Prove that graphs build by buildCfg are sufficient

That is, if we have a (semantic) trace, we can
find a corresponding path through the CFG.

Signed-off-by: Danila Fedorin <danila.fedorin@gmail.com>

Language/Graphs.agda

@@ -79,19 +79,13 @@ _↦_ g₁ g₂ = record
     }
 
 loop : Graph → Graph
-loop g = record g
-    { edges = Graph.edges g List.++
-              List.cartesianProduct (Graph.outputs g) (Graph.inputs g)
-    }
-
-optional : Graph → Graph
-optional g = record
+loop g = record
     { size = 2 Nat.+ Graph.size g
     ; nodes = [] ∷ [] ∷ Graph.nodes g
     ; edges = (2 ↑ʳᵉ Graph.edges g) List.++
               List.map (zero ,_) (2 ↑ʳⁱ Graph.inputs g) List.++
               List.map (_, suc zero) (2 ↑ʳⁱ Graph.outputs g) List.++
-              ((zero , suc zero) ∷ [])
+              ((suc zero , zero) ∷ (zero , suc zero) ∷ [])
     ; inputs = zero ∷ []
     ; outputs = (suc zero) ∷ []
     }
@@ -122,4 +116,4 @@ buildCfg : Stmt → Graph
 buildCfg ⟨ bs₁ ⟩ = singleton (bs₁ ∷ [])
 buildCfg (s₁ then s₂) = buildCfg s₁ ↦ buildCfg s₂
 buildCfg (if _ then s₁ else s₂) = singleton [] ↦ (buildCfg s₁ ∙ buildCfg s₂) ↦ singleton []
-buildCfg (while _ repeat s) = optional (loop (buildCfg s))
+buildCfg (while _ repeat s) = loop (buildCfg s)