By Yde Venema

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A σ ≤ (sA ◦ uA , . . 15(ii), and find that tA = (sA )σ ◦ uA n 1 , . . , un 1 A σ (t ) , as desired. ✷ As a sample application, we show how Sahlqvist canonicity is an easy consequence of the previous theorem. 21 Sahlqvist equations are canonical over the class of all Boolean algebras with τ -operators. Proof. First we treat inequalities of the form ϕ(β1 , . . , βn ) ≤ ψ, where ϕ only uses ∧, ∨ and modalities, all βi are boxed atoms, and ψ is positive. But then it is immediate by the previous proposition that ϕ(β1 , .

12 As suggested by notation, the topology σ is closely connected to the kind of inclusion of B in Bσ . Let us just mention a couple of salient facts here. First, it is easy to see that the set {[a, b] | a, b ∈ B} is a basis for σ. This reveals that the set B is topologically dense in σ, in the sense that every σ-open set contains an element of B. But also, B constitutes the collection of isolated points of σ — recall that a point x is isolated in a topology if the singleton {x} is open. It is the latter two properties that make it possible to extend arbitrary maps between Boolean algebras to their extensions; we will come back to this at the end of this section.

29) Now suppose that we have developed a canonical way to extend an n-ary map f : An → A to an n-ary map f σ : (Aσ )n → Aσ ; it then immediately follows from (28) that A |= s ≈ t only if (sA )σ = (tA )σ . Aσ (30) Aσ Hence, in case s and t are stable on A, that is, if (sA )σ = s and (tA )σ = t , then we may infer from σ A |= s ≈ t that Aσ |= s ≈ t. This motivates a careful analysis of the relation between the functions sA (the term function of s in Aσ ) and (sA )σ (the extension to Aσ of the term function sA ).