YES

1 decompositions
 #1 -----------
   1: c() -> b()
   4: f(f(a())) -> c()

@Knuth and Bendix' criterion
--- R
   1: c() -> b()
   4: f(f(a())) -> c()

--- S
   1: c() -> b()
   4: f(f(a())) -> c()


NOTE: input TRS is reduced

original is
   1: c() -> b()
   2: a() -> a()
   3: b() -> b()
   4: f(f(a())) -> c()

reduced to
   1: c() -> b()
   4: f(f(a())) -> c()