"... In PAGE 15: ... We shall write E1 vCP E2 (E1 =CP E2) to indicate that E1 v E2 (E1 = E2) can be derived within CP. Table6 contains the standard inequations for testing from [20, 28]. The laws in Table 7 state that process is less de ned than every PAL process (UND1) and assert the strictness of all binary operators (strictness of follows from IC4).... ..."

"... In PAGE 15: ... We shall write E1 vCP E2 (E1 =CP E2) to indicate that E1 v E2 (E1 = E2) can be derived within CP. Table6 contains the standard inequations for testing from [21, 30]. The laws in Table 7 state that process is less de ned than every PAL process (UND1) and assert the strictness of all binary operators (strictness of follows from IC4).... ..."

"... In PAGE 7: ...Table1 . Note that the symmetric versions of rules (Sum) and (Par1) have been omitted.... In PAGE 7: ... Thus, by rule (Act2) we have (P; fr1; r2g) f(r1;2);(r2;3)g ?! (Q; ;), whereas (P; fr1; r2g), (P; fr1; r2g), and (P; fr1; r2g) have no transitions. Except for the appearance of worlds in con- gurations, the rules of Table1 are essentially identical to the ones for ACSR. It is worth pointing out that all processes in a parallel composition need to... ..."

"... In PAGE 5: ... The results obtained for the equivalences are extended to the associated preorders as well. Acknowledgment My thanks to Tony Hoare for suggesting that the axioms of Table2 could be... In PAGE 61: ...61 17.2 Axiomatizing the equivalences In Table2 , complete axiomatizations can be found for twelve of the fteen semantic equivalences of this paper that di er on BCCSP. Axioms for singleton-failures, 2-nested simulation and possible- futures semantics are more cumbersome, and the corresponding testing notions are less plausible.... In PAGE 61: ... I is a unary operator that calculates the set of initial actions of a process expression, coded as a process expression again. Theorem 8 For each of the semantics O 2 fT; S; CT; CS; F; R; F T; RT; P W; RS; Bg two process expressions p; q 2 T(BCCSP) are O-equivalent i they can be proved equal from the axioms marked with \+ quot; in the column for O in Table2 . The axioms marked with \v quot; or \! quot; are valid in O-semantics but not needed for the proof.... In PAGE 62: ... So assume p vO q and (3) has been proven for all pairs of smaller expres- sions p0; q0 2 T(BCCSP). Provided TO contains at least the rst four axioms of Table2 , one has TO ` q = q + p i TO ` q = q + ap0 for every summand ap0 of p. Take O = B, so p vB q.... In PAGE 63: ... This law falls outside conditional equational logic, but it can be reformulated equationally by considering the two cases I(x) = 0 = I(y) and I(x) 6 = 0 6 = I(y). In the rst case it must be that TB ` x = 0 = y and hence the law follows from the third and fourth axiom of Table2 . In the second, observe that I(p) 6 = 0 i p has the form bq +r with b 2 Act.... In PAGE 65: ...2 gives TB ` == nnhnn == = == nnU(h)nn == for h 2 IH. 2 In Theorem 8 the fth and seventh axioms of Table2 may be replayed by a n X i=1 (bixi + biyi) = a n X i=1 (bixi + biyi) + a n X i=1 biyi and a n X i=1 bixi + a n X i=1 biyi = a n X i=1 (bixi + biyi): These laws derive the same closed substitution instances. Thus none of the axiomatizations require the operator I, or conditional equations.... In PAGE 66: ...The linear time { branching time spectrum I from the axioms marked with `+ apos; or `! apos; in the column for O in Table2 . It follows that the axioms marked with `v apos; are derivable.... In PAGE 66: ... 17.3 Axiomatizing the preorders In Table 3, complete axiomatizations can be found for the eleven preorders corresponding to the equivalences axiomatized in Table2 (there is no preorder for tree semantics (U)). This time prov- ability is de ned according to the standards of either rst-order logic with inequality or conditional inequational logic, i.... In PAGE 66: ... In the latter case, the axioms of Table 3 also constitute complete axiomatizations of the equivalences. The three axioms in Table 3 in which the inequality is written \v quot; represent strengthenings of the corresponding axioms in Table2 . The axioms in which the inequality is written \w quot; are merely slick reformulations of the corresponding axioms in Table 2, and could be replaced by them.... In PAGE 66: ... The three axioms in Table 3 in which the inequality is written \v quot; represent strengthenings of the corresponding axioms in Table 2. The axioms in which the inequality is written \w quot; are merely slick reformulations of the corresponding axioms in Table2 , and could be replaced by them. Unlike in Table 2, the characteristic axiom for the readiness preorder (the ninth) is now a substitution instance of the characteristic axiom for the failures preorder (the tenth).... In PAGE 66: ... The axioms in which the inequality is written \w quot; are merely slick reformulations of the corresponding axioms in Table 2, and could be replaced by them. Unlike in Table2 , the characteristic axiom for the readiness preorder (the ninth) is now a substitution instance of the characteristic axiom for the failures preorder (the tenth). Note that the characteristic axiom for the ready simulation preorder (the fth) derives all closed instances of I(x) = I(y) ) ax v a(x + y), which gives the fth axiom of Table 2.... In PAGE 66: ... Unlike in Table 2, the characteristic axiom for the readiness preorder (the ninth) is now a substitution instance of the characteristic axiom for the failures preorder (the tenth). Note that the characteristic axiom for the ready simulation preorder (the fth) derives all closed instances of I(x) = I(y) ) ax v a(x + y), which gives the fth axiom of Table2 . Hence all closed instances of the characteristic axiom for the ready trace preorder (the seventh) are derivable from the fth and eighth axioms.... In PAGE 68: ... Thus requirements 1 and 2(a) are ful lled. As (the closed instances of) the axioms for the respective equivalences from Table2 are easily derivable from the ones for the corresponding preorders from Table 3, requirement 2(b) is ful lled as well. Requirement 3, which used to follow from Theorem 6 and Propositions 16.... In PAGE 71: ...1 and the soundness of the axioms for BCCSP. \First ) quot; (completeness of the axioms for the equivalences): Let T 0 O be the set of axioms marked with \+ quot; in the column for O in Table2 , but using a Pn i=1 bixi + a Pn i=1 biyi = a Pn i=1(bixi + biyi) and a Pn i=1(bixi + biyi) = a Pn i=1(bixi + biyi) + a Pn i=1 biyi instead of the axioms involving the operator I. As Theorem 8 establishes completeness for closed terms only, it holds for T 0 O as well.... In PAGE 71: ... As Theorem 8 establishes completeness for closed terms only, it holds for T 0 O as well. Claim: If T 0 O ` p = Pm j=1 aqj for p; qj 2 T(BCCSP), then, modulo applications of the rst three axioms of Table2 , p has the form p = Pn i=1 api. Proof of the claim: As all axioms in T 0 O are equations, I may use induction on the proof of p = Pm j=1 aqj in equational logic.... In PAGE 78: ... Furthermore, the predicate p T(BCCSP quot;) is generated by the rules of Table 6. Now the complete axiomatizations of Table2 apply to BCCSP quot; p( quot;) p(p) p(p + q) p(q) p(p + q) Table 6: Rules for the termination predicate as well, provided that the occurrences of 0 are changed into , an axiom I( quot;) = quot; is added, and the characteristic axioms for CS and CT also get variants in which by + z resp. cy + v is replaced by quot;.... ..."

"... In PAGE 3: ... An LTS graph is shown in Figure 1. Given an LTS S = lt; S; ; ; s0 gt;, let p; q 2 S and 2 [ f g, the conventional nota- tions shown in Table1 are relevant to a given LTS, as introduced in [3]. In this paper we use M; P; S; : : : to represent LTSs; M; P; Q; : : :, for sets of states; a; b; c; : : :, for actions; and i; p; q; s : : :, for states.... ..."

"... In PAGE 84: ... Correspondingly, the expression betray a man to every enemy of him (45) would by our grammar be assigned the denotation dom((B; A) \ (D3 I)) : M: (46) This term is equivalent to the set fxj(9y)(y 2 M ^ (8z)(zAy ! Bxyz))g; (47) which is in line with our intuition about the meaning of (45). In Table1 a grammar was given for a fragment of English that is large enough to derive many of Peirce apos;s English examples to illustrate his operations and their use to construct complex terms. In the next section we use our fragment to check Peirce apos;s constructions.... ..."

"... In PAGE 2: ... But we provide methods to test for epsilon1-transitions and to convert an epsilon1-NDFA to a NDFA. See Table4 , Appendix A, for more details. 2.... ..."

"... In PAGE 9: ... But it propagates events like the Complementarity and the Redundancy components. So as the other components, it has problems with the content of input events and the modification of parameters when it is running ( Table2... In PAGE 10: ... The processing of the component requires concurrent access to the data structure. This brings execution errors (Java errors in Table2 (a)) and an abnormal behavior of the component in lazy strategy. In some cases, calculations are made using accurate information on the state of the data structure, but the result of the calculations is wrong because the state has changed since the event arrival.... In PAGE 10: ... The eager and lazy strategies are based on the concept of temporal window. Even if boundary tests with only one temporal window ( Table2 (c)) do not bring error, many test failures come from temporal windows (Tables 2(a)). In some cases, an event is not processed at the end of its temporal window.... ..."

"... In PAGE 9: ... (13), it makes sense for a Markov net (N, P) and for a reachable marking m, to use the notations: Pm, Em, (14) to respectively denote the probability Pv and the expectation Ev for any finite configuration v with positive likelihood, and such that m = m(v), provided that such a v exists. Table1 summarizes and compares the definitions introduced so far, and em- phasizes the symmetry between sequential and concurrent systems. Table 1.... ..."