YES

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

@Rule Labeling
--- R
   2: a() -> b()
   4: b() -> a()
   5: b() -> c()

--- S
   2: a() -> b()
   4: b() -> a()
   5: b() -> c()


NOTE: input TRS is reduced

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

reduced to
   2: a() -> b()
   4: b() -> a()
   5: b() -> c()