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Sm ∈ / J}. −1 (V(I)), where s is the Stone Then b = ni=1 bti · m j=1 −bsj = 0. So, u ∈ s(b) ⊆ θ representation mapping see [9, p. 99]. p (T ). Thus, θ is a homeomorphism. Case 2. )-poset. 1. There is u0 ∈ Ult(B(T )) so that θu0 = ∅. Indeed, set V0 = {−bs : s ∈ T }. For m m all m = 0, for all s1 , . . )-poset. Thus, V0 has ﬁnite intersection property. So, there is u0 in Ult(B(T )) so that bs ∈ / u0 for all s ∈ T . So, θu0 = ∅. 2. p (T ). ). 3. ). 4. p (T ) ⊆ θ(Ult(B(T ))). Upper Semi-lattice Algebras and Combinatorics 33 Now, by 1.

Set µ+ (b) := |supp+ (b)| and µ+ (0) = 0. Note that the results of this section, appear in [16], extending the work done in [15]. We shall reproduce, here, the main steps of the proof. Details are left to the reader. 8. Let e, e be disjoint elements of E. Then: i) e⊥e implies µ+ (e + e ) = 2. ii) e e implies µ+ (e + e ) = 1. 50 M. Bekkali and D. Zhani Proof. Set b = e + e . Case 1. e⊥e . By [1] this form of b is unique and then µ+ (b) = 2. e . Say that Case 2. e n e = bt · − m bsi , e = bt · − bτj .

Put A = supp(x) = {t1 , . . , tn } and B = supp(y) = {s1 , . . , sm }. We have: x y = bt1 · · · btn bs1 · · · bsm . Note that bti bsj = 0 if ti = sj , thus x y = {bu : u ∈ (A∪B)\(A∩B)}. Hence supp(x y) = A B = supp(x) supp(y). 3. i) For any x, y, z pairwise distinct we have: µ(x y z) = µ(x y) + µ(y z) + µ(z x) − µ(x) + µ(y) + µ(z) + 4S(x, y, z). ii) For any free set {x, y, z} in B(T ) we have: µ(x y z) = µ(x y) + µ(y z) + µ(z x) − µ(x) − µ(y) − µ(z). 4. If n ≥ 3 and {x1 , . . , xn } ⊆ B(T ) then: a) S(x1 , .