YES

1 decompositions
 #0 -----------
   1: a() -> c()
   2: c() -> a()
   3: c() -> h(a(),c())

@Rule Labeling
--- R
   1: a() -> c()
   2: c() -> a()
   3: c() -> h(a(),c())

--- S
   1: a() -> c()
   2: c() -> a()
   3: c() -> h(a(),c())