YES

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

@Strongly Commuting
--- R
   1: h(a(),b()) -> a()
   2: f(c()) -> f(c())
   3: c() -> h(a(),h(b(),b()))

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