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Leta ∗ := e ⊕ a ⊕ a 2 ⊕··· (32)Interesting, and easy to check, properties of this star operation areLet also a + := a ⊗ a ∗ and notice thatTheorem 9(i) a ∗ b is the least solution of (31).(a ∗ ) p = a ∗ , ∀p ∈ N and (a ∗ ) ∗ = a ∗ (33)(ii) For all solution x, one has that x = a ∗ x.Proofa ∗ = e ⊕ a + and a ∗ ≥ a + (34)(i) Clearly, if x is a solution, x ≥ b and x ≥ ax. On the one hand, a p x ≥a p b. On the other hand, x ≥ ax ≥ ... ≥ a p x. Hence, x ≥ a p b,thusx ≥ ⊕ ∞p=0 ap b = a ∗ b. Finally, it is straightforward to check that a ∗ b is alsoa solution.(ii) From (31), one gets that x = a(ax ⊕ b) ⊕ b = ...= a p x ⊕ (e ⊕···⊕a p−1 )bfor all p. Summing up all these equalities for p ∈ N yields x = a ∗ x ⊕ a ∗ b.But from (i) above, a ∗ x ≥ a ∗ b or a ∗ x ⊕ a ∗ b = a ∗ x which completes theproof.□Theorem 9 answers the issue of existence of solutions to (31) in complete dioids.About uniqueness, we observe that if the homogeneous equationx = ax (35)has a solution y ≠ ε, yz is also a solution of (35), ∀z ∈D, and obviously a ∗ b⊕yzis a solution of the nonhomogeneous equation (31). In usual linear algebra, it iswell known that all solutions of the nonhomogeneous equation are obtained byadding all solutions of the homogeneous equation to a particular solution of thenonhomogeneous. Here, for such a result to hold true, the “particular solution”has to be the least one (namely a ∗ b) because “adding” also means “increasing”.Theorem 10 Let D be a distributive dioid. Then, if x is a solution of (31), itcan be written as x = y ⊕ a ∗ b where y is a solution of (35).Proof For any given solution x of (31), we consider the subset C x := {c | x =c ⊕ a ∗ b}. Ifc ∈C x , then ac ∈C x too. As a matter of fact, x = c ⊕ a ∗ b ⇒ ax =ac ⊕ a + b ⇒ x = ax ⊕ b = ac ⊕ a ∗ b from the equality in (34). Also, if c, d ∈C x ,then c ⊕ d ∈ C x since x = x ⊕ x = c ⊕ d ⊕ a ∗ b. Therefore, if c ∈C x , then a ∗ c ∈C xtoo.14