YES

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

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

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