YES

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

@Mutually Orthogonal
--- R
   1: h(f(f(f(h(c(),h(b(),c()))))),h(f(f(b())),b())) -> f(h(b(),f(a())))
   2: f(c()) -> b()

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