Heterogeneity in φ-Correspondence Sample Clauses

Heterogeneity in φ-Correspondence. I now move on to φ-Correspondence and show that the relation is also heterogeneous. The argument is straightforward. φ-Correspondence occurs between head feature nodes and non-head feature nodes. An element is either a head or a dependent: it cannot be both, and it cannot be neither (see footnote 9). For this reason, for any φ-Correspondence relation ℛφ, the intersection of the domain X of ℛφ, and of the codomain Y of ℛφ is always empty. X = {x | x ∈ φ ∧ x ∈ O ∧ x ∈ H} Y = {x | x ∈ φ ∧ x ∈ O ∧ x ∈ }, where φ ∈ Φ. In fact, φ-Correspondence is very similar to B/R-Correspondence. Both relations act on elements in the output. In the case of B/R-Correspondence, they are output root nodes, in the case of φ-Correspondence they are output feature nodes. In both cases, there is an optional, cumulative, phonetically silent property that creates an asymmetry and allows 12 Base and Reduplicant are not properties of our model. However, they can be easily formalized as cumulative properties, much alike H and . us to distinguish the two sets. In B/R-Correspondence it is the property of being in the base (or in the reduplicant); in φ-Correspondence it is property of being a head. Heterogeneity is one of the axioms that allow us to establish that φ-Correspondence and I/O-Correspondence are the same kind of relations. In the next two sections, I discuss two further restrictions that heterogeneity imposes on the theory: a reduction on the space of candidates and the constraints correspondence can have on the type of axioms.

Heterogeneity in φ-Correspondence. I now move on to φ-Correspondence and show that the relation is also heterogeneous. The argument is straightforward. φ-Correspondence occurs between head feature nodes and non-head feature nodes. An element is either a head or a dependent: it cannot be both, and it cannot be neither (see footnote 9). For this reason, for any φ-Correspondence relation ℛφ, the intersection of the domain X of ℛφ, and of the codomain Y of ℛφ is always empty. X = {x | x ∈ φ ∧ x ∈ O ∧ x ∈ H} Y = {x | x ∈ φ ∧ x ∈ O ∧ x ∈ 𝐻}, where φ ∈ Φ.