Step 2 We fix t$_{i}$ / ∈ V ( D$_{i}$ ) and construct a reflexive directed graph D$_{i}$$_{+1}$ such that V ( D$_{i}$$_{+1}$ ) = V ( D$_{i}$ ) ∪ { t$_{i}$ } , and A $^{(}$D ∗ i +1 ) = σ$_{i}$ ∪ τ$_{i}$ where σ$_{i}$ and τ$_{i}$ are defined using construction A or construction B. In both constructions τ$_{i}$ = A $^{(}$D ∗ i +1 ) \ σ$_{i}$ and | τ$_{i}$ | = | A ( D ∗ $_{i}$) \ σ$_{i}$ | so | A ( D ∗ $_{i}$) | = ∣ $_{∣}$A $^{(}$D ∗ i +1 ) ∣ $_{∣}$. The arrows in τ$_{i}$ will all contain t$_{i}$ . If an arrow does not contain x$_{i}$ then it will be in σ$_{i}$ . Moreover σ$_{i}$ consists of the arrows belonging to both D$_{i}$$_{+1}$ and D$_{i}$ . Depending on the construction, an arrow may be contained in σ$_{i}$ even if it contains x$_{i}$ . Use construction B if there exist a$_{i}$, b$_{i}$ ∈ A$_{i}$ such that ( b$_{i}$, x$_{i}$, y$_{i}$ ) ∈ Trans ( D$_{i}$ ) and a$_{i}$y$_{i}$ / ∈ A ( D$_{i}$ ) for some y$_{i}$ ∈ Y$_{i}$ . Otherwise use construction A. Construction A. • B$_{i}$ = { b ∈ V ( D$_{i}$ ) \ { x$_{i}$ } : ( x$_{i}$, b, y ) ∈ Trans ( D$_{i}$ ) or ( x$_{i}$, y, b ) ∈ Trans ( D$_{i}$ ) for some y ∈ Y$_{i}$ } • σ$_{i}$ = A ( D ∗ $_{i}$) \ ( { ax$_{i}$ : a ∈ A$_{i}$ \ { x$_{i}$ }} ∪ { x$_{i}$b : b ∈ B$_{i}$ } ) • τ$_{i}$ = { at$_{i}$ : a ∈ A$_{i}$ } ∪ { t$_{i}$b : b ∈ B$_{i}$ } If y ∈ Y$_{i}$ then y ∈ B$_{i}$ since ( x$_{i}$, y, y ) ∈ Trans ( D$_{i}$ ). Therefore Y$_{i}$ ⊆ B$_{i}$ . Construction B. • T$_{i}$ = { cz : c ∈ V ( D$_{i}$ ) , z ∈ V ( D$_{i}$ ) \ { x$_{i}$ } , ( c, x$_{i}$, z ) ∈ Trans ( V ( D$_{i}$ )) , and cy$_{i}$ / ∈ A ( D$_{i}$ ) } • σ$_{i}$ = A ( D ∗ $_{i}$) \ { cx$_{i}$, x$_{i}$z : c, z ∈ V ( D$_{i}$ ) and cz ∈ T$_{i}$ } 13