YES

1 decompositions
 #0 -----------
   1: f(a()) -> b()
   2: f(a()) -> f(c())
   3: a() -> d()
   4: f(d()) -> b()
   5: f(c()) -> b()
   6: d() -> c()

@Strongly Commuting
--- R
   1: f(a()) -> b()
   2: f(a()) -> f(c())
   3: a() -> d()
   4: f(d()) -> b()
   5: f(c()) -> b()
   6: d() -> c()

--- S
   1: f(a()) -> b()
   2: f(a()) -> f(c())
   3: a() -> d()
   4: f(d()) -> b()
   5: f(c()) -> b()
   6: d() -> c()