YES

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

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

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