Countermodel to "Theorem" 4

To develop the countermodel, we shall identify objects b and d, and a property P, such that:

IsTheFormOf(b,P) & ParticipatesPH(d,b) & ¬dP

Consider the necessarily empty property being-Q-and-not-Q (for some arbitrarily chosen property Q). Call this property P. I.e., P = [λz Qz & ¬Qz]. Consider a distinct necessarily empty property, say, the property of being-round-and-square, and call this property T. I.e., T = [λz Rz & Sz]. Note that in object theory one can consistently assert that P ≠ T even though □∀x(Px ≡ Tx). The reason is that identity for properties (‘F = G’) is defined as □∀x(xF ≡ xG). So properties may be distinct even though necessarily equivalent.

Now the following two facts are theorems of quantified modal logic:

[□∀y¬Fy & □∀y¬Gy] → □∀x(Fx ≡ Gx)

□∀x(Fx ≡ Gx) → ∀H(F⇒H ≡ G⇒H)

The first asserts that necessarily empty properties are necessarily equivalent. (Clearly, if neither F nor G are exemplified by any objects at any possible world, then F and G are exemplified by all and only the same objects at every possible world.) The second asserts that necessarily equivalent properties necessarily imply the same properties. Clearly, both of these theorems apply to P and T as we have defined them; so from the fact that P and T are necessarily empty, it follows that they necessarily imply the same properties.

Now choose b to be the abstract object that encodes all and only the properties implied by P. Then, it is straighforward to verify that IsTheFormOf(b,P). And choose c to be the abstract object that encodes all and only the properties implied by T. So we also know that IsTheFormOf(c,T). Now even though P ≠ T, it follows that b = c. To see this, note that from the facts:

  1. b encodes all and only the properties necessarily implied by P,
  2. c encodes all and only the properties necessarily implied by T, and
  3. P and T necessarily imply the same properties,

it follows that b and c encode the same properties. Thus b = c, by the definition of identity for abstract objects.

We can complete the countermodel by considering the abstract object d which encodes just one property, namely, T. We can now establish (i) ParticipatesPH(d,b), and (ii) ¬dP. To establish (i), we have to show:

(A) ∃F(IsTheFormOf(b,F) & dF)

So, if we can show IsTheFormOf(b,T) & dT, we can generalize to get (A). To show IsTheFormOf(b,T), we simply note that we've already established both IsTheFormOf(c,T) and b=c. So IsTheFormOf(b,T). And clearly, by definition of d, we know dT. So, by existential generalization, we have established (i) ParticipatesPH(d,b). To establish (ii), note that by definition, d encodes only a single property, namely, T. Since P ≠ T, it follows that ¬dP.

So we've identified objects b and d and property P such that:
IsTheFormOf(b,P) & ParticipatesPH(d,b) & ¬dP

Q.E.D.