YES 1 decompositions #0 ----------- 1: g(a()) -> f(g(a())) 2: g(b()) -> c(a()) 3: a() -> b() 4: f(x) -> h(x) 5: h(x) -> c(b()) @Rule Labeling --- R 1: g(a()) -> f(g(a())) 2: g(b()) -> c(a()) 3: a() -> b() 4: f(x) -> h(x) 5: h(x) -> c(b()) --- S 1: g(a()) -> f(g(a())) 2: g(b()) -> c(a()) 3: a() -> b() 4: f(x) -> h(x) 5: h(x) -> c(b())