YES

2 decompositions
 #1 -----------
   4: a() -> b()

 #2 -----------
   1: f(b()) -> c()
   3: f(c()) -> b()

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

--- S
   4: a() -> b()


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

--- S
   1: f(b()) -> c()
   3: f(c()) -> b()


NOTE: input TRS is reduced

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

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