Example 3.8 Let D be reflexive directed graph (i) shown in Figure 4. The clasps are 2 and 4 and we set x$_{1}$ = 2 . Figure 4: A construction using the proof of Theorem 3.2. Step 1 Y$_{1}$ = { 4 , 6 } and A$_{1}$ is empty. Step 2 Use construction A since A$_{1}$ is empty. Let D$_{2}$ be the reflexive directed graph with V ( D$_{2}$ ) = V ( D$_{1}$ ) ∪ { t$_{1}$ } and A ( D ∗ $_{2}$) = σ$_{1}$ ∪ τ$_{1}$ where B$_{1}$ = { 4 , 6 } , σ$_{1}$ = A ( D ∗ $_{1}$) \ { 24 , 26 } , and τ$_{1}$ = { t$_{1}$ 4 , t$_{1}$ 6 } . This gives (ii) in Figure 4. Step 3 Define θ$_{1}$ : V ( D$_{2}$ ) → V ( D$_{1}$ ) by θ$_{1}$ ( t$_{1}$ ) = 2 and θ$_{1}$ ( u ) = u for all u ∈ V ( D$_{1}$ ) . Step 4 We go back to step 1 with x$_{2}$ = 4 since 4 is the only clasp in D$_{2}$ . Step 1 We have Y$_{2}$ = { 6 , 7 } and A$_{2}$ = { t$_{1}$, 3 } . Step 2 Use construction B with a$_{2}$ = 3 , b$_{2}$ = t$_{1}$ , and y$_{2}$ = 6 . Let D$_{3}$ be the reflexive directed graph with V ( D$_{3}$ ) = V ( D$_{2}$ ) ∪ { t$_{2}$ } and A ( D ∗ $_{3}$) = σ$_{2}$ ∪ τ$_{2}$ where σ$_{2}$ = A ( D ∗ $_{2}$) \ { 34 , 47 } , τ$_{2}$ = { 3 t$_{2}$, t$_{2}$ 7 } . This gives (iii) in Figure 4. 11