YES

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

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

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