YES

1 decompositions
 #0 -----------
   1: a() -> b()
   2: b() -> a()
   3: a() -> d()

@Rule Labeling
--- R
   1: a() -> b()
   2: b() -> a()
   3: a() -> d()

--- S
   1: a() -> b()
   2: b() -> a()
   3: a() -> d()