YES

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

 #2 -----------
   7: +(x2,x1) -> +(x1,x2)
   9: +(+(x2,x1),x3) -> +(x2,+(x1,x3))
  10: +(0(),x2) -> x2
  11: +(s(x2),x1) -> s(+(x2,x1))

@Rule Labeling
--- R
   1: a() -> b()
   2: b() -> c()
   3: c() -> a()
   4: f(s(a())) -> a()

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


@Jouannaud and Kirchner's criterion
--- R
   7: +(x2,x1) -> +(x1,x2)
   9: +(+(x2,x1),x3) -> +(x2,+(x1,x3))
  10: +(0(),x2) -> x2
  11: +(s(x2),x1) -> s(+(x2,x1))

--- S
   7: +(x2,x1) -> +(x1,x2)
   9: +(+(x2,x1),x3) -> +(x2,+(x1,x3))
  10: +(0(),x2) -> x2
  11: +(s(x2),x1) -> s(+(x2,x1))


NOTE: input TRS is reduced

original is
   1: a() -> b()
   2: b() -> c()
   3: c() -> a()
   4: f(s(a())) -> a()
   5: f(s(b())) -> f(s(a()))
   6: f(s(c())) -> f(s(b()))
   7: +(x2,x1) -> +(x1,x2)
   8: +(x2,+(x1,x3)) -> +(+(x2,x1),x3)
   9: +(+(x2,x1),x3) -> +(x2,+(x1,x3))
  10: +(0(),x2) -> x2
  11: +(s(x2),x1) -> s(+(x2,x1))

reduced to
   1: a() -> b()
   2: b() -> c()
   3: c() -> a()
   4: f(s(a())) -> a()
   7: +(x2,x1) -> +(x1,x2)
   9: +(+(x2,x1),x3) -> +(x2,+(x1,x3))
  10: +(0(),x2) -> x2
  11: +(s(x2),x1) -> s(+(x2,x1))