YES

1 decompositions
 #1 -----------
   1: f(f(f(x1))) -> a()
   3: f(a()) -> a()
   4: f(f(g(g(x2)))) -> f(a())
   6: g(a()) -> a()

@Knuth and Bendix' criterion
--- R
   1: f(f(f(x1))) -> a()
   3: f(a()) -> a()
   4: f(f(g(g(x2)))) -> f(a())
   6: g(a()) -> a()

--- S
   1: f(f(f(x1))) -> a()
   3: f(a()) -> a()
   4: f(f(g(g(x2)))) -> f(a())
   6: g(a()) -> a()


NOTE: input TRS is reduced

original is
   1: f(f(f(x1))) -> a()
   2: f(f(a())) -> a()
   3: f(a()) -> a()
   4: f(f(g(g(x2)))) -> f(a())
   5: g(f(a())) -> a()
   6: g(a()) -> a()

reduced to
   1: f(f(f(x1))) -> a()
   3: f(a()) -> a()
   4: f(f(g(g(x2)))) -> f(a())
   6: g(a()) -> a()