YES

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

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

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