB-

this

follows

immediately

from

-LRB-

i

-RRB-

ACP?drt?ID

has

the

elimination

property

for

BPA?drt?ID

-LRB-

see

Theorem

-RRB-

-LRB-

ii

-RRB-

ACP?drt?ID

is

a

conservative

extension

of

BPA?drt?ID

-LRB-

see

Theorem

-RRB-

Remark

-LRB-

Completeness

of

ACP?drt?ID

-RRB-

Completeness

of

ACP?drt?ID

is

also

claimed

-LRB-

without

proof

-RRB-

in

Theorem

of

-LSB-

-RSB-

Soundness

and

Completeness

of

Definition

-LRB-

Signature

of

-RRB-

The

signature

of

is

identical

to

the

signature

of

ACP?drt?ID

as

given

in

Definition

it

consists

of

the

undelayable

atomic

actions

faja

Ag

the

undelayable

deadlock

constant

ffi

the

alternative

composition

operator

the

sequential

composition

operator

the

time

unit

delay

operator

oerel

the

now

operator

rel

the

-LRB-

communicating

-RRB-

merge

operator

k

the

left

merge

operator

k

and

the

communication

merge

operator

j

Definition

-LRB-

Axioms

of

-RRB-

The

process

algebra

is

axiomatized

by

the

axioms

of

given

in

Definition

on

page

minus

Axiom

plus

Axioms

and

shown

in

Table

on

page

and

Axioms

shown

in

Table

on

the

next

page

Definition

-LRB-

Semantics

of

-RRB-

The

semantics

of

are

given

by

the

term

deduction

system

which

is

identical

to

the

term

deduction

system

T?ACP?drt?ID

given

in

Definition

on

page

Definition

-LRB-

Bisimulation

and

Bisimulation

Model

for

-RRB-

Bisimulation

for

and

the

corresponding

bisimulation

model

are

defined

in

the

same

way

as

for

BPA?drt?ffi

and

BPA

respectively

Replace

BPA?drt?ffi

by

in

Definition

on

page

and

BPA

by

in

Definition

on

page

a

j

oerel?x

ffi

oerel?x

j

a

ffi

a

x

j

oerel?y

ffi

oerel?x

j

a

y

ffi

Table

Additional

axioms

for

Definition

-LRB-

Basic

Terms

of

-RRB-

If

we

speak

of

basic

terms

in

the

context

of

we

mean

oe

ffi

basic

terms

as

defined

in

Definition

on

page

Definition

-LRB-

Number

of

Symbols

of

an

Term

-RRB-

We

define

n?x

the

number

of

symbols

of

x

inductively

as

follows

-LRB-

i

-RRB-

For

a

Affi

we

define

n?a

-LRB-

ii

-RRB-

for

closed

terms

x

and

y

we

define

n?x

y

n?x

y

n?x

k

y

n?x

k

y

n?x

j

y

n?x

n?y

-LRB-

iii

-RRB-

for

a

closed

term

x

we

define

n?oerel?x

n

rel?x

n?x

Theorem

-LRB-

Elimination

for

-RRB-

Let

t

be

a

closed

term

Then

there

is

a

closed

BPA?drt?ID

term

s

such

that

s

t.

Proof

First

a

term

rewriting

system

is

given

which

is

proven

to

be

strongly

normalizing

Thereafter

it

is

shown

that

every

normal

form

of

a

closed

term

is

a

closed

BPA?drt?ID

term

The

term

rewriting

system

used

is

given

in

Table

on

the

following

page

Note

that

we

have

added

natural

number

subscripts

n

to

the

merge

operators

in

order

to

deal

with

the

mutually

recursive

nature

of

definition

of

these

operators

For

a

description

of

this

technique

and

a

rigorous

formal

justification

of

its

use

see

Theorem

of

-LSB-

-RSB-

and

the

references

given

there

Using

the

method

of

the

lexicographical

path

ordering

we

prove

that

the

term

rewriting

system

associated

with

is

strongly

normalizing

Thereto

the

operator

is

assigned

the

lexicographical

status

of

the

first

argument

and

the

following

well-founded

ordering

on

constant

and

function

symbols

is

used

a

oerel

k

k

k

n

jn

kn

k

n

jn

That

the

left-hand

side

of

every

rewriting

rule

is

bigger

than

the

right-hand

side

with

respect

to

the

ordering

Olpo

is

shown

by

the

following

reductions

x

kn

y

Olpo

x

k?n

y

Olpo

x

k?n

y

x

k?n

y

Olpo

x

k?n

y

x

k?n

y

x

k?n

y

Olpo

x

k?n

y

k

n

x

k?n

y

x

k?n

y

k

n

x

k?n

y

x

k?n

y

jn?x

k?n

y

Olpo

x

k

n

y

y

k

n

x

x

jny

x

kn

y

x

k

n

y

y

k

n

x

x

jny

a

k

n

x

a

x

a

x

k

n

y

a

x

kn

y

x

y

k

n

z

x

k

n

z

y

k

n

z

oerel?x

k

n

a

ffi

oerel?x

k

n

a

y

ffi

oerel?x

k

n

a

y

oerel?x

k

n

y

oerel?x

k

n

a

y

z

oerel?x

k

n

z

oerel?x

k

n

oerel?y

oerel?x

k

n

y

a

jnb

c

if

fl?a

b

c

ffi

a

jnb

ffi

if

fl?a

b

ffi

a

jnb

x

a

jnb

x

a

x

jnb

a

jnb

x

a

x

jn

b

y

a

jn

x

kn

y

oerel?x

jnoerel?y

oerel?x

jny

a

jnoerel?x

ffi

oerel?x

jna

ffi

a

x

jnoerel?y

ffi

oerel?x

jna

y

ffi

x

y

jnz

x

jnz

y

jnz

x

jn?y

z

x

jny

x

jnz

Table

Term

rewriting

system

for

a

jnb

Olpo

a

j?n

b

Olpo

c

a

jnb

Olpo

a

j?n

b

Olpo

ffi

a

jnb

x

Olpo

a

j?n

b

x

Olpo

a

j?n

b

x

a

j?n

b

x

Olpo

a

jnb

x

b

x

Olpo

a

jnb

b

x

Olpo

a

jnb

b

x

Olpo

a

jnb

b

x

Olpo

a

jnb

x

a

x

jnb

Olpo

a

x

j?n

b

Olpo

a

x

j?n

b

a

x

j?n

b

Olpo

a

x

jnb

a

x

Olpo

a

jnb

a

x

Olpo

a

jnb

a

x

Olpo

a

jnb

a

x

Olpo

a

jnb

x

a

x

jn

y

Olpo

a

x

j?n

b

y

Olpo

a

x

j?n

b

y

a

x

j?n

b

y

Olpo

a

x

j?n

b

y

a

x

j?n

b

y

kn

a

x

j?n

b

y

Olpo

a

x

jn

b

y

a

x

j?n

b

y

kn

a

x

j?n

b

y

Olpo

a

jn

a

x

kn

b

y

Olpo

a

jn

a

x

kn

b

y

Olpo

a

jn

x

kn

y

oerel?x

jnoerel?y

Olpo

oerel?x

j?n

oerel?y

Olpo

oerel?oerel?x

j?n

oerel?y

Olpo

oerel?oe?rel?x

jnoe?rel?y

Olpo

oerel?x

jny

a

jnoerel?x

Olpo

a

j?n

oerel?x

Olpo

ffi

oerel?x

jna

Olpo

oerel?x

j?n

a

Olpo

ffi

a

x

jnoerel?y

Olpo

a

x

j?n

oerel?y

Olpo

ffi

oerel?x

jna

y

Olpo

oerel?x

j?n

a

y

Olpo

ffi

x

y

jnz

Olpo

x

y

j?n

z

Olpo

x

y

j?n

z

x

y

j?n

z

Olpo

x

y

jnz

x

y

jnz

Olpo

x

jnz

y

jnz

x

jn?y

z

Olpo

x

j?n

y

z

Olpo

x

j?n

y

z

x

j?n

y

z

Olpo

x

jn?y

z

x

jn?y

z

Olpo

x

jn?y

x

jn?z

Note

that

we

do

not

give

reductions

for

as

these

already

have

been

given

in

the

proof

of

Theorem

and

since

the

new

ordering

is

a

proper

extension

of

the

old

one

these

proofs

remain

valid

It

remains

to

prove

that

every

normal

form

of

a

closed

term

is

a

closed

BPA?drt?ID

term

We

prove

this

as

follows

suppose

that

s

is

a

normal

form

of

a

closed

term

and

furthermore

suppose

that

s

is

not

a

closed

BPA?drt?ID

term

Now

consider

the

smallest

subterm

of

s

that

is

not

a

closed

BPA?drt?ID

term

Then

must

be

of

the

form

k

of

the

form

k

or

of

the

form

j

for

closed

BPA?drt?ID

terms

and

By

the

elimination

theorem

for

BPA?drt?ID

Theorem

we

may

assume

that

and

are

basic

terms

Now

in

the

first

case

k

clearly

rewriting

rule

is

applicable

This

contradicts

the

assumption

that

s

is

a

normal

form

so

this

case

can

not

occur

That

the

second

case

k

can

not

occur

is

proven

in

the

same

way

as

already

done

in

the

proof

of

the

elimination

theorem

for

Theorem

So

it

remains

to

derive

a

contradiction

for

the

third

case

j

The

following

cases

can

be

considered

for

basic

terms

and

-LRB-

for

some

a

b

Affi

and

basic

terms

-RRB-

-LRB-

i

-RRB-

If

a

and

b

we

can

apply

if

fl?a

b

ffi

if

not

we

can

apply

-LRB-

ii

-RRB-

If

a

and

b

we

can

apply

-LRB-

iii

-RRB-

If

a

and

we

can

apply

-LRB-

iv

-RRB-

If

a

and

b

we

can

apply

-LRB-

v

-RRB-

If

a

and

b

we

can

apply

-LRB-

vi

-RRB-

If

a

and

we

can

apply

-LRB-

vii

-RRB-

If

and

b

we

can

apply

-LRB-

viii

-RRB-

If

and

b

we

can

apply

-LRB-

ix

-RRB-

If

and

we

can

apply

-LRB-

x

-RRB-

If

and

is

of

an

arbitrary

form

we

can

apply

-LRB-

xi

-RRB-

If

is

of

an

arbitrary

form

and

we

can

apply

This

sums

up

all

possible

sixteen

cases

-LRB-

with

seven

cases

thrown

together

in

-LRB-

x

-RRB-

and

-LRB-

xi

-RRB-

-RRB-

In

all

of

these

cases

we

can

apply

one

of

the

rewriting

rules

so

is

not

a

normal

form

This

contradicts

the

assumption

that

s

is

a

normal

form

From

this

contradiction

we

conclude

that

s

does

not

contain

a

merge

operator

Therefore

s

must

be

closed

BPA?drt?ID

term

which

had

to

be

proven

Corollary

-LRB-

Elimination

for

-RRB-

Let

t

be

a

closed

term

Then

there

is

a

basic

term

s

such

that

s

t.

Proof

This

follows

immediately

from

-LRB-

i

-RRB-

The

elimination

theorem

for

-LRB-

see

Theorem

-RRB-

-LRB-

ii

-RRB-

the

elimination

theorem

for

BPA?drt?ID

-LRB-

see

Theorem

-RRB-

-LRB-

iii

-RRB-

the

fact

that

all

axioms

of

BPA?drt?ID

are

also

contained

in

Remark

-LRB-

Elimination

for

-RRB-

Elimination

for

a

slightly

different

version

of

is

also

claimed

-LRB-

without

proof

-RRB-

in

Theorem

of

-LSB-

-RSB-

-LRB-

where

is

called

ACPdt

-RRB-

Theorem

-LRB-

Ground

Equivalence

of

ACP?drt?ID

and

-RRB-

For

all

closed

ACP?drt?ID

terms

s

and

t

we

have

ACP?drt?ID

s

t

if

and

only

if

s

t.

Proof

It

suffices

to

show

that

for

closed

terms

every

axiom

of

is

derivable

from

the

axioms

of

ACP?drt?ID

and

vice

versa

that

for

closed

terms

every

axiom

of

ACP?drt?ID

is

derivable

from

the

axioms

of

We

can

restrict

ourselves

to

the

axioms

that

are

not

shared

by

both

theories

Furthermore

the

proofs

regarding

Axioms

that

are

given

in

Theorem

on

page

with

respect

to

PA?drt?ID

and

remain

valid

with

respect

to

ACP?drt?ID

and

Part

I

First

we

show

that

Axioms

of

are

derivable

in

ACP?drt?ID

Axiom

ACP?drt?ID

a

j

oerel?x

ffi

Consider

the

following

derivation

ACP?drt?ID

a

j

oerel?x

rel?a

j

oerel?x

ffi

Axiom

ACP?drt?ID

oerel?x

j

a

ffi

This

axiom

is

treated

symmetrically

to

the

previous

axiom

Axiom

ACP?drt?ID

a

x

j

oerel?y

ffi

Consider

the

following

derivation

ACP?drt?ID

a

x

j

oerel?y

rel?a

x

j

oerel?y

rel?a

x

j

oerel?y

ffi

Axiom

ACP?drt?ID

oerel?x

j

a

y

ffi

This

axiom

is

treated

symmetrically

to

the

previous

axiom

Part

II

Secondly

we

show

that

Axioms

of

ACP?drt?ID

are

also

derivable

in

Axiom

oerel?x

j

rel?y

ffi

Use

the

general

form

of

basic

term

y.

Take

y

X

i

m

ai

ti

X

j

n

bj

X

k

p

oerel?uk

for

m

n

p

N

ai

bj

Affi

and

basic

terms

ti

and

uk

Then

we

have

oerel?x

j

rel?y

oerel?x

j

rel

@X

i

m

ai

ti

X

j

n

bj

X

k

p

oerel?uk

A

oerel?x

j

@X

i

m

rel?ai

ti

X

j

n

rel?bj

X

k

p

rel?oerel?uk

A

oerel?x

j

@X

i

m

rel?ai

ti

X

j

n

rel?bj

X

k

p

rel?oerel?uk

A

oerel?x

j

@X

i

m

ai

ti

X

j

n

bj

X

k

p

ffi

A

@oerel

x

j

X

i

m

ai

ti

A

@oerel

x

j

X

j

n

bj

A

@oerel

x

j

X

k

p

ffi

A

@X

i

m

oerel?x

j

ai

ti

A

@X

j

n

oerel?x

j

bj

A

@X

k

p

oerel?x

j

ffi

A

X

i

m

ffi

X

j

n

ffi

X

k

p

ffi

ffi

Axiom

rel?x

j

oerel?y

ffi

This

axiom

is

treated

symmetrically

to

the

previous

axiom

Theorem

-LRB-

Ground

Equivalence

of

T?ACP?drt?ID

and

-RRB-

The

term

deduction

systems

T?ACP?drt?ID

and

are

ground

equivalent

Proof

Both

term

deduction

systems

have

the

same

signature

and

the

same

set

of

deduction

rules

Then

it

is

trivial

that

the

same

equalities

hold

between

closed

terms

Corollary

-LRB-

Soundness

of

-RRB-

The

set

of

all

closed

terms

modulo

bisimulation

equivalence

is

a

model

of

Proof

This

follows

immediately

from

the

following

observations

and

Theorem

-LRB-

i

-RRB-

ACP?drt?ID

and

are

ground

equivalent

equational

specifications

-LRB-

see

Theorem

-RRB-

-LRB-

ii

-RRB-

T?ACP?drt?ID

and

are

ground

equivalent

term

deduction

systems

-LRB-

see

Theorem

-RRB-

-LRB-

iii

-RRB-

Soundness

of

ACP?drt?ID

-LRB-

see

Theorem

-RRB-

Remark

-LRB-

Soundness

of

-RRB-

Soundness

of

a

slightly

different

version

of

is

also

claimed

-LRB-

without

proof

-RRB-

in

Theorem

of

-LSB-

-RSB-

-LRB-

where

is

called

ACPdt

-RRB-

Corollary

-LRB-

Completeness

of

-RRB-

The

equational

specification

is

a

complete

axiomatization

of

the

set

of

closed

terms

modulo

bisimulation

equivalence

Proof

This

follows

immediately

from

the

following

observations

and

Theorem

-LRB-

i

-RRB-

ACP?drt?ID

and

are

ground

equivalent

equational

specifications

-LRB-

see

Theorem

-RRB-

-LRB-

ii

-RRB-

T?ACP?drt?ID

and

are

ground

equivalent

term

deduction

systems

-LRB-

see

Theorem

-RRB-

-LRB-

iii

-RRB-

Completeness

of

ACP?drt?ID

-LRB-

see

Theorem

-RRB-

Remark

-LRB-

Completeness

of

-RRB-

Completeness

of

a

slightly

different

version

of

is

also

claimed

-LRB-

without

proof

-RRB-

in

Theorem

of

-LSB-

-RSB-

-LRB-

where

is

called

ACPdt

-RRB-

Soundness

and

Completeness

of

PA?drt

Definition

-LRB-

Signature

of

PAdrt

-RRB-

The

signature

of

PAdrt

consists

of

the

undelayable

atomic

actions

faja

Ag

the

delayable

atomic

actions

faja

Ag

the

undelayable

deadlock

constant

ffi

the

delayable

deadlock

constant

ffi

the

immediate

deadlock

constant

ffi

the

alternative

composition

operator

the

sequential

composition

operator

the

time

unit

delay

operator

oerel

the

now

operator

rel

the

unbounded

start

delay

operator

b

c

the

-LRB-

free

-RRB-

merge

operator

k

and

the

left

merge

operator

k

Definition

-LRB-

Axioms

of

PAdrt

-RRB-

The

process

algebra

PAdrt

is

axiomatized

by

the

axioms

of

BPAdrt

given

in

Definition

on

page

Axioms

and

shown

in

Table

on

page

Axiom

shown

in

Table

on

page

and

finally

Axioms

and

shown

in

Table

PAdrt

DRTSID

DCSID

ATS

USD

a

k

x

ffi

a

x

ffi

a

x

k

y

ffi

a

x

k

y

ffi

oerel?x

k

rel?y

ffi

ffi

x

k

ffi

ffi

ffi

k

x

ffi

Table

Additional

axioms

for

PAdrt

Definition

-LRB-

Semantics

of

PAdrt

-RRB-

The

semantics

of

PAdrt

are

given

by

the

term

deduction

system

T?PAdrt

induced

by

the

deduction

rules

for

BPAdrt

given

in

Definition

on

page

and

the

deduction

rules

for

the

-LRB-

free

-RRB-

merge

with

immediate

deadlock

shown

in

Table

on

the

next

page

x

a

ID?y

x

k

y

a

k

y

y

a

ID?x

x

k

y

a

x

k

x

a

ID?y

x

k

y

a

k

y

x

a

p

ID?y

x

k

y

a

y

y

a

p

ID?x

x

k

y

a

x

x

a

p

ID?y

x

k

y

a

y

x

oe

y

oe

x

k

y

oe

k

x

oe

y

oe

x

k

y

oe

k

ID?x

ID?x

k

y

ID?y

ID?x

k

y

ID?x

ID?x

k

y

ID?y

ID?x

k

y

Table

Deduction

rules

for

the

-LRB-

free

-RRB-

merge

with

immediate

deadlock

Definition

-LRB-

Bisimulation

and

Bisimulation

Model

for

PAdrt

-RRB-

Bisimulation

for

PAdrt

and

the

corresponding

bisimulation

model

are

defined

in

the

same

way

as

for

BPA?drt

and

BPA

respectively

Replace

BPA?drt

by

PAdrt

in

Definition

on

page

and

BPA

by

PAdrt

in

Definition

on

page

Definition

-LRB-

Basic

Terms

of

PAdrt

-RRB-

If

we

speak

of

basic

terms

in

the

context

of

PAdrt

we

mean

oe

ffi

ffi

ffi

basic

terms

as

defined

in

Definition

on

page

Definition

-LRB-

Number

of

Symbols

of

a

PAdrt

term

-RRB-

We

define

n?x

the

number

of

symbols

of

x

inductively

as

follows

-LRB-

i

-RRB-

We

define

n

ffi

-LRB-

ii

-RRB-

for

a

Affi

we

define

n?a

n?a

-LRB-

iii

-RRB-

for

closed

PAdrt

terms

x

and

y

we

define

n?x?y

n?x?y

n?x

k

y

n?x

k

y

n?x

n?y

-LRB-

iv

-RRB-

for

a

closed

PAdrt

term

x

we

define

n?oerel?x

n

rel?x

n?bxc

n?x

Proposition

-LRB-

Properties

of

PA?drt

-RRB-

For

PAdrt

terms

x

and

y

and

any

a

Affi

we

have

the

following

equalities

-LRB-

i

-RRB-

PA?drt

bac

a

-LRB-

ii

-RRB-

PA?drt

bx

yc

bxc

y

-LRB-

iii

-RRB-

PA?drt

bx

yc

bxc

byc

-LRB-

iv

-RRB-

PA?drt

boerel?x?c

ffi

-LRB-

v

-RRB-

PA?drt

b?ffic

ffi

-LRB-

vi

-RRB-

PA?drt

a

k

bxc

a

bxc

-LRB-

vii

-RRB-

PA?drt

a

x

k

byc

a

x

k

byc

-LRB-

viii

-RRB-

PAdrt

rel?a

a

-LRB-

ix

-RRB-

PAdrt

bxc

ffi

bxc

Proof

The

proofs

for

equality

-LRB-

i

-RRB-

-LRB-

v

-RRB-

and

-LRB-

viii

-RRB-

-LRB-

ix

-RRB-

given

in

Proposition

on

page

with

respect

to

BPAdrt

remain

valid

in

the

setting

of

PAdrt

as

can

be

easily

checked

Equality

-LRB-

vi

-RRB-

and

-LRB-

vii

-RRB-

do

not

appear

in

Proposition

Consider

the

following

computation

for

equality

-LRB-

vi

-RRB-

PAdrt

a

k

bxc

bac

k

bxc

rel?a

oerel?bac

k

bxc

a

oerel?a

k

bxc

a

k

bxc

oerel?a

k

bxc

a

k

bxc

ffi

oerel?a

k

rel?x

oerel?bxc

a

bxc

ffi

oerel?a

k

bxc

rel?a

bxc

oerel?a

k

bxc

rel?a

bxc

oerel?a

k

bxc

Using

RSP

-LRB-

USD

-RRB-

we

obtain

PA?drt

a

k

bxc

a

bxc

bac

bxc

a

bxc

Finally

consider

the

following

computation

for

equality

-LRB-

vii

-RRB-

PAdrt

a

x

k

byc

bac

x

k

byc

rel?a

oerel?bac

x

k

byc

a

oerel?a

x

k

byc

a

x

oerel?a

x

k

byc

a

x

k

byc

oerel?a

x

k

byc

a

x

k

byc

ffi

oerel?a

x

k

rel?y

oerel?byc

a

x

k

byc

ffi

oerel?a

x

k

byc

rel?a

x

k

byc

oerel?a

x

k

byc

rel?a

x

k

byc

oerel?a

x

k

byc

Using

RSP

-LRB-

USD

-RRB-

we

obtain

PA?drt

a

x

k

byc

a

x

k

byc

bac

x

k

byc

a

x

k

byc

Theorem

-LRB-

Elimination

for

PA?drt

-RRB-

Let

t

be

a

closed

PAdrt

term

Then

there

is

a

closed

BPAdrt

term

s

such

that

PA?drt

t

s.

Proof

Let

t

be

a

closed

PAdrt

term

The

theorem

is

proven

by

induction

on

n?t

and

case

distinction

on

the

general

structure

of

t.

-LRB-

i

-RRB-

t

ffi

Then

t

is

a

closed

BPAdrt

term

-LRB-

ii

-RRB-

t

a

for

some

a

Affi

Then

t

is

a

closed

BPAdrt

term

-LRB-

iii

-RRB-

t

a

for

some

a

Affi

Then

t

is

a

closed

BPAdrt

term

-LRB-

iv

-RRB-

t

for

closed

PAdrt

terms

and

By

induction

there

are

closed

BPAdrt

terms

and

such

that

PA?drt

and

PA?drt

But

then

also

PA?drt

and

is

a

closed

BPAdrt

term

-LRB-

v

-RRB-

t

for

closed

PAdrt

terms

and

This

case

is

treated

analogously

to

case

-LRB-

ii

-RRB-

-LRB-

vi

-RRB-

t

for

a

closed

PAdrt

term

This

case

is

treated

analogously

to

case

-LRB-

ii

-RRB-

-LRB-

vii

-RRB-

t

for

a

closed

PAdrt

term

This

case

is

treated

analogously

to

case

-LRB-

ii

-RRB-

-LRB-

viii

-RRB-

t

for

a

closed

PAdrt

term

This

case

is

treated

analogously

to

case

-LRB-

ii

-RRB-

-LRB-

ix

-RRB-

t

k

for

closed

PAdrt

terms

and

By

induction

there

are

closed

BPAdrt

terms

and

such

that

PA?drt

and

PA?drt

By

Theorem

the

elimination

theorem

for

BPAdrt

there

are

basic

terms

and

such

that

BPA?drt

and

BPA?drt

But

then

also

PA?drt

PA?drt

and

PA?drt

k

k

We

proceed

by

induction

on

the

structure

of

basic

terms

and

distinguish

all

possible

cases

for

basic

term

-LRB-

a

-RRB-

ffi

Then

PA?drt

k

k

ffi

k

ffi

and

ffi

is

a

closed

BPAdrt

term

-LRB-

b

-RRB-

a

for

some

a

Affi

Using

Lemma

we

distinguish

two

cases

i.

ffi

Then

we

have

PA?drt

k

k

k

ffi

ffi

and

ffi

is

a

closed

BPAdrt

term

ii

ffi

Then

we

have

PA?drt

k

k

a

k

ffi

a

ffi

a

and

a

is

a

closed

BPAdrt

term

-LRB-

c

-RRB-

a

for

some

a

Affi

Using

Lemma

we

distinguish

four

cases

i.

ffi

Then

we

have

PA?drt

k

k

k

ffi

ffi

and

ffi

is

a

closed

BPAdrt

term

ii

ffi

Then

we

have

PA?drt

k

k

a

k

bac

k

rel?a

oerel?bac

k

a

oerel?a

k

a

k

oerel?a

k

a

k

ffi

oerel?a

k

ffi

a

ffi

ffi

a

ffi

a

ffi

a

ffi

a

and

a

is

a

closed

BPAdrt

term

iii

Then

using

Proposition

-LRB-

vi

-RRB-

we

have

PA?drt

k

k

a

k

a

a

and

a

is

a

closed

BPAdrt

term

iv

for

a

basic

term

such

that

Then

we

have

PA?drt

k

k

a

k

bac

k

rel?a

oerel?bac

k

a

oerel?a

k

a

k

oerel?a

k

a

k

oerel?a

k

a

k

ffi

oerel?a

k

a

ffi

oerel?a

k

a

oerel?a

k

a

oerel?a

k

By

the

induction

hypothesis

there

exists

a

closed

BPAdrt

term

p

such

that

PA?drt

a

k

p.

Then

PA?drt

k

a

oerel?a

k

a

oerel?p

and

a

oerel?p

is

a

closed

BPAdrt

term

-LRB-

d

-RRB-

a

for

some

a

Affi

and

basic

term

Using

Lemma

we

distinguish

two

cases

i.

ffi

Then

we

have

PA?drt

k

k

k

ffi

ffi

and

ffi

is

a

closed

BPAdrt

term

ii

ffi

Then

we

have

PA?drt

k

k

a

k

ffi

a

k

ffi

a

k

By

the

induction

hypothesis

there

exists

a

closed

BPAdrt

term

p

such

that

PA?drt

k

p.

Then

PA?drt

k

a

k

a

p

and

a

p

is

a

closed

BPAdrt

term

-LRB-

e

-RRB-

a

for

some

a

Affi

and

basic

term

Using

Lemma

we

distinguish

four

cases

i.

ffi

Then

we

have

PA?drt

k

k

k

ffi

ffi

and

ffi

is

a

closed

BPAdrt

term

ii

ffi

Then

we

have

PA?drt

k

k

a

k

bac

k

rel?a

oerel?bac

k

a

oerel?a

k

a

oerel?a

k

a

oerel?a

k

a

k

oerel?a

k

a

k

ffi

oerel?a

k

ffi

a

k

ffi

ffi

a

k

ffi

a

k

ffi

k

a

ffi

k

a

k

By

the

induction

hypothesis

there

exists

a

closed

BPAdrt

term

p

such

that

PA?drt

k

p.

Then

PA?drt

k

a

k

a

p

and

a

p

is

a

closed

BPAdrt

term

iii

Then

using

Proposition

-LRB-

vii

-RRB-

we

have

PA?drt

k

k

a

k

a

k

a

k

By

the

induction

hypothesis

there

exists

a

closed

BPAdrt

term

p

such

that

PA?drt

k

p.

Then

PA?drt

k

a

k

a

p

and

a

p

is

a

closed

BPAdrt

term

iv

for

a

basic

term

such

that

Then

we

have

PA?drt

k

k

a

k

bac

k

rel?a

oerel?bac

k

a

oerel?a

k

a

oerel?a

k

a

oerel?a

k

a

k

oerel?a

k

a

k

oerel?a

k

a

k

ffi

oerel?a

k

a

k

ffi

oerel?a

k

a

k

k

a

k

k

By

the

induction

hypothesis

there

exist

closed

BPAdrt

terms

and

such

that

PA?drt

k

and

PA?drt

k

Then

PA?drt

k

a

k

k

a

and

a

is

a

closed

BPAdrt

term

-LRB-

f

-RRB-

for

basic

terms

and

Then

PA?drt

k

k

k

k

k

By

induction

there

exist

closed

BPAdrt

terms

and

such

that

PA?drt

k

and

PA?drt

k

Then

also

PA?drt

k

k

k

and

is

a

closed

BPAdrt

term

-LRB-

g

-RRB-

for

a

basic

term

Using

Lemma

we

distinguish

four

cases

i.

ffi

Then

we

have

PA?drt

k

k

a

k

ffi

ffi

and

ffi

is

a

closed

BPAdrt

term

ii

ffi

Then

PA?drt

k

k

k

ffi

ffi

and

ffi

is

a

closed

BPAdrt

term

iii

Then

we

have

PA?drt

k

k

k

k

k

k

By

the

induction

hypothesis

there

exists

a

closed

BPAdrt

term

p

such

that

PA?drt

k

p.

Then

PA?drt

k

k

oerel?p

and

oerel?p

is

a

closed

BPAdrt

term

iv

for

a

basic

term

such

that

Then

we

have

PA?drt

k

k

k

k

By

the

induction

hypothesis

there

is

a

closed

BPAdrt

term

p

such

that

PA?drt

k

p.

Then

PA?drt

k

k

oerel?p

and

oerel?p

is

a

closed

BPAdrt

term

-LRB-

x

-RRB-

t

k

for

closed

PAdrt

terms

and

Then

PA?drt

k

k

k

By

-LRB-

ix

-RRB-

there

are

closed

BPAdrt

terms

and

such

that

PA?drt

k

and

PA?drt

k

But

then

also

PA?drt

k

k

k

and

is

a

closed

BPAdrt

term

Corollary

-LRB-

Elimination

for

PA?drt

-RRB-

Let

t

be

a

closed

PAdrt

term

Then

there

is

a

basic

term

s

such

that

PA?drt

s

t.

Proof

This

follows

immediately

from

-LRB-

i

-RRB-

The

elimination

theorem

for

PA?drt

-LRB-

see

Theorem

-RRB-

-LRB-

ii

-RRB-

the

elimination

theorem

for

BPA?drt

-LRB-

see

Theorem

-RRB-

-LRB-

iii

-RRB-

the

fact

that

all

axioms

of

BPA?drt

are

also

contained

in

PA?drt

Remark

-LRB-

Elimination

for

PAdrt

-RRB-

Elimination

for

a

somewhat

different

version

of

PAdrt

is

also

claimed

-LRB-

without

proof

-RRB-

in

Section

of

-LSB-

-RSB-

Theorem

-LRB-

Soundness

of

PA?drt

-RRB-

The

set

of

closed

PAdrt

terms

modulo

bisimulation

equivalence

is

a

model

of

PA?drt

Proof

We

only

prove

soundness

for

the

axioms

of

PAdrt

that

have

not

been

treated

in

earlier

soundness

proofs

Note

that

to

extend

these

proofs

to

PAdrt

we

have

to

check

that

the

bisimulations

given

in

previous

soundness

proofs

respect

the

ID

predicate

-LRB-

as

required

by

transfer

condition

-LRB-

iv

-RRB-

in

Definition

on

page

-RRB-

However

as

the

fact

that

they

do

can

be

easily

checked

we

will

not

give

details

Axiom

Take

the

relation

R

f?s

s

a

k

s

ffi

a

s

ffi

js

C?PAdrt?g

We

look

at

the

transitions

of

both

sides

at

the

same

time

The

only

transition

of

the

left-hand

side

is

a

k

s

ffi

a

s

ffi

and

the

only

transition

of

the

right-hand

side

is

a

s

ffi

a

s

ffi

and

note

that

s

ffi

s

ffi

R.

Finally

neither

side

satisfies

the

ID

predicate

ID?a

k

s

ffi

and

ID?a

s

ffi

-LRB-

note

that

ID?s

ffi

even

if

ID?s

-RRB-

Axiom

Take

the

relation

R

f?s

s

a

s

k

t

ffi

a

s

k

t

ffi

js

t

C?PAdrt?g

We

look

at

the

transitions

of

both

sides

at

the

same

time

The

only

transition

of

the

left-hand

side

is

a?s

k

t?ffi

a

s

k

t?ffi

and

the

only

transition

of

the

right-hand

side

is

a

s

k

t?ffi

a

s

k

t?ffi

and

note

that

s

k

t?ffi

s

k

t?ffi

R.

Finally

neither

side

satisfies

the

ID

predicate

ID?a?s

k

t

ffi

and

ID?a

s

k

t

ffi

Axiom

Take

the

relation

R

f?oerel?s

k

rel?t

ffi

ffi?js

t

C?PAdrt?g

We

look

at

the

transitions

of

both

sides

at

the

same

time

Observe

that

there

are

no

transitions

possible

on

the

left-hand

side

oerel?s

k

rel?t

ffi

Also

for

the

right-hand

side

there

are

no

transitions

possible

ffi

Finally

neither

side

satisfies

the

ID

predicate

ID?oerel?s

k

rel?t

ffi

and

ID?ffi

-LRB-

note

that

ID

rel?t

ffi

even

if

ID?t

-RRB-

Axiom

Take

the

relation

R

f?s

k

ffi

ffi?js

C?PAdrt?g

We

look

at

the

transitions

of

both

sides

at

the

same

time

Observe

that

there

are

no

transitions

possible

on

the

left-hand

side

s

k

ffi

Also

for

the

right-hand

side

there

are

no

transitions

possible

ffi

Finally

both

sides

satisfy

the

ID

predicate

ID?s

k

ffi

and

ID

ffi

-LRB-

note

that

ID?s

k

ffi

even

if

ID?s

-RRB-

Axiom

Take

the

relation

R

f

ffi

s

k

ffi?js

C?PAdrt?g

This

case

is

treated

symmetrically

to

the

previous

case

Remark

-LRB-

Soundness

of

PAdrt

-RRB-

Soundness

of

a

somewhat

different

version

of

PAdrt

is

also

claimed

-LRB-

without

proof

-RRB-

in

Section

of

-LSB-

-RSB-

Theorem

-LRB-

Conservativity

of

PA?drt

with

respect

to

BPA?drt

-RRB-

The

equational

specification

PA?drt

is

a

conservative

extension

of

the

equational

specification

BPA?drt

Proof

In

order

to

prove

conservativity

it

is

sufficient

to

verify

that

the

following

conditions

are

satisfied

-LRB-

i

-RRB-

Bisimulation

equivalence

is

definable

in

terms

of

predicate

and

relation

symbols

only

-LRB-

ii

-RRB-

BPA?drt

is

a

complete

axiomatization

with

respect

to

the

bisimulation

equivalence

model

induced

by

T?BPAdrt

-LRB-

see

Theorem

-RRB-

-LRB-

iii

-RRB-

PA?drt

is

a

sound

axiomatization

with

respect

to

the

bisimulation

equivalence

model

induced

by

T?PAdrt

-LRB-

see

Theorem

-RRB-

-LRB-

iv

-RRB-

T?PAdrt

is

an

operationally

conservative

extension

of

T?BPAdrt

And

in

order

for

T?PAdrt

to

be

an

operationally

conservative

extension

of

T?BPAdrt

we

must

verify

the

following

conditions

-LRB-

i

-RRB-

T?BPAdrt

is

a

pure

well-founded

term

deduction

system

in

path

format

-LRB-

ii

-RRB-

T?PAdrt

is

a

term

deduction

system

in

path

format

-LRB-

iii

-RRB-

T?BPAdrt

T?PAdrt

is

defined

That

the

above

properties

hold

can

be

trivially

checked

from

the

relevant

definitions

Theorem

-LRB-

Completeness

of

PA?drt

-RRB-

The

equational

specification

PA?drt

is

a

complete

axiomatization

of

the

set

of

closed

PAdrt

terms

modulo

bisimulation

equivalence

Proof

By

Verhoef

's

General

Completeness

Theorem

-LRB-

see

-LSB-

-RSB-

or

Theorem

of

-LSB-

-RSB-

-RRB-

this

follows

immediately

from

-LRB-

i

-RRB-

PA?drt

has

the

elimination

property

for

BPAdrt

-LRB-

see

Theorem

-RRB-

-LRB-

ii

-RRB-

PA?drt

is

a

conservative

extension

of

BPA?drt

-LRB-

see

Theorem

-RRB-

Remark

-LRB-

Completeness

of

PAdrt

-RRB-

Completeness

of

a

somewhat

different

version

of

PAdrt

is

also

claimed

-LRB-

without

proof

-RRB-

in

Section

of

-LSB-

-RSB-

Definition

-LRB-

Axioms

for

the

Ultimate

Start

Delay

and

Merge

-RRB-

We

define

Axioms

and

for

the

ultimate

start

delay

as

given

in

Table

on

the

next

page

Note

that

they

precisely

correspond

to

the

equalities

of

Proposition

-LRB-

vi

-RRB-

and

-LRB-

vii

-RRB-

a

k

bxc

a

bxc

a

x

k

byc

a

x

k

byc

Table

Additional

axioms

for

the

ultimate

start

delay

and

merge

Corollary

-LRB-

Soundness

of

PAdrt

-RRB-

The

set

of

closed

PAdrt

terms

modulo

bisimulation

equivalence

is

a

model

of

PAdrt

Proof

This

now

follows

directly

from

the

soundness

of

PA?drt

-LRB-

see

Theorem

on

page

-RRB-

and

the

fact

that

Axioms

are

derivable

in

PA?drt

-LRB-

see

Proposition

on

page

-RRB-

Corollary

-LRB-

Completeness

of

PAdrt

-RRB-

If

we

add

Axioms

of

Table

on

page

Axiom

of

Table

on

page

and

Axioms

of

Table

to

PAdrt

we

again

have

a

complete

axiomatization

of

the

set

of

closed

PAdrt

terms

modulo

bisimulation

equivalence

Proof

Careful

inspection

of

the

dependencies

between

the

proofs

in

this

section

reveals

that

the

proof

of

Theorem

only

relies

upon

RSP

-LRB-

USD

-RRB-

to

ensure

Proposition

-LRB-

i

-RRB-

-LRB-

vii

-RRB-

So

we

obviously

do

not

need

RSP

-LRB-

USD

-RRB-

anymore

if

we

add

the

corresponding

Axioms

Note

that

in

this

way

we

get

a

purely

equational

axiomatization

-LRB-

i.e.

without

conditional

axioms

or

principles

-RRB-

Soundness

and

Completeness

of

ACP?drt

Definition

-LRB-

Signature

of

ACPdrt

-RRB-

The

signature

of

ACPdrt

consists

of

the

undelayable

atomic

actions

faja

Ag

the

delayable

atomic

actions

faja

Ag

the

undelayable

deadlock

constant

ffi

the

delayable

deadlock

constant

ffi

the

immediate

deadlock

constant

ffi

the

alternative

composition

operator

the

sequential

composition

operator

the

time

unit

delay

operator

oerel

the

now

operator

rel

the

unbounded

start

delay

operator

b

c

the

-LRB-

communicating

-RRB-

merge

operator

k

the

left

merge

operator

k

and

the

communication

merge

operator

j

Definition

-LRB-

Axioms

of

ACPdrt

-RRB-

The

process

algebra

ACPdrt

is

axiomatized

by

the

axioms

of

PAdrt

given

in

Definition

on

page

minus

Axiom

plus

Axioms

and

shown

in

Table

on

page

Axioms

shown

in

Table

on

page

and

Axioms

and

shown

in

Table

on

the

following

page

ACPdrt

DRTSID

DCSID

ATS

USD

oerel?x

j

rel?y

ffi

ffi

rel?x

ffi

j

oerel?y

ffi

x

j

ffi

ffi

ffi

j

x

ffi

Table

Additional

axioms

for

ACPdrt

Definition

-LRB-

Semantics

of

ACPdrt

-RRB-

The

semantics

of

ACPdrt

are

given

by

the

term

deduction

system

T?ACPdrt

induced

by

the

deduction

rules

for

PAdrt

given

in

Definition

on

page

the

deduction

rules

for

the

communication

merge

shown

in

Table

on

page

and

the

additional

deduction

rules

for

ACPdrt

shown

in

Table

ID?x

ID?x

j

y

ID?y

ID?x

j

y

Table

Additional

deduction

rules

for

ACPdrt

Definition

-LRB-

Bisimulation

and

Bisimulation

Model

for

ACPdrt

-RRB-

Bisimulation

for

ACPdrt

and

the

corresponding

bisimulation

model

are

defined

in

the

same

way

as

for

BPA?drt

and

BPA

respectively

Replace

BPA?drt

by

ACPdrt

in

Definition

on

page

and

BPA

by

ACPdrt

in

Definition

on

page

Definition

-LRB-

Basic

Terms

of

ACPdrt

-RRB-

If

we

speak

of

basic

terms

in

the

context

of

ACPdrt

we

mean

oe

ffi

ffi

ffi

basic

terms

as

defined

in

Definition

on

page

Definition

-LRB-

Number

of

Symbols

of

an

ACPdrt

term

-RRB-

We

define

n?x

the

number

of

symbols

of

x

inductively

as

follows

-LRB-

i

-RRB-

We

define

n

ffi

-LRB-

ii

-RRB-

for

a

Affi

we

define

n?a

n?a

-LRB-

iii

-RRB-

for

closed

ACPdrt

terms

x

and

y

we

define

n?x

y

n?x

y

n?x

k

y

n?x

k

y

n?x

j

y

n?x

n?y

-LRB-

iv

-RRB-

for

a

closed

ACPdrt

term

x

we

define

n?oerel?x

n

rel?x

n?bxc

n?x

Proposition

-LRB-

Properties

of

ACP?drt

Part

I

-RRB-

For

ACPdrt

terms

x

and

y

and

any

a

Affi

we

have

the

following

equalities

-LRB-

i

-RRB-

ACP?drt

bac

a

-LRB-

ii

-RRB-

ACP?drt

bx

yc

bxc

y

-LRB-

iii

-RRB-

ACP?drt

bx

yc

bxc

byc

-LRB-

iv

-RRB-

ACP?drt

boerel?x?c

ffi

-LRB-

v

-RRB-

ACP?drt

b?ffic

ffi

-LRB-

vi

-RRB-

ACP?drt

a

k

bxc

a

bxc

-LRB-

vii

-RRB-

ACP?drt

a

x

k

byc

a

x

k

byc

-LRB-

viii

-RRB-

ACPdrt

rel?a

a

-LRB-

ix

-RRB-

ACPdrt

bxc

ffi

bxc

Proof

The

proofs

for

these

equalities

given

in

Proposition

on

page

with

respect

to

PAdrt

remain

valid

in

the

setting

of

ACPdrt

as

can

be

easily

checked

Proposition

-LRB-

Properties

of

ACP?drt

Part

II

-RRB-

For

ACPdrt

terms

x

and

y

and

any

a

b

c

Affi

we

have

the

following

equalities

-LRB-

i

-RRB-

ACP?drt

a

j

b

c

if

fl?a

b

c

ffi

-LRB-

ii

-RRB-

ACP?drt

a

j

b

ffi

if

fl?a

b

ffi

-LRB-

iii

-RRB-

ACP?drt

a

j

b

x

a

j

b

x

-LRB-

iv

-RRB-

ACP?drt

a

x

j

b

a

j

b

x

-LRB-

v

-RRB-

ACP?drt

a

x

j

b

y

a

j

b

x

k

y

Proof

-LRB-

i

-RRB-

Consider

the

following

computation

ACPdrt

a

j

b

bac!j

bbc

rel?a

oerel?bac

j

rel?b

oerel?bbc

a

oerel?a

j

b

oerel?b

a

j

b

a

j

oerel?b

oerel?a

j

b

oerel?a

j

oerel?b

a

j

b

rel?a

j

oerel?b

oerel?a

j

rel?b

oerel?a

j

oerel?b

c

ffi

ffi

oerel?a

j

b

c

oerel?a

j

b

rel?c

oerel?a

j

b

Using

RSP

-LRB-

USD

-RRB-

we

obtain

ACP?drt

a

j

b

bcc

c

-LRB-

ii

-RRB-

This

case

is

treated

like

the

previous

one

but

with

c

replaced

by

ffi

-LRB-

iii

-RRB-

Consider

the

following

computation

ACPdrt

a

j

b

x

bac

j

bbc

x

rel?a

oerel?bac

j

rel?b

oerel?bbc

x

a

oerel?a

j

b

oerel?b

x

a

oerel?a

j

b

x

oerel?b

x

a

j

b

x

a

j

oerel?b

x

oerel?a

j

b

x

oerel?a

j

oerel?b

x

a

j

b

x

rel?a

j

oerel?b

x

oerel?a

j

rel?b

x

oerel?a

j

oerel?b

x

a

j

b

x

rel?a

j

oerel?b

x

oerel?a

j

rel?b

x

oerel?a

j

oerel?b

x

fl?a

b

x

ffi

ffi

oerel?a

j

b

x

fl?a

b

x

oerel?a

j

b

x

rel?fl?a

b

x

oerel?a

j

b

x

rel?fl?a

b

x

oerel?a

j

b

x

Using

RSP

-LRB-

USD

-RRB-

we

obtain

ACP?drt

a

j

b

x

bfl?a

b

xc

bfl?a

b?c

x

fl?a

b

x

a

j

b

x

-LRB-

iv

-RRB-

This

case

is

treated

symmetrically

to

the

previous

case

-LRB-

v

-RRB-

Consider

the

following

computation

ACPdrt

a

x

j

b

y

bac

x

j

bbc

y

rel?a

oerel?bac

x

j

rel?b

oerel?bbc

y

a

oerel?a

x

j

b

oerel?b

y

a

x

oerel?a

x

j

b

y

oerel?b

y

a

x

j

b

y

a

x

j

oerel?b

y

oerel?a

x

j

b

y

oerel?a

x

j

oerel?b

y

a

x

j

b

y

rel?a

x

j

oerel?b

y

oerel?a

x

j

rel?b

y

oerel?a

x

j

oerel?b

y

a

x

j

b

y

rel?a

x

j

oerel?b

y

oerel?a

x

j

rel?b

y

oerel?a

x

j

oerel?b

y

fl?a

b

x

k

y

ffi

ffi

oerel?a

x

j

b

y

fl?a

b

x

k

y

oerel?a

x

j

b

y

rel?fl?a

b

x

k

y

oerel?a

x

j

b

y

rel?fl?a

b

x

k

y

oerel?a

x

j

b

y

Using

RSP

-LRB-

USD

-RRB-

we

obtain

ACP?drt

a

x

j

b

y

bfl?a

b

x

k

y?c

bfl?a

b?c

x

k

y

fl?a

b

x

k

y

a

j

b

x

k

y

Remark

-LRB-

Properties

of

ACP?drt

Part

II

-RRB-

Note

that

the

equalities

of

Proposition

on

page

are

in

a

sense

the

delayable

reformulations

of

Axioms

and

for

the

communication

merge

j

Such

reformulations

are

however

not

possible

for

the

axioms

for

the

left

merge

k

Take

for

example

although

we

do

have

ACP?drt

a

k

x

ffi

a

x

ffi

the

delayable

reformulation

does

not

hold

ACP?drt

a

k

x

ffi

a

x

ffi

as

can

be

seen

by

instantiating

x

with

any

x

such

that

x

In

that

case

namely

a

x

ffi

can

delay

while

a

k

x

ffi

can

not

because

x

can

not

Proposition

-LRB-

Properties

of

ACPdrt

Part

III

-RRB-

For

ACPdrt

terms

x

and

y

and

any

a

b

Affi

we

have

the

following

equalities

-LRB-

i

-RRB-

ACPdrt

a

j

b

a

j

b

-LRB-

ii

-RRB-

ACPdrt

a

j

b

a

j

b

-LRB-

iii

-RRB-

ACPdrt

a

j

b

x

a

j

b

x

-LRB-

iv

-RRB-

ACPdrt

a

x

j

b

a

j

b

x

-LRB-

v

-RRB-

ACPdrt

a

j

b

x

a

j

b

x

-LRB-

vi

-RRB-

ACPdrt

a

x

j

b

a

j

b

x

-LRB-

vii

-RRB-

ACPdrt

a

x

j

b

y

a

j

b

x

k

y

-LRB-

viii

-RRB-

ACPdrt

a

x

j

b

y

a

j

b

x

k

y

Proof

-LRB-

i

-RRB-

Consider

the

following

computation

ACPdrt

a

j

b

a

j

bbc

a

j

rel?b

oerel?bbc

a

j

b

oerel?b

a

j

b

a

j

oerel?b

a

j

b

rel?a

j

oerel?b

a

j

b

ffi

a

j

b

-LRB-

ii

-RRB-

This

case

is

treated

symmetrically

to

the

previous

case

-LRB-

iii

-RRB-

Consider

the

following

computation

ACPdrt

a

j

b

x

a

j

bbc

x

a

j

rel?b

oerel?bbc

x

a

j

b

oerel?b

x

a

j

b

x

oerel?b

x

a

j

b

x

a

j

oerel?b

x

a

j

b

x

rel?a

j

oerel?b

x

a

j

b

x

ffi

a

j

b

x

a

j

b

x

-LRB-

iv

-RRB-

This

case

is

treated

symmetrically

to

the

previous

case

-LRB-

v

-RRB-

Consider

the

following

computation

ACPdrt

a

j

b

x

bac

j

b

x

rel?a

oerel?bac

j

b

x

a

oerel?a

j

b

x

a

j

b

x

oerel?a

j

b

x

a

j

b

x

oerel?a

j

rel?b

x

a

j

b

x

oerel?a

j

rel?b

x

a

j

b

x

ffi

a

j

b

x

a

j

b

x

-LRB-

vi

-RRB-

This

case

is

treated

symmetrically

to

the

previous

case

-LRB-

vii

-RRB-

Consider

the

following

computation

ACPdrt

a

x

j

b

y

a

x

j

bbc

y

a

x

j

rel?b

oerel?bbc

y

a

x

j

b

oerel?b

y

a

x

j

b

y

oerel?b

y

a

x

j

b

y

a

x

j

oerel?b

y

a

x

j

b

y

rel?a

x

j

oerel?b

y

a

x

j

b

y

rel?a

x

j

oerel?b

y

a

x

j

b

y

ffi

a

x

j

b

y

a

j

b

x

k

y

-LRB-

viii

-RRB-

This

case

is

treated

symmetrically

to

the

previous

case

Theorem

-LRB-

Elimination

for

ACP?drt

-RRB-

Let

t

be

a

closed

ACPdrt

term

Then

there

is

closed

BPAdrt

term

s

such

that

ACP?drt

t

s.

Proof

Let

t

be

a

closed

ACPdrt

term

The

theorem

is

proven

by

induction

on

n?t

and

case

distinction

on

the

general

structure

of

t.

-LRB-

i

-RRB-

t

ffi

Then

t

is

a

closed

BPAdrt

term

-LRB-

ii

-RRB-

t

a

for

some

a

Affi

Then

t

is

a

closed

BPAdrt

term

-LRB-

iii

-RRB-

t

a

for

some

a

Affi

Then

t

is

a

closed

BPAdrt

term

-LRB-

iv

-RRB-

t

for

closed

ACPdrt

terms

and

By

induction

there

are

closed

BPAdrt

terms

and

such

that

ACP?drt

and

ACP?drt

But

then

also

ACP?drt

and

is

a

closed

BPAdrt

term

-LRB-

v

-RRB-

t

for

closed

ACPdrt

terms

and

This

case

is

treated

analogously

to

case

-LRB-

ii

-RRB-

-LRB-

vi

-RRB-

t

for

a

closed

ACPdrt

term

This

case

is

treated

analogously

to

case

-LRB-

ii

-RRB-

-LRB-

vii

-RRB-

t

for

a

closed

ACPdrt

term

This

case

is

treated

analogously

to

case

-LRB-

ii

-RRB-

-LRB-

viii

-RRB-

t

for

a

closed

ACPdrt

term

This

case

is

treated

analogously

to

case

-LRB-

ii

-RRB-

-LRB-

ix

-RRB-

t

k

for

closed

ACPdrt

terms

and

By

induction

there

are

closed

BPAdrt

terms

and

such

that

ACP?drt

and

ACP?drt

By

Theorem

the

elimination

theorem

for

BPAdrt

there

are

basic

terms

and

such

that

BPA?drt

and

BPA?drt

But

then

also

ACP?drt

ACP?drt

and

ACP?drt

k

k

We

proceed

by

induction

on

the

structure

of

basic

terms

and

distinguish

all

possible

cases

for

basic

term

-LRB-

a

-RRB-

ffi

Then

ACP?drt

k

k

ffi

k

ffi

and

ffi

is

a

closed

BPAdrt

term

-LRB-

b

-RRB-

a

for

some

a

Affi

Using

Lemma

we

distinguish

two

cases

i.

ffi

Then

we

have

ACP?drt

k

k

k

ffi

ffi

and

ffi

is

a

closed

BPAdrt

term

ii

ffi

Then

we

have

ACP?drt

k

k

a

k

ffi

a

ffi

a

and

a

is

a

closed

BPAdrt

term

-LRB-

c

-RRB-

a

for

some

a

Affi

Using

Lemma

we

distinguish

four

cases

i.

ffi

Then

we

have

ACP?drt

k

k

k

ffi

ffi

and

ffi

is

a

closed

BPAdrt

term

ii

ffi

Then

we

have

ACP?drt

k

k

a

k

bac

k

rel?a

oerel?bac

k

a

oerel?a

k

a

k

oerel?a

k

a

k

ffi

oerel?a

k

ffi

a

ffi

ffi

a

ffi

a

ffi

a

ffi

a

and

a

is

a

closed

BPAdrt

term

iii

Then

using

Proposition

-LRB-

vi

-RRB-

we

have

ACP?drt

k

k

a

k

a

a

and

a

is

a

closed

BPAdrt

term

iv

for

a

basic

term

such

that

Then

we

have

ACP?drt

k

k

a

k

bac

k

rel?a

oerel?bac

k

a

oerel?a

k

a

k

oerel?a

k

a

k

oerel?a

k

a

k

ffi

oerel?a

k

a

ffi

oerel?a

k

a

oerel?a

k

a

oerel?a

k

By

the

induction

hypothesis

there

exists

a

closed

BPAdrt

term

p

such

that

ACP?drt

a

k

p.

Then

ACP?drt

k

a

oerel?a

k

a

oerel?p

and

a

oerel?p

is

a

closed

BPAdrt

term

-LRB-

d

-RRB-

a

for

some

a

Affi

and

basic

term

Using

Lemma

we

distinguish

two

cases

i.

ffi

Then

we

have

ACP?drt

k

k

k

ffi

ffi

and

ffi

is

a

closed

BPAdrt

term

ii

ffi

Then

we

have

ACP?drt

k

k

a

k

ffi

a

k

ffi

a

k

By

the

induction

hypothesis

there

exists

a

closed

BPAdrt

term

p

such

that

ACP?drt

k

p.

Then

ACP?drt

k

a

k

a

p

and

a

p

is

a

closed

BPAdrt

term

-LRB-

e

-RRB-

a

for

some

a

Affi

and

basic

term

Using

Lemma

we

distinguish

four

cases

i.

ffi

Then

we

have

ACP?drt

k

k

k

ffi

ffi

and

ffi

is

a

closed

BPAdrt

term

ii

ffi

Then

we

have

ACP?drt

k

k

a

k

bac

k

rel?a

oerel?bac

k

a

oerel?a

k

a

oerel?a

k

a

oerel?a

k

a

k

oerel?a

k

a

k

ffi

oerel?a

k

ffi

a

k

ffi

ffi

a

k

ffi

a

k

ffi

k

a

ffi

k

a

k

By

the

induction

hypothesis

there

exists

a

closed

BPAdrt

term

p

such

that

ACP?drt

k

p.

Then

ACP?drt

k

a

k

a

p

and

a

p

is

a

closed

BPAdrt

term

iii

Then

using

Proposition

-LRB-

vii

-RRB-

we

have

ACP?drt

k

k

a

k

a

k

a

k

By

the

induction

hypothesis

there

exists

a

closed

BPAdrt

term

p

such

that

ACP?drt

k

p.

Then

ACP?drt

k

a

k

a

p

and

a

p

is

a

closed

BPAdrt

term

iv

for

a

basic

term

such

that

Then

we

have

ACP?drt

k

k

a

k

bac

k

rel?a

oerel?bac

k

a

oerel?a

k

a

oerel?a

k

a

oerel?a

k

a

k

oerel?a

k

a

k

oerel?a

k

a

k

ffi

oerel?a

k

a

k

ffi

oerel?a

k

a

k

k

a

k

k

By

the

induction

hypothesis

there

exist

closed

BPAdrt

terms

and

such

that

ACP?drt

k

and

ACP?drt

k

Then

ACP?drt

k

a

k

k

and

is

a

closed

BPAdrt

term

-LRB-

f

-RRB-

for

basic

terms

and

Then

ACP?drt

k

k

k

k

k

By

induction

there

exist

closed

BPAdrt

terms

and

such

that

ACP?drt

k

and

ACP?drt

k

Then

also

ACP?drt

k

k

k

and

is

a

closed

BPAdrt

term

-LRB-

g

-RRB-

for

a

basic

term

Using

Lemma

we

distinguish

four

cases

i.

ffi

Then

we

have

ACP?drt

k

k

a

k

ffi

ffi

and

ffi

is

a

closed

BPAdrt

term

ii

ffi

Then

ACP?drt

k

k

k

ffi

ffi

and

ffi

is

a

closed

BPAdrt

term

iii

Then

we

have

ACP?drt

k

k

k

k

k

k

By

the

induction

hypothesis

there

exists

a

closed

BPAdrt

term

p

such

that

ACP?drt

k

p.

Then

ACP?drt

k

k

oerel?p

and

oerel?p

is

a

closed

BPAdrt

term

iv

for

a

basic

term

such

that

Then

we

have

ACP?drt

k

k

k

k

By

the

induction

hypothesis

there

is

a

closed

BPAdrt

term

p

such

that

ACP?drt

k

p.

Then

ACP?drt

k

k

oerel?p

and

oerel?p

is

a

closed

BPAdrt

term

-LRB-

x

-RRB-

t

j

for

closed

ACPdrt

terms

and

By

induction

there

are

closed

BPAdrt

terms

and

such

that

ACP?drt

and

ACP?drt

By

Theorem

the

elimination

theorem

for

BPAdrt

there

are

basic

terms

and

such

that

BPA?drt

and

BPA?drt

But

then

also

ACP?drt

ACP?drt

and

ACP?drt

j

j

We

prove

this

case

by

simultaneous

induction

on

the

structure

of

basic

terms

and

We

examine

all

possible

cases

-LRB-

of

which

there

are

in

total

some

of

which

can

be

treated

simultaneously

reducing

our

task

to

just

cases

-RRB-

-LRB-

a

-RRB-

ffi

and

is

of

arbitrary

form

Then

ACP?drt

j

j

ffi

j

ffi

and

ffi

is

a

closed

BPAdrt

term

-LRB-

b

-RRB-

is

of

arbitrary

form

and

ffi

This

case

is

treated

symmetrically

to

the

previous

case

-LRB-

c

-RRB-

a

and

b

for

some

a

b

Affi

Suppose

that

fl?a

b

c

Then

we

have

ACP?drt

j

j

a

j

b

c

and

c

is

a

closed

BPAdrt

term

-LRB-

d

-RRB-

a

and

b

for

some

a

b

Affi

Suppose

that

fl?a

b

c

Then

we

have

ACP?drt

j

j

a

j

b

c

and

c

is

a

closed

BPAdrt

term

-LRB-

e

-RRB-

a

and

b

for

some

a

b

Affi

This

case

is

treated

symmetrically

to

the

previous

case

-LRB-

f

-RRB-

a

and

b

for

some

a

b

Affi

Suppose

that

fl?a

b

c

Then

we

have

ACP?drt

j

j

a

j

b

c

and

c

is

a

closed

BPAdrt

term

-LRB-

g

-RRB-

a

and

for

some

a

b

Affi

and

some

basic

term

Suppose

that

fl?a

b

c

Then

we

have

ACP?drt

j

j

a

j

b

c

and

c

is

a

closed

BPAdrt

term

-LRB-

h

-RRB-

a

and

b

for

some

a

b

Affi

and

some

basic

term

This

case

is

treated

symmetrically

to

the

previous

case

-LRB-

i

-RRB-

a

and

for

some

a

b

Affi

and

some

basic

term

Suppose

that

fl?a

b

c

Then

we

have

ACP?drt

j

j

a

j

b

c

and

c

is

a

closed

BPAdrt

term

-LRB-

j

-RRB-

a

and

b

for

some

a

b

Affi

and

some

basic

term

This

case

is

treated

symmetrically

to

the

previous

case

-LRB-

k

-RRB-

a

and

for

some

a

b

Affi

and

some

basic

term

Suppose

that

fl?a

b

c

Then

we

have

ACP?drt

j

j

a

j

b

c

and

c

is

a

closed

BPAdrt

term

-LRB-

l

-RRB-

a

and

b

for

some

a

b

Affi

and

some

basic

term

This

case

is

treated

symmetrically

to

the

previous

case

-LRB-

m

-RRB-

a

and

for

some

a

b

Affi

and

some

basic

term

Suppose

that

fl?a

b

c

Then

we

have

ACP?drt

j

j

a

j

b

c

and

c

is

a

closed

BPAdrt

term

-LRB-

n

-RRB-

a

and

b

for

some

a

b

Affi

and

some

basic

term

This

case

is

treated

symmetrically

to

the

previous

case

-LRB-

o

-RRB-

a

and

b

for

some

a

b

Affi

and

some

basic

terms

and

Suppose

that

fl?a

b

c

Then

we

have

ACP?drt

c

k

By

the

induction

hypothesis

there

exists

a

closed

BPAdrt

term

such

that

ACP?drt

k

So

ACP?drt

j

c

k

c

and

c

is

a

closed

BPAdrt

term

-LRB-

p

-RRB-

a

and

b

for

some

a

b

Affi

and

some

basic

terms

and

Suppose

that

fl?a

b

c

Then

we

have

ACP?drt

c

k

By

the

induction

hypothesis

there

exists

a

closed

BPAdrt

term

such

that

ACP?drt

k

So

ACP?drt

j

c

k

c

and

c

is

a

closed

BPAdrt

term

-LRB-

q

-RRB-

a

and

b

for

some

a

b

Affi

and

some

basic

terms

and

This

case

is

treated

symmetrically

to

the

previous

case

-LRB-

r

-RRB-

a

and

b

for

some

a

b

Affi

and

some

basic

terms

and

Suppose

that

fl?a

b

c

Then

we

have

ACP?drt

c

k

By

the

induction

hypothesis

there

exists

a

closed

BPAdrt

term

such

that

ACP?drt

k

So

ACP?drt

j

c

k

c

and

c

is

a

closed

BPAdrt

term

-LRB-

s

-RRB-

for

some

basic

terms

and

and

is

of

arbitrary

form

Then

ACP?drt

j

j

j

j

j

By

the

induction

hypothesis

there

exist

closed

BPAdrt

terms

and

such

that

ACP?drt

and

ACP?drt

So

we

have

ACP?drt

and

is

a

closed

BPAdrt

term

-LRB-

t

-RRB-

is

of

arbitrary

form

and

for

some

basic

terms

and

This

case

is

treated

symmetrically

to

the

previous

case

-LRB-

u

-RRB-

and

a

basic

term

and

is

of

arbitrary

form

Using

Lemma

we

distinguish

four

cases

i.

ffi

Then

we

have

ACP?drt

j

j

j

ffi

ffi

and

ffi

is

a

closed

BPAdrt

term

ii

ffi

Then

ACP?drt

ffi

ffi

and

ffi

is

a

closed

BPAdrt

term

iii

Then

we

have

ACP?drt

j

j

j

j

j

j

By

the

induction

hypothesis

there

is

a

closed

BPAdrt

term

p

such

that

ACP?drt

j

p

But

then

also

ACP?drt

j

j

oerel?p

and

oerel?p

is

a

closed

BPAdrt

term

iv

for

a

basic

term

such

that

Then

we

have

ACP?drt

j

j

j

j

j

By

the

induction

hypothesis

there

is

a

closed

BPAdrt

term

p

such

that

ACP?drt

j

p

But

then

also

ACP?drt

j

j

oerel?p

and

oerel?p

is

a

closed

BPAdrt

term

-LRB-

v

-RRB-

is

of

arbitrary

form

and

and

a

basic

term

This

case

is

treated

symmetrically

to

the

previous

case

-LRB-

xi

-RRB-

t

k

for

closed

ACPdrt

terms

and

Then

ACP?drt

k

k

k

j

By

-LRB-

ix

-RRB-

and

-LRB-

x

-RRB-

there

are

closed

BPAdrt

terms

and

such

that

ACP?drt

k

ACP?drt

k

and

ACP?drt

j

But

then

also

ACP?drt

k

k

k

j

and

is

a

closed

BPAdrt

term

Corollary

-LRB-

Elimination

for

ACP?drt

-RRB-

Let

t

be

a

closed

ACPdrt

term

Then

there

is

a

basic

term

s

such

that

ACP?drt

s

t.

Proof

This

follows

immediately

from

-LRB-

i

-RRB-

The

elimination

theorem

for

ACP?drt

-LRB-

see

Theorem

-RRB-

-LRB-

ii

-RRB-

the

elimination

theorem

for

BPA?drt

-LRB-

see

Theorem

-RRB-

-LRB-

iii

-RRB-

the

fact

that

all

axioms

of

BPA?drt

are

also

contained

in

ACP?drt

Remark

-LRB-

Elimination

for

ACPdrt

-RRB-

Elimination

for

a

somewhat

different

version

of

ACPdrt

is

also

claimed

-LRB-

without

proof

-RRB-

in

Section

of

-LSB-

-RSB-

Theorem

-LRB-

Soundness

of

ACP?drt

-RRB-

The

set

of

closed

ACPdrt

terms

modulo

bisimulation

equivalence

is

a

model

of

ACP?drt

Proof

We

only

prove

soundness

for

the

axioms

of

ACPdrt

that

have

not

been

treated

in

earlier

soundness

proofs

Note

that

to

extend

these

proofs

to

ACPdrt

we

have

to

check

that

the

bisimulations

given

in

previous

soundness

proofs

respect

the

ID

predicate

-LRB-

as

required

by

transfer

condition

-LRB-

iv

-RRB-

in

on

page

-RRB-

However

as

the

fact

that

they

do

can

be

easily

checked

we

will

not

give

details

Axiom

Take

the

relation

R

f?oerel?s

j

rel?t

ffi

ffi?js

t

C?ACPdrt?g

We

look

at

the

transitions

of

both

sides

at

the

same

time

Observe

that

there

are

no

transitions

possible

on

the

left-hand

side

oerel?s

j

rel?t

ffi

Also

for

the

right-hand

side

there

are

no

transitions

possible

ffi

Finally

neither

side

satisfies

the

ID

predicate

ID?oerel?s

j

rel?t

ffi

and

ID?ffi

Axiom

Take

the

relation

R

f

rel?s

ffi

j

oerel?t

ffi?js

t

C?ACPdrt?g

This

case

is

treated

symmetrically

to

the

previous

case

Axiom

Take

the

relation

R

f?s

j

ffi

ffi?js

C?ACPdrt?g

We

look

at

the

transitions

of

both

sides

at

the

same

time

Observe

that

there

are

no

transitions

possible

on

the

left-hand

side

s

j

ffi

Also

for

the

right-hand

side

there

are

no

transitions

possible

ffi

Finally

both

sides

satisfy

the

ID

predicate

ID?s

j

ffi

and

ID

ffi

Axiom

Take

the

relation

R

f

ffi

j

s

ffi?js

C?ACPdrt?g

This

case

is

treated

symmetrically

to

the

previous

case

Remark

-LRB-

Soundness

of

ACPdrt

-RRB-

Soundness

of

a

somewhat

different

version

of

ACPdrt

is

also

claimed

-LRB-

without

proof

-RRB-

in

Section

of

-LSB-

-RSB-

Theorem

-LRB-

Conservativity

of

ACP?drt

with

respect

to

BPA?drt

-RRB-

The

equational

specification

ACP?drt

is

a

conservative

extension

of

the

equational

specification

BPA?drt

Proof

In

order

to

prove

conservativity

it

is

sufficient

to

verify

that

the

following

conditions

are

satisfied

-LRB-

i

-RRB-

Bisimulation

equivalence

is

definable

in

terms

of

predicate

and

relation

symbols

only

-LRB-

ii

-RRB-

BPA?drt

is

a

complete

axiomatization

with

respect

to

the

bisimulation

equivalence

model

induced

by

T?BPAdrt

-LRB-

see

Theorem

-RRB-

-LRB-

iii

-RRB-

ACP?drt

is

a

sound

axiomatization

with

respect

to

the

bisimulation

equivalence

model

induced

by

T?ACPdrt

-LRB-

see

Theorem

-RRB-

-LRB-

iv

-RRB-

T?ACPdrt

is

an

operationally

conservative

extension

of

T?BPAdrt

And

in

order

for

T?ACPdrt

indeed

to

be

an

operationally

conservative

extension

of

T?BPAdrt

we

must

verify

the

following

conditions

-LRB-

i

-RRB-

T?BPAdrt

is

a

pure

well-founded

term

deduction

system

in

path

format

-LRB-

ii

-RRB-

T?ACPdrt

is

a

term

deduction

system

in

path

format

-LRB-

iii

-RRB-

T?BPAdrt

T?ACPdrt

is

defined

That

the

above

properties

hold

can

be

trivially

checked

from

the

relevant

definitions

Theorem

-LRB-

Completeness

of

ACP?drt

-RRB-

The

equational

specification

ACP?drt

is

a

complete

axiomatization

of

the

set

of

closed

ACPdrt

terms

modulo

bisimulation

equivalence

Proof

By

Verhoef

's

General

Completeness

Theorem

-LRB-

see

-LSB-

-RSB-

or

Theorem

of

-LSB-

-RSB-

-RRB-

this

follows

immediately

from

-LRB-

i

-RRB-

ACP?drt

has

the

elimination

property

for

BPAdrt

-LRB-

see

Theorem

-RRB-

-LRB-

ii

-RRB-

ACP?drt

is

a

conservative

extension

of

BPA?drt

-LRB-

see

Theorem

-RRB-

Remark

-LRB-

Completeness

of

ACPdrt

-RRB-

Completeness

of

a

somewhat

different

version

of

ACPdrt

is

also

claimed

-LRB-

without

proof

-RRB-

in

Section

of

-LSB-

-RSB-

Definition

-LRB-

Axioms

for

the

Communication

Merge

and

Delayable

Actions

-RRB-

We

define

the

Axioms

and

for

the

ultimate

start

delay

with

respect

to

the

communication

merge

as

shown

in

Table

on

the

following

page

Note

that

they

precisely

correspond

to

the

equalities

of

Proposition

Corollary

-LRB-

Soundness

of

ACPdrt

-RRB-

The

set

of

closed

ACPdrt

terms

modulo

bisimulation

equivalence

is

a

model

of

ACPdrt

a

j

b

c

if

fl?a

b

c

ffi

a

j

b

ffi

if

fl?a

b

ffi

a

j

b

x

a

j

b

x

a

x

j

b

a

j

b

x

a

x

j

b

y

a

j

b

x

k

y

Table

Axioms

for

the

communication

merge

and

delayable

actions

Proof

This

follows

directly

from

the

soundness

of

ACP?drt

-LRB-

see

Theorem

on

page

-RRB-

and

the

fact

that

Axiom

and

are

derivable

in

ACP?drt

-LRB-

see

Proposition

on

page

and

Proposition

on

page

-RRB-

Corollary

-LRB-

Completeness

of

ACPdrt

-RRB-

If

we

add

Axioms

of

Table

on

page

Axiom

of

Table

on

page

Axioms

of

Table

on

page

and

Axioms

and

of

Table

to

ACPdrt

we

again

have

a

complete

axiomatization

of

the

set

of

closed

ACPdrt

terms

modulo

bisimulation

equivalence

Proof

Careful

inspection

of

the

dependencies

between

the

proofs

in

this

section

reveals

that

the

proof

of

Theorem

only

relies

upon

RSP

-LRB-

USD

-RRB-

to

ensure

Proposition

-LRB-

i

-RRB-

-LRB-

vii

-RRB-

and

Proposition

So

we

obviously

do

not

need

RSP

-LRB-

USD

-RRB-

anymore

if

we

add

the

corresponding

Axioms

and

Note

that

in

this

way

we

get

a

purely

equational

axiomatization

-LRB-

i.e.

without

conditional

axioms

or

principles

-RRB-

Note

also

that

Axioms

and

are

not

conditional

axioms

but

axiom

schemes

as

are

all

axioms

that

contain

an

atomic

action

Definition

-LRB-

Axiom

for

the

Ultimate

Start

Delay

and

Communication

Merge

-RRB-

We

define

the

Axiom

for

the

ultimate

start

delay

of

a

communication

merge

as

shown

in

Table

bx

j

yc

bxc!j

byc

Table

Axiom

for

bx

j

yc

Theorem

-LRB-

Soundness

of

-RRB-

The

set

of

closed

ACPdrt

terms

modulo

bisimulation

equivalence

is

a

model

of

Proof

Take

the

relation

R

s

s

bs

j

tc

bsc!j

btc

js

C?ACPdrt

First

we

look

at

the

transitions

of

the

left-hand

side

-LRB-

i

-RRB-

Suppose

bs

j

tc

a

p.

By

inspection

of

the

deduction

rules

we

distinguish

the

following

cases

-LRB-

a

-RRB-

s

b

t

c

fl?b

c

a

and

p

k

Then

bsc

b

and

btc

c

so

bsc!j

btc

a

k

and

k

k

R.

-LRB-

b

-RRB-

s

b

p

t

c

fl?b

c

a

and

p

Then

bsc

b

p

and

btc

c

so

bsc!j

btc

a

and

R.

-LRB-

c

-RRB-

s

b

t

c

p

fl?b

c

a

and

p

Then

bsc

b

and

btc

c

p

so

bsc!j

btc

a

and

R.

-LRB-

ii

-RRB-

Suppose

bs

j

tc

a

p.

By

inspection

of

the

deduction

rules

we

can

conclude

that

s

b

p

t

c

p

and

fl?b

c

a

Then

bsc

b

p

and

btc

c

p

so

bsc!j

btc

a

p.

-LRB-

iii

-RRB-

Suppose

bs

j

tc

oe

p.

By

inspection

of

the

deduction

rules

we

can

conclude

that

p

bs

j

tc

We

also

have

bsc!j

btc

oe

bsc!j

btc

and

bs

j

tc

bsc!j

btc

R.

Secondly

we

look

at

the

transition

of

the

right-hand

side

-LRB-

i

-RRB-

Suppose

bsc

j

btc

a

p.

By

inspection

of

the

deduction

rules

we

distinguish

the

following

cases

-LRB-

a

-RRB-

s

b

t

c

fl?b

c

a

and

p

k

Then

bs

j

tc

c

k

and

k

k

R.

-LRB-

b

-RRB-

s

b

p

t

c

fl?b

c

a

and

p

Then

bs

j

tc

a

and

R.

-LRB-

c

-RRB-

s

b

t

c

p

fl?b

c

a

and

p

Then

bs

j

tc

a

and

R.

-LRB-

ii

-RRB-

Suppose

bsc!jbtc

a

p.

By

inspection

of

the

deduction

rules

we

can

conclude

that

s

b

p

t

c

p

and

fl?b

c

a

Then

bs

j

tc

a

p

and

we

are

done

-LRB-

iii

-RRB-

Suppose

bsc!j

btc

oe

p.

By

inspection

of

the

deduction

rules

we

can

conclude

that

p

bsc!j

btc

We

also

have

bs

j

tc

oe

bs

j

tc

and

bs

j

tc

bsc!j

btc

R.

Finally

we

look

at

the

immediate

deadlock

predicate

Neither

side

has

immediate

deadlock

ID?bs

j

tc

and

ID?bsc!j

btc

-LRB-

note

that

ultimate

start

delay

removes

immediate

deadlock

see

Remark

on

page

-RRB-

Proposition

-LRB-

Properties

of

ACPdrt

Part

IV

-RRB-

For

ACPdrt

terms

x

and

y

and

any

a

b

c

Affi

we

have

the

following

equalities

-LRB-

i

-RRB-

ACPdrt

a

j

b

c

if

fl?a

b

c

ffi

-LRB-

ii

-RRB-

ACPdrt

a

j

b

ffi

if

fl?a

b

ffi

-LRB-

iii

-RRB-

ACPdrt

a

j

b

x

a

j

b

x

-LRB-

iv

-RRB-

ACPdrt

a

x

j

b

a

j

b

x

-LRB-

v

-RRB-

ACPdrt

a

x

j

b

y

a

j

b

x

k

y

Proof

-LRB-

i

-RRB-

ACPdrt

a

j

b

bac

j

bbc

ba

j

bc

bcc

c

-LRB-

ii

-RRB-

ACPdrt

a

j

b

bac

j

bbc

ba

j

bc

bffic

ffi

-LRB-

iii

-RRB-

ACPdrt

a

j

b

x

bac!j

bbc

x

bac

j

bb

xc

ba

j

b

xc

b?a

j

b

xc

b?a

j

b?c

x

bac!j

bbc

x

a

j

b

x

-LRB-

iv

-RRB-

ACPdrt

a

x

j

b

bac

x

j

bbc

ba

xc

j

bbc

ba

x

j

bc

b?a

j

b

xc

b?a

j

b?c

x

bac!j

bbc

x

a

j

b

x

-LRB-

v

-RRB-

ACPdrt

a

x

j

b

y

bac

x

j

bbc

y

ba

xc

j

bb

yc

ba

x

j

b

yc

b?a

j

b

x

k

y?c

b?a

j

b?c

x

k

y

bac!jbbc

x

k

y

a

j

b

x

k

y

Remark

-LRB-

Properties

of

ACPdrt

Part

IV

-RRB-

Note

that

the

equalities

of

Proposition

on

the

preceding

page

correspond

precisely

to

the

equalities

of

Proposition

on

page

and

Axioms

and

from

Definition

on

page

Corollary

-LRB-

Soundness

of

ACPdrt

-RRB-

The

set

of

closed

ACPdrt

terms

modulo

bisimulation

equivalence

is

a

model

of

ACPdrt

Proof

This

follows

directly

from

the

soundness

of

ACP?drt

-LRB-

see

Theorem

on

page

-RRB-

and

the

facts

that

Axioms

are

derivable

in

ACP?drt

-LRB-

see

Proposition

on

page

-RRB-

and

that

Axiom

is

sound

-LRB-

see

Theorem

on

page

-RRB-

Corollary

-LRB-

Completeness

of

ACPdrt

-RRB-

If

we

add

Axioms

of

Table

on

page

Axiom

of

Table

on

page

Axioms

of

Table

on

page

and

Axiom

of

Table

on

page

to

ACPdrt

we

again

have

a

complete

axiomatization

of

the

set

of

closed

ACPdrt

terms

modulo

bisimulation

equivalence

Proof

As

show

in

Proposition

if

we

have

the

add

Axioms

and

to

the

axioms

of

ACPdrt

we

can

-LRB-

for

closed

terms

-RRB-

derive

Axiom

and

Using

Corollary

on

page

we

can

now

trivially

derive

the

desired

result

Conclusions

To

begin

with

we

are

reasonably

confident

that

the

axiomatizations

listed

in

this

paper

are

sound

and

complete

An

overview

of

the

main

theorems

is

listed

in

Table

on

page

Before

we

started

this

paper

we

were

also

confident

of

the

soundness

and

completeness

of

these

axiomatizations

but

at

that

time

wrongly

so

We

discovered

that

the

axiomatizations

we

started

out

with

most

of

which

have

been

published

and

claimed

sound

and

complete

before

were

neither

sound

nor

complete

We

highlight

two

characteristic

cases

ffl

Weakening

to

brings

the

need

to

introduce

something

we

did

not

realize

at

first

This

left

all

interesting

theories

incomplete

Only

when

was

needed

in

the

proof

of

Lemma

-LRB-

iv

-RRB-

we

found

out

about

this

mistake

ffl

Introducing

ffi

in

a

context

that

supports

communication

brings

the

need

to

weaken

to

which

we

did

not

realize

due

to

the

intuitive

and

obvious

nature

of

This

left

some

theories

unsound

We

found

out

this

problem

after

we

could

not

complete

the

last

few

trivial

details

of

the

proof

of

Theorem

Both

these

problems

were

discovered

when

we

were

writing

out

all

details

of

trivial

proofs

proofs

which

we

had

originally

not

planned

to

do

at

all

So

we

eventually

decided

to

give

as

much

and

as

detailed

proofs

as

reasonably

manageable

And

as

to

be

expected

we

found

some

more

mistakes

like

the

ones

listed

above

As

a

side-result

we

gained

insight

into

the

various

aspects

of

axiomatizations

Firstly

when

we

weigh

the

merits

of

the

rel/oerel

axiomatization

style

of

-LSB-

-RSB-

-LRB-

with

theories

like

PA?drt?ID

and

ACP?drt?ID

-RRB-

against

those

of

the

classic

axiomatization

style

of

-LSB-

-RSB-

-LRB-

with

theories

like

-RRB-

we

conclude

that

the

rel/oerel

style

is

better

suited

towards

practical

applications

as

it

makes

calculations

easier

However

from

a

theoretical

viewpoint

it

's

troublesome

it

does

not

lend

itself

well

to

term

rewriting

system

analysis

and

worse

it

does

not

seem

to

be

compatible

with

the

addition

of

the

empty

step

as

is

shown

in

-LSB-

-RSB-

On

the

other

hand

the

classic

style

is

not

ideal

either

It

appears

less

intuitive

and

it

needs

more

axioms

Compare

for

example

Axioms

of

Table

on

page

with

Axioms

of

Table

on

page

Here

the

classic

style

needs

five

axioms

to

do

what

the

rel/oerel

style

can

do

much

clearer

in

two

axioms

Consequently

calculations

in

the

classic

style

are

much

longer

too

Second

we

have

shown

how

to

eliminate

the

recursion

principle

RSP

-LRB-

USD

-RRB-

from

the

theories

that

contain

it

As

shown

in

Corollaries

and

one

can

straightforwardly

derive

unconditional

axioms

to

replace

the

conditional

principle

RSP

-LRB-

USD

-RRB-

The

recipe

is

always

the

same

identify

in

the

correctness

proof

of

the

conditional

theory

the

places

where

RSP

-LRB-

USD

-RRB-

is

used

put

those

applications

in

a

separate

lemma

and

introduce

an

axiom

for

every

clause

of

that

lemma

Using

this

recipe

we

introduced

Axioms

and

The

advantage

is

clear

having

a

fully

unconditional

theory

enables

us

to

reason

fully

algebraically

giving

us

a

fuller

apparatus

of

methods

to

work

with

On

the

other

hand

the

principle

RSP

-LRB-

USD

-RRB-

is

clean

neat

and

simple

and

can

be

applied

in

any

theory

while

the

USD

axiomatization

style

requires

new

axioms

for

every

new

theory

That

this

can

lead

to

unwieldy

theories

can

be

observed

from

-LSB-

-RSB-

We

feel

that

the

RSP

-LRB-

USD

-RRB-

axiomatization

style

which

till

now

has

only

appeared

in

the

rather

obscure

papers

-LSB-

-RSB-

deserves

a

wider

audience

Third

we

find

that

the

absence

of

the

empty

step

-LRB-

a

constant

such

that

x

x

x

see

-LSB-

-RSB-

for

details

-RRB-

is

a

major

nuisance

To

begin

with

the

empty

step

would

allow

us

to

express

our

axioms

much

more

compact

For

example

Axiom

would

be

just

an

instance

of

Axiom

if

the

x

in

the

latter

could

take

the

value

Similarly

Axioms

could

also

be

collapsed

to

just

one

axiom

with

the

help

of

-LRB-

and

see

-LSB-

-RSB-

for

a

theory

in

which

it

is

hard

to

find

any

axiom

that

could

not

be

formulated

better

with

the

help

of

the

empty

step

-RRB-

The

absence

of

is

even

felt

worse

when

doing

calculations

If

we

look

for

example

at

the

proof

of

Theorem

we

see

we

have

to

distinguish

cases

-LRB-

-RRB-

when

doing

simultaneous

induction

on

two

variables

as

a

basic

term

in

ACPdrt

can

take

seven

essentially

different

forms

With

the

help

of

we

could

reduce

this

to

five

forms

and

only

cases

would

have

to

be

considered

when

doing

induction

on

two

variables

Similar

considerations

hold

for

the

proof

of

Theorem

on

page

where

the

absence

of

for

some

case

even

forces

us

into

a

sixteen-fold

-LRB-

sic

-RRB-

increase

in

proof

obligations

We

conclude

that

there

is

a

clear

need

for

the

empty

step

in

discrete-time

process

algebra

Then

we

hope

that

this

paper

can

serve

as

a

reference

point

to

our

knowledge

this

is

the

first

paper

that

extensively

lists

all

important

discrete-time

process

algebra

theories

together

with

all

relevant

definitions

and

elementary

theorems

Furthermore

as

all

proofs

in

this

paper

are

constructive

it

should

now

be

easy

to

develop

a

tool

that

can

automatically

rewrite

two

bisimilar

ACPdrt

terms

into

one

another

Finally

note

that

we

have

surveyed

several

distinct

methods

for

proving

soundness

and

completeness

To

prove

soundness

we

have

used

ffl

the

direct

method

-LRB-

see

Remark

on

page

-RRB-

ffl

the

indirect

method

-LRB-

see

Remark

on

page

-RRB-

and

ffl

the

ground

equivalence

method

-LRB-

see

Remark

on

page

-RRB-

To

prove

completeness

we

have

used

ffl

the

direct

method

-LRB-

see

Remark

on

page

-RRB-

ffl

the

indirect

method

-LRB-

see

Remark

on

page

-RRB-

ffl

Verhoef

's

method

-LRB-

see

Remark

on

page

-RRB-

and

ffl

the

ground

equivalence

method

-LRB-

see

Remark

on

page

-RRB-

We

believe

that

this

spectrum

of

methods

provides

a

convenient

starting

point

to

prove

soundness

and

completeness

of

most

-LRB-

timed

-RRB-

process

algebras

that

have

been

described

in

the

literature

with

the

exception

of

theories

that

support

abstraction

As

far

as

future

research

is

concerned

we

would

like

to

generalize

our

results

to

a

setting

that

includes

abstraction

This

seems

however

not

at

all

trivial

and

may

require

a

substantial

effort

Furthermore

as

noted

above

further

research

on

the

empty

step

is

justified

This

work

is

currently

in

progress

and

will

be

published

as

-LSB-

-RSB-

Acknowledgments

Note

that

some

of

the

results

presented

here

were

taken

from

-LSB-

-RSB-

our

Lemma

is

their

Lemma

-LRB-

but

with

a

more

detailed

proof

-RRB-

our

Theorem

is

their

Theorem

-LRB-

with

a

more

detailed

proof

-RRB-

our

Theorem

is

their

Theorem

and

finally

our

Theorem

is

their

Theorem

-LRB-

also

with

a

more

detailed

proof

-RRB-

We

like

to

thank

Jos

Baeten

Roel

Bloo

Sjouke

Mauw

and

Hans

Mulder

for

their

various

useful

suggestions

A

Nomenclature

and

notational

issues

In

this

section

we

present

a

quick

overview

of

the

nomenclature

used

in

ACP-style

discrete-time

process

algebras

A.

Signatures

In

Table

we

give

the

signature

of

some

process

algebras

among

which

all

those

treated

in

this

paper

a

ffi

b

c

oerel

rel

a

ffi

ffi

k

k

j

BPA

ffl

ffl

ffl

BPAffi

ffl

ffl

ffl

ffl

PA

ffl

ffl

ffl

ffl

ffl

PAffi

ffl

ffl

ffl

ffl

ffl

ffl

ACP

ffl

ffl

ffl

ffl

ffl

ffl

ffl

BPA?drt?ffi

ffl

ffl

ffl

ffl

BPA?drt?ID

ffl

ffl

ffl

ffl

ffl

ffl

BPA?drt

ffl

ffl

ffl

ffl

ffl

ffl

ffl

BPAdrt?ID

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

BPAdrt

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

PA?drt?ID

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

PA?drt

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

PAdrt?ID

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

PAdrt

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ACP?drt?ID

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ACP?drt

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ACPdrt?ID

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ACPdrt

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

Table

Signatures

of

some

process

algebras

In

naming

discrete-time

process

algebras

we

have

among

others

the

following

conventions

ffl

The

subscript

drt

signifies

discrete

relative

time

so

we

have

the

relative-time

time-unit

delay

-LRB-

oerel

-RRB-

and

furthermore

either

non-delayable

actions

-LRB-

double

underlined

a

-RRB-

or

delayable

actions

-LRB-

no

special

notation

a

-RRB-

or

both

If

we

have

the

undelayable

deadlock

constant

-LRB-

ffi

-RRB-

we

also

have

the

now

operator

-LRB-

rel

-RRB-

and

vice

versa

ffl

The

superscript

signifies

that

we

do

not

have

delayable

actions

-LRB-

a

-RRB-

so

only

the

non-delayable

actions

remain

-LRB-

a

-RRB-

ffl

The

postfix

ffi

signifies

the

absence

of

all

deadlock

constants

non-delayable

-LRB-

ffi

-RRB-

delayable

-LRB-

ffi

-RRB-

or

immediate

-LRB-

ffi

-RRB-

ffl

The

postfix

ID

signifies

the

absence

of

the

immediate

deadlock

constant

-LRB-

ffi

-RRB-

A.

Comparison

with

other

notations

In

the

-LRB-

older

-RRB-

literature

about

ACP-style

discrete-time

process

algebras

we

sometimes

find

slightly

different

notations

Below

we

list

the

most

important

differences

ffl

We

find

the

notation

BPAdt

for

BPA?drt?ffi

BPAffidt

for

BPA?drt?ID

and

PAffidt

for

PA?drt?ID

-LSB-

-RSB-

ffl

We

find

the

notation

ACPdt

for

ACP?drt?ID

-LSB-

-RSB-

ffl

We

find

the

notations

oed

and

oerel

for

oerel

-LSB-

-RSB-

ffl

We

find

the

notation

cts?a

-LRB-

current

time

slice

-RRB-

for

the

undelayable

action

a

and

ats?a

-LRB-

any

time

slice

-RRB-

for

the

delayable

action

a

-LSB-

-RSB-

ffl

We

find

the

notation

a

for

the

undelayable

action

a

and

a

for

the

delayable

action

a

-LSB-

-RSB-

ffl

It

has

been

suggested

to

refer

to

the

delayable

deadlock

constant

ffi

as

livelock

and

to

the

immediate

deadlock

constant

ffi

as

the

immediate

time

stop

full

time

stop

or

catastrophic

deadlock

-LSB-

-RSB-

ffl

The

unbounded

start

delay

operator

b

c

was

first

described

by

Nicollin

and

Sifakis

and

is

therefore

also

known

as

one

of

the

Nicollin-Sifakis

operators

-LSB-

-RSB-

ffl

The

naming

scheme

for

axioms

-LRB-

etc.

-RRB-

has

become

quite

problematic

Naming

is

generally

not

applied

consistently

within

ACP

articles

published

let

alone

across

them

We

have

no

illusions

this

article

is

any

different

although

and

because

we

sort

of

tried

to

adhere

to

the

latest

fashions

ffl

Often

the

convention

is

used

that

the

variables

x

y

and

z

refer

to

open

process

terms

and

the

variables

s

t

and

u

to

closed

process

terms

Although

some

parts

of

this

article

conform

to

this

convention

we

shamefully

admit

that

in

large

parts

it

is

blatantly

violated

B

Overview

B.

Axioms

The

list

below

shows

all

axioms

mentioned

in

this

paper

in

alphabetical

order

Behind

the

name

of

the

axioms

the

number

is

listed

of

the

page

where

the

axiom

is

introduced

x

y

y

x

-LRB-

page

-RRB-

x

y

z

x

y

z

-LRB-

page

-RRB-

x

x

x

-LRB-

page

-RRB-

x

y

z

x

z

y

z

-LRB-

page

-RRB-

x

y

z

x

y

z

-LRB-

page

-RRB-

x

ffi

x

-LRB-

page

-RRB-

a

ffi

a

-LRB-

page

-RRB-

x

ffi

x

-LRB-

page

-RRB-

ffi

x

ffi

-LRB-

page

-RRB-

ffi

x

ffi

-LRB-

page

-RRB-

a

bac

ATS

-LRB-

page

-RRB-

rel?a

a

-LRB-

page

-RRB-

rel?x

y

rel?x

rel?y

-LRB-

page

-RRB-

rel?x

y

rel?x

y

-LRB-

page

-RRB-

rel?oerel?x

ffi

-LRB-

page

-RRB-

rel

ffi

ffi

DCSID

-LRB-

page

-RRB-

oerel?x

oerel?y

oerel?x

y

-LRB-

page

-RRB-

oerel?x

y

oerel?x

y

-LRB-

page

-RRB-

ffi

x

ffi

-LRB-

page

-RRB-

a

ffi

a

-LRB-

page

-RRB-

x

ffi

x

-LRB-

page

-RRB-

oerel?x

ffi

oerel?x

-LRB-

page

-RRB-

a

j

b

c

if

fl?a

b

c

ffi

-LRB-

page

-RRB-

a

j

b

ffi

if

fl?a

b?ffi

-LRB-

page

-RRB-

x

k

y

x

k

y

y

k

x

x

j

y

-LRB-

page

-RRB-

a

j

b

x

a

j

b

x

-LRB-

page

-RRB-

a

x

j

b

a

j

b

x

-LRB-

page

-RRB-

a

x

j

b

y

a

j

b

x

k

y

-LRB-

page

-RRB-

oerel?x

j

oerel?y

oerel?x

j

y

-LRB-

page

-RRB-

oerel?x

j

rel?y

ffi

-LRB-

page

-RRB-

oerel?x

j

rel?y

ffi

ffi

-LRB-

page

-RRB-

rel?x

j

oerel?y

ffi

-LRB-

page

-RRB-

rel?x

ffi

j

oerel?y

ffi

-LRB-

page

-RRB-

a

j

oerel?x

ffi

-LRB-

page

-RRB-

oerel?x

j

a

ffi

-LRB-

page

-RRB-

a

x

j

oerel?y

ffi

-LRB-

page

-RRB-

oerel?x

j

a

y

ffi

-LRB-

page

-RRB-

x

y

j

z

x

j

z

y

j

z

-LRB-

page

-RRB-

x

j

y

z

x

j

y

x

j

z

-LRB-

page

-RRB-

x

k

y

x

k

y

y

k

x

-LRB-

page

-RRB-

a

k

x

a

x

-LRB-

page

-RRB-

a

k

x

ffi

a

x

ffi

-LRB-

page

-RRB-

a

x

k

y

a

x

k

y

-LRB-

page

-RRB-

a

x

k

y

ffi

a

x

k

y

ffi

-LRB-

page

-RRB-

x

y

k

z

x

k

z

y

k

z

-LRB-

page

-RRB-

oerel?x

k

rel?y

ffi

-LRB-

page

-RRB-

oerel?x

k

rel?y

ffi

ffi

-LRB-

page

-RRB-

oerel?x

k

rel?y

oerel?z

oerel?x

k

z

-LRB-

page

-RRB-

oerel?x

k

a

ffi

-LRB-

page

-RRB-

oerel?x

k

a

y

ffi

-LRB-

page

-RRB-

oerel?x

k

a

y

oerel?x

k

y

-LRB-

page

-RRB-

oerel?x

k

a

y

z

oerel?x

k

z

-LRB-

page

-RRB-

oerel?x

k

oerel?y

oerel?x

k

y

-LRB-

page

-RRB-

x

k

ffi

ffi

-LRB-

page

-RRB-

ffi

k

x

ffi

-LRB-

page

-RRB-

x

j

ffi

ffi

-LRB-

page

-RRB-

ffi

j

x

ffi

-LRB-

page

-RRB-

oerel

ffi

ffi

DRTSID

-LRB-

page

-RRB-

bxc

rel?x

oerel?bxc

USD

-LRB-

page

-RRB-

bac

a

-LRB-

page

-RRB-

bx

yc

bxc

y

-LRB-

page

-RRB-

bx

yc

bxc

byc

-LRB-

page

-RRB-

boerel?x?c

ffi

-LRB-

page

-RRB-

b?ffic

ffi

-LRB-

page

-RRB-

a

k

bxc

a

bxc

-LRB-

page

-RRB-

a

x

k

byc

a

x

k

byc

-LRB-

page

-RRB-

bx

j

yc

bxc!j

byc

-LRB-

page

-RRB-

a

j

b

c

if

fl?a

b

c

ffi

-LRB-

page

-RRB-

a

j

b

ffi

if

fl?a

b

ffi

-LRB-

page

-RRB-

a

j

b

x

a

j

b

x

-LRB-

page

-RRB-

a

x

j

b

a

j

b

x

-LRB-

page

-RRB-

a

x

j

b

y

a

j

b

x

k

y

-LRB-

page

-RRB-

B.

Theories

The

list

below

gives

for

every

theory

we

defined

the

axioms

it

contains

ffl

BPA

ffl

BPAffi

ffl

BPA?drt?ffi

ffl

BPA?drt?ID

ffl

BPAdrt?ID

ATS

USD

ffl

BPA?drt

DRTSID

DCSID

ffl

BPAdrt

DRTSID

DCSID

ATS

USD

ffl

PA?drt?ID

ffl

ffl

ACP?drt?ID

ffl

ffl

PAdrt

DRTSID

DCSID

ATS

USD

ffl

ACPdrt

DRTSID

DCSID

ATS

USD

In

Table

on

the

next

page

we

give

an

overview

of

the

axioms

of

the

-LRB-

basic

-RRB-

process

algebras

treated

in

Section

for

the

purpose

of

comparing

the

respective

theories

with

each

other

We

have

the

following

legend

ffl

A

ffl

indicates

that

the

axiom

is

present

in

the

theory

ffl

A

indicates

that

the

axiom

is

not

present

but

can

be

derived

-LRB-

for

closed

terms

-RRB-

from

the

other

axioms

in

the

theory

ffl

A

indicates

that

the

axiom

does

not

hold

in

the

theory

ffl

A

indicates

that

the

axiom

is

meaningless

in

the

theory

as

there

is

a

signature

conflict

In

Table

on

page

we

give

an

overview

of

the

axioms

pertaining

to

merge

operators

of

the

-LRB-

concurrent

-RRB-

process

algebras

treated

in

Section

The

legend

is

the

same

as

for

the

previous

table

BPA

BPAffi

BPA?drt?ffi

BPA?drt?ID

BPA?drt?ID

BPA?drt

BPA?drt

Main

Axioms

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ATS

USD

ffl

ffl

ID

Axioms

ffl

ffl

DCSID

DRTSID

ffl

ffl

Auxiliary

Axioms

ffl

ffl

USD-Style

Axioms

Table

Overview

of

axioms

of

basic

process

algebras

PA?drt?ID

ACP?drt?ID

PAdrt

ACPdrt

Main

Axioms

for

k

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

Main

Axioms

for

j

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ID

Axioms

ffl

ffl

ffl

ffl

ffl

ffl

ffl

ffl

USD-Style

Axioms

Table

Overview

of

axioms

of

concurrent

process

algebras

B.

Theorems

In

Table

we

give

an

overview

of

the

definitions

of

the

axioms

and

semantics

of

the

process

algebras

listed

in

this

paper

with

corresponding

theorems

regarding

elimination

soundness

and

completeness

Axioms

Semantics

Elimination

Soundness

Completeness

Basic

Process

Algebras

BPA

Def

Def

Prop.

Thm

Thm

BPAffi

Def

Def

Prop.

Thm

Thm

BPA?drt?ffi

Def

Def

Thm

Thm

Thm

BPA?drt?ID

Def

Def

Thm

Thm

Thm

BPA?drt?ID

Def

Def

Thm

Thm

Thm

BPA?drt

Def

Def

Thm

Thm

Thm

BPA?drt

Def

Def

Thm

Thm

Thm

Concurrent

Process

Algebras

PA?drt?ID

Def

Def

Cor

Thm

Thm

Def

Def

Cor

Cor

Cor

ACP?drt?ID

Def

Def

Cor

Thm

Thm

Def

Def

Cor

Cor

Cor

PA?drt

Def

Def

Cor

Thm

Thm

ACP?drt

Def

Def

Cor

Thm

Thm

Table

Overview

of

definitions

and

theorems

Please

take

note

of

the

following

ffl

PA?drt?ID

and

are

two

slightly

different

axiomatizations

of

the

same

theory

with

the

same

semantics

taken

from

-LSB-

-RSB-

and

-LSB-

-RSB-

respectively

We

prove

both

axiomatizations

sound

and

complete

ffl

The

same

for

ACP?drt?ID

and

List

of

Tables

Axioms

of

BPA

Deduction

rule

for

a.

Deduction

rules

for

and

Axioms

for

ffi

Alternative

for

Axiom

Axioms

for

oerel

Deduction

rules

for

a

and

oerel

Term

Rewriting

System

for

BPA?drt?ffi

Axioms

for

ffi

Alternative

for

Axioms

Axioms

for

rel

Deduction

rules

for

rel

Term

Rewriting

System

for

BPA?drt?ID

Axioms

for

delayable

actions

Recursive

Specification

Principle

for

the

Unbounded

Start

Delay

Deduction

rules

for

delayable

actions

Term

rewriting

system

for

BPAdrt?ID

Additional

axioms

for

the

ultimate

start

delay

Axioms

for

ffi

Deduction

rules

for

ID

Term

rewriting

system

for

BPA?drt

Term

rewriting

system

for

BPAdrt

Axiom

for

b?ffic

Axioms

for

the

-LRB-

free

-RRB-

merge

Additional

axioms

for

PA?drt?ID

Deduction

rules

for

the

-LRB-

free

-RRB-

merge

Additional

axioms

for

Additional

rewriting

rules

for

Axioms

for

the

-LRB-

communicating

-RRB-

merge

Additional

axioms

for

ACP?drt?ID

Deduction

rules

for

the

-LRB-

communicating

-RRB-

merge

Additional

axioms

for

Term

rewriting

system

for

Additional

axioms

for

PAdrt

Deduction

rules

for

the

-LRB-

free

-RRB-

merge

with

immediate

deadlock

Additional

axioms

for

the

ultimate

start

delay

and

merge

Additional

axioms

for

ACPdrt

Additional

deduction

rules

for

ACPdrt

Axioms

for

the

communication

merge

and

delayable

actions

Axiom

for

bx

j

yc

Signatures

of

some

process

algebras

Overview

of

axioms

of

basic

process

algebras

Overview

of

axioms

of

concurrent

process

algebras

Overview

of

definitions

and

theorems

Bibliography

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-RSB-

E.

H.

L.

AARTS

et

al.

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Simplex

Sigillum

Veri

Eindhoven

University

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Liber

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prof.

dr.

F.

E.

J.

Kruseman

Aretz

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-RSB-

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ABRAMSKY

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GABBAY

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MAIBAUM

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Logic

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Computational

Structures

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University

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GABBAY

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BAETEN

AND

J.

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BERGSTRA

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algebra

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Discrete

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In

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C.

M.

BAETEN

AND

J.

A.

BERGSTRA

Some

simple

calculations

in

discrete

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algebra

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-RSB-

J.

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Discrete

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-LSB-

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AND

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RENIERS

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University

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Process

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CSR

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University

of

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Concrete

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University

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-LSB-

-RSB-

T.

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Branching

bisimilarity

is

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University

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as

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-LSB-

-RSB-

S.

H.

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BOS

AND

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in

discrete-time

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algebra

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Programming

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appear

-LSB-

-RSB-

B.

JOHNSSON

AND

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PARROW

editors

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International

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on

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August

-LSB-

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-RSB-

pages

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Concepts

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editor

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International

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on

Fundamentals

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Computation

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Notes

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-RSB-

J.

J.

VEREIJKEN

Fischer

's

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algebra

Technical

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CSR

Eindhoven

University

of

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Computing

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for

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In

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C.

VERHOEF

A

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conservative

extension

theorem

in

process

algebra

In

-LSB-

-RSB-

pages

Typeset

by

LATEX

in

Lucida

Bright

acp
acpdrt
acpdrtid
bpa
bpadrt
bpadrtffi
bpadrtid
bpaffi
completeness
discrete time
elimination
padrt
padrtid
process algebra
relative time
rsp
soundness
usd