YES

1 decompositions
 #0 -----------
   1: g(f(a())) -> f(g(f(a())))
   2: g(f(a())) -> f(f(a()))
   3: f(f(a())) -> f(a())

@--testing feature--
--- R
   2: g(f(a())) -> f(f(a()))
   3: f(f(a())) -> f(a())

--- S
   2: g(f(a())) -> f(f(a()))
   3: f(f(a())) -> f(a())