YES

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

@Jouannaud and Kirchner's criterion
--- R
   1: f(f(g(g(x)))) -> f(a())
   2: g(a()) -> a()
   3: f(a()) -> a()
   4: f(f(f(x))) -> a()
   5: f(f(a())) -> a()

--- S
   1: f(f(g(g(x)))) -> f(a())
   2: g(a()) -> a()
   3: f(a()) -> a()
   4: f(f(f(x))) -> a()
   5: f(f(a())) -> a()