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S = 1 t * 0 and thus since b B c A , we have in a , x a x t of s = -it iff s -> (x6y) = 1 , xxy iff t -» (x&y) = 1 . a in B , and x&-*x = 0 . s by and It is nontrivial x -» 0 = -«x is not related to t are B , s -* (s6t) = s - > 0 = t * l hence that are nontrivial congruences requires some work. A ; therefore in identities of B xay It suffices to do the proofs only for since B SAt = 0 . Now define two congruences To show , which is contradic- -it = -»-*s = s * 1 ; similarly, t * 1 By additional Boolean identities in svt = 1 , and s * 1 , 8 b6c Also Now by ; or .

Which are meet-irreducible in the lattice of all subgroups of £ . For any given h the number of nonisomorphic is finite, but unknown in general. irreducibles It should be pointed out that normality and meet-irreducibility are distinct notions; hence, each equational subclass does not necessarily have a unique irreducible as a generator. This is now summarized and extended to h-adic relations in general, where qualitatively the results are similar, but the specific irreducibles A are more involved.

Where x(c) = > (c G C) , x(d) = <«>,d> (d G D) , x (•) = <<*>, °°> . Clearly, C and D are nontrivial, and B is subdirectly reducible. We may now safely assume that implies that B B\» is transitive. has one generator (remember that constants and comes for free). on one generator, Since A «> This is all the is the free algebra B must be a homomorphic image of it. To determine the irreducible factor algebras, we need only consider congruences of of A^ . Now it is easy to see that any congruence A/a, is determined by its restriction to J O ZL h .

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