By Gonzalo Rodriguez-Pereyra

Gonzalo Rodriguez-Pereyra bargains a clean philosophical account of homes. How is it that various things (such as crimson roses) can proportion an identical estate (redness)? based on resemblance nominalism, issues have their houses in advantage of comparable to different issues. This retro view is championed with readability and rigor.

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Extra resources for Resemblance nominalism: a solution to the problem of universals

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The truth that the conclusion is meant to express is: Sx Sy (x has red hair ∧ y is an individual concept of x ∧ you know that y was born in Megara). But if this is the conclusion of the argument, it would still seem to be invalid. 11 But the point of this book is not to discuss other views in detail: it is simply to paint one particular view. So, is there a semantics that preserves inferences such as Particular Generalization? As we will see, there is. 9 Similar problems beset a paratactic account of knowing that (of the kind developed by Boër and Lycan (1986)) and any other account which makes it impossible for a variable to bind inside and outside an epistemic context simultaneously in the natural way.

The case for S is similar. The other case is when w ∈ O. In this case, there is no recursion at all, and the argument for each formula is the same as that for → and modal operators at impossible worlds. 30 · Semantics for Intentionality Lemma 2 is exactly what one needs to show that quantifiers behave properly. In particular: Corollary 3 If t is free when substituted for x in A(x) then: 1. AxA(x) |= A(t) 2. A(t) |= SxA(x) + Proof For 1, suppose that @ + s AxA(x). Then for all d ∈ D, @ s(x/d) + A(x).

But: @ + t Pa which is not true. ⇔ w + Pa ⇔ δ(a) ∈ δ + (Px1 , w) 24 · Semantics for Intentionality The semantics also solve the problems concerning the Barcan formulas. For example, t AxPx Ax t Px. To see this, consider an interpretation where D = {a} C = {@} O = {w} @R δs (t) w (and nothing else) δ + (AxPx, w) δ + (Px, w) ={ } =φ (where Px is a matrix). @ + s t AxPx ⇔ w + s AxPx ⇔ ∈ δ + (AxPx, w) which is true. But: @ + s Ax t Px ⇔ for all d ∈ D, @ + s(x/d) t Px ⇔ for all d ∈ D, w + s(x/d) Px ⇔ for all d ∈ D, d ∈ δ + (Px, w) which is not true.

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