By Stanley Burris

"As a graduate textbook, the paintings is a yes winner. With its transparent, leisurely exposition and beneficiant number of routines, the ebook attains its pedagogical ambitions stylishly. additionally, the paintings will serve good as a study tool…[offering] a wealthy collection of vital new effects that have been formerly scattered through the technical literature. commonly, the proofs within the ebook are tidier than the unique arguments." —

*Mathematical Reviews*of the yank Mathematical Society.

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This publication constitutes the refereed court cases of the Joint Workshop on approach Algebra and function Modeling and Probabilistic equipment in Verification, PAPM-PROBMIV 2001, held in Aachen, Germany in September 2001. The 12 revised complete papers offered including one invited paper have been rigorously reviewed and chosen from 23 submissions.

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Ik ∈ I, k < ∞}. 8. Let θ be a member of Eq(A). For a ∈ A, the equivalence class (or coset) of a modulo θ is the set a/θ = {b ∈ A : b, a ∈ θ}. The set {a/θ : a ∈ A} is denoted by A/θ. §4. 9. For θ ∈ Eq(A) and a, b ∈ A we have (a) A = a∈A a/θ. (b) a/θ = b/θ implies a/θ ∩ b/θ = ∅. Proof. (Exercise). 9. 10. A partition π of a set A is a family of nonempty pairwise disjoint subsets of A such that A = π. The sets in π are called the blocks of π. The set of all partitions of A is denoted by Π(A). For π in Π(A), let us deﬁne an equivalence relation θ(π) by θ(π) = { a, b ∈ A2 : {a, b} ⊆ B for some B in π}.

It is not diﬃcult to see that the right-hand side of the above equation is indeed an equivalence relation, and also that each of the relational products in parentheses is contained in θ1 ∨ θ2 . ✷ If {θi }i∈I is a subset of Eq(A) then it is also easy to see that i∈I θi is just i∈I θi . The following straightforward generalization of the previous theorem describes arbitrary sups in Eq(A). 7. If θi ∈ Eq(A) for i ∈ I, then θi = {θi0 ◦ θi1 ◦ · · · ◦ θik : i0 , . . , ik ∈ I, k < ∞}. 8. Let θ be a member of Eq(A).

If A is congruence-permutable, then A is congruence-modular. Proof. Let θ1 , θ2 , θ3 ∈ Con A with θ1 ⊆ θ2 . We want to show that θ2 ∩ (θ1 ∨ θ3 ) ⊆ θ1 ∨ (θ2 ∩ θ3 ), so suppose a, b is in θ2 ∩ (θ1 ∨ θ3 ). 9 there is an element c such that aθ1 c θ3 b holds as θ1 ∨ θ3 = θ1 ◦ θ3 . 45 §5. Congruences and Quotient Algebras By symmetry c, a ∈ θ1 ; hence c, a ∈ θ2 , and then by transitivity c, b ∈ θ2 . Thus c, b ∈ θ2 ∩ θ3 , so from aθ1 c(θ2 ∩ θ3 )b follows a, b ∈ θ1 ◦ (θ2 ∩ θ3 ); hence a, b ∈ θ1 ∨ (θ2 ∩ θ3 ).