since (M, w) |= [X1 = 1](X3 **= 0)**, we must have have that (M, w′) |= X3 = 0. This gives a contradiction. A similar contradiction arises if (M, w′) |= X2 = 1 or if (M, w′) |= X3 = 1. Since
(M, w′) |= X1 = 1 ∨ X2 = 1 ∨ X3 = 1 by construction, this gives a contradiction to the assumption that (**M, w**) |= ϕ.

Finally, I must show that there is a causal model T ∈ Tuniq and a context ~u such that (**T, ~u**) |= ϕ. Let V = {X1, X2, X3} and let U = {U}. Take R(X1) = R(X2) = R(X3) = {0, 1} and R(U) = {0}. In defining F, I write Fiinstead of FXi
, and omit the U argument (since it is always 0). Thus, F1(0, **0) = 0**
means that when X2 = X3 = 0, then X1 = 0. Define
- F1(0, 0) = 0; F1(0, 1) = 1; F1(1, 0) = 0; F1(1, **1) = 0**; - F2(0, 0) = 0; F2(0, 1) = 0; F2(1, 0) = 1; F2(1, **1) = 0**; - F3(0, 0) = 0; F3(0, 1) = 1; F3(1, 0) = 0; F3(1, **1) = 0**.

I must now verify that T = (S, F) ∈ Tuniq, and that (T, 0) |= ϕ. This is straightforward, although tedious. First observe that (0, 0, 0) is the only solution to all the equations in the basic causal model. It is easy to see that (0, 0, 0) is a solution. To see that it is the only solution, observe that (0, 0, 1) cannot be a solution because F3(0, **0) = 0**; similarly, (0, 1, 0) and (1, 0, 0) cannot be solutions; (0, 1, 1) cannot be a solution because F2(0, **1) = 0**; (1, 1, 0) cannot be a solution because F1(1, **0) = 0**; (1, 0, 1) cannot be a solution because F3(1, **0) = 0**; finally, (1, 1, 1) cannot be a solution because F1(1, **1) = 0**.

It is clear that there must be a unique solution if we set two of the three variables (forced by the equation for the third variable); for example, if X1 = 1 and X2 = 0, then the unique solution is (1, 0, 0). If X1, X2, or X3 is set to 0, then one solution is (0, 0, 0). The solution is unique for the same reasons that (0, 0, 0) was the unique solution to the original collection of equations. Finally, if X1 = 1, then (1, 1, 0) is a solution; if X2 = 1, then (0, 1, 1) is a solution; and if X3 = 1, then (1, 0, 1) is a solution. We must show that there are no other solutions in all three cases. If X1 = 1, (1,0,0) is not a solution since F2(1, **0) = 1**; (1, 0, 1) is not a solution because F3(1, **0) = 0**; and (1, 1, 1) is not a solution because F3(1, **1) = 0**. If X2 = 1, (0, 1, 0) is not a solution because F3(0, **1) = 1**; (1, 1, 0) is not a solution because F1(1, **0) = 0**; (1, 1, 1) is not a solution because F1(1, **1) = 0**. Finally, if X3 = 1,
(0, 0, 1) is not a solution because F1(0, **1) = 1**; (0, 1, 1) is not a solution because F2(0, **1) = 0**; and
(1, 1, 1) is not a solution because F1(1, **1) = 0**. Thus, T ∈ Tuniq.

It is now straightforward to check that (**T, u**) |= ϕ.

In Section 2.1, I argued that the choice of language was significant. All the results of this section are stated for the language Luniq. In light of Lemma 2.1, Theorems 3.4 and 3.5 apply without change to the language L
+. Moreover, since Tuniq ⊆ T and Luniq ⊆ L+, the formula ¬ϕ from Theorem 3.7 (which is in Luniq, and hence also in L
+) continues not to be valid in T , while being valid in counterfactual structures. While C5 is not valid in T , the generalization of C5 mentioned at the end of Section 2.1 is valid in T [Halpern 2000], and is not valid in counterfactual structures. So the full class T of causal models is incomparable in expressive power to counterfactual structures with respect the language L
+
(which is the language arguably most appropriate for T ). Formally, we have:
Corollary 3.8 T is incomparable in expressive power to M *with respect to the language* L
+.