By G. Buskes, A. van Rooij (auth.), Jorge Martínez (eds.)

From the twenty eighth of February during the third of March, 2001, the dep. of Math ematics of the collage of Florida hosted a convention at the many points of the sphere of Ordered Algebraic buildings. formally, the name was once "Conference on Lattice Ordered teams and I-Rings", yet its subject material advanced past the constraints one may well go along with any such label. This quantity is formally the lawsuits of that convention, even though, likewise, it really is extra actual to view it as a supplement to that occasion. The convention was once the fourth in wh at has changed into aseries of comparable meetings, on Ordered Algebraic buildings, held in consecutive years. the 1st, held on the collage of Florida in Spring, 1998, was once a modest and casual affair. The 5th is within the ultimate making plans phases at this writing, for March 7-9, 2002, at Vanderbilt college. And even though those occasions stay modest and fairly casual, their scope has broadened, as they've got succeeded in attracting mathematicians from different, similar fields, as weIl as from extra far away lands.

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From the twenty eighth of February during the third of March, 2001, the dept of Math ematics of the collage of Florida hosted a convention at the many facets of the sector of Ordered Algebraic constructions. formally, the name was once "Conference on Lattice Ordered teams and I-Rings", yet its subject material developed past the constraints one could go along with the sort of label.

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**Additional info for Ordered Algebraic Structures: Proceedings of the Gainesville Conference Sponsored by the University of Florida 28th February — 3rd March, 2001**

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6. [BT] The convex normal subalgebra generated by a subset 5 in a residuated lattice L is cn(5) = {a E L: x:S a:S x\e for some x E rrfß(5)}. Proof. Let (5) denote the right hand side above. Since this is clearly a subset of cn(5), we need to show that 5 ~ (5) and that (5) is a convex normal subalgebra. Suppose s E 5, and consider x = s 1\ eis 1\ e E rrf ß(5). Then x :S sand s:s (els)\e:S x\e, hence sE (5). This shows 5 ~ (5). Now let a,b E (5). Then there are x,y E rrfß(5) such that x :S a :S x\e and y :S b :S y\e.

For any al, ... , an E L, t _L (al, ... ,an)=t L- (all\e, ... ,anl\e). Proof. By definition this is true for variables and the constant term e. Assurne the statement holds for terms sand t. Since (s/t)- = s- /t- 1\ e, we have (S/t)-L(al,"" an) (s-L(al,"" L- an)/LrL(al, .. , an)) 1\ e L L- (s (a 1 1\ e, ... , an 1\ e) / t (s/t) L- (a 1 1\ e, ... , an 1\ e)) 1\ e (all\e, ... , V, 1\ complete the proof. 10. For any L E Re, L - F s = t if and only ifL F s- = r. 38 P. JIPSEN AND C. TSINAKIS Proof. Suppose L - F s = t, and let al, ...

Therefore h(s\t) <; [s\t]. Suppose h(t) <; [q,q',r]. Then t E h(t) implies qtq':::; r is Gentzen provable. For any s' E h(s) <; [s] we have that s' :::; s is Gentzen provable, so from (\ left) we get A 49 SURVEY OF RESIDUATED LATTICES that S'(S\t) E [q,q',r]. By (*) it follows that S'(S\t) E h(t) whenever s' E h(s), hence h(s){s\t} S; h(t). This implies S\t E h(s)\h(t) = h(s\t). The case for s/t E h(s/t) S; [s/t] is similar. Since we are assuming that h has been extended to a homomorphism from TM to L, we have h(e) = eL = {e}G.