Step 3 • τ$_{i}$ = { ct$_{i}$, t$_{i}$z : c, z ∈ V ( D$_{i}$ ) and cz ∈ T$_{i}$ } There exists z ∈ Y$_{i}$ such that ( a$_{i}$, x$_{i}$, z ) ∈ Trans ( D$_{i}$ ) since a$_{i}$ ∈ A$_{i}$ . Moreover a$_{i}$y$_{i}$ / ∈ A ( D$_{i}$ ), z ∈ x$_{i}$ , and x$_{i}$ / ∈ Y$_{i}$ so a$_{i}$z ∈ T$_{i}$ . Therefore T$_{i}$ is nonempty. Define θ$_{i}$ : V ( D$_{i}$$_{+1}$ ) → V ( D$_{i}$ ) so that θ$_{i}$ ( t$_{i}$ ) = x$_{i}$ and θ$_{i}$ ( u ) = u for all u ∈ V ( D$_{i}$ ). Before moving on we prove θ$_{i}$ is a compression. Routine calculations show part (1) of Definition 1.1 hold and there is a well-defined map θ ∗ i : A $^{(}$D ∗ i +1 ) → A ( D ∗ $_{i}$) given by θ ∗ i ( uv ) = θ$_{i}$ ( u ) θ$_{i}$ ( v ) for all u, v ∈ V ( D$_{i}$ ) such that uv ∈ A $^{(}$D ∗ i +1 $^{)}$. It is easy to see θ ∗ i ( σ$_{i}$ ) ∪ θ ∗ i ( τ$_{i}$ ) = A ( D ∗ $_{i}$) hence θ ∗ i is surjective. We have already shown ∣ $_{∣}$A $^{(}$D ∗ i +1 ) ∣ = | A ( D ∗ $_{i}$) | so θ ∗ i is a bijection. The only condition left is part 2 of Definition 1.1. Suppose d$_{1}$, d$_{2}$, d$_{3}$ ∈ V ( D$_{i}$ ) and ( d$_{1}$, d$_{2}$, d$_{3}$ ) ∈ Trans ( D$_{i}$ ). We must show ( d$_{1}$, d$_{2}$, d$_{3}$ ) is the image of a transitive triple in D$_{i}$$_{+1}$ . This is easy if d$_{1}$, d$_{2}$, d$_{3}$ are not distinct since every arrow in D$_{i}$ is the image of an arrow in D$_{i}$$_{+1}$ . Assume d$_{1}$, d$_{2}$, d$_{3}$ are distinct. We check every possible case and make repeated use of the fact uv ∈ σ$_{i}$ if and only if uv ∈ A ( D ∗ $_{i}$) for all u, v ∈ V ( D$_{i}$ ) \ { x$_{i}$ } . In cases 2, 3, and 4 we have x$_{i}$ = d$_{r}$ for some r ∈ { 1 , 2 , 3 } . We will show either ( d$_{1}$, d$_{2}$, d$_{3}$ ) ∈ Trans ( D$_{i}$ ) so that θ$_{i}$ ( d$_{j}$ ) = d$_{j}$ for j = 1 , 2 , 3 or the desired transitive triple is obtained by replacing d$_{r}$ with t$_{i}$ so that θ$_{i}$ ( d$_{j}$ ) = d$_{j}$ for j = r , and θ$_{i}$ ( t$_{i}$ ) = d$_{r}$ . We split cases between construction B and construction A when necessary. Note that x$_{i}$ is a soloist by part 1 of Lemma 3.6. Case 1 d$_{1}$, d$_{2}$, d$_{3}$ ∈ V ( D$_{i}$ ) \ { x$_{i}$ } 14