YES

1 decompositions
 #1 -----------
   1: a(x1) -> x1
   2: a(a(x2)) -> b(c(x2))
   3: b(x3) -> x3
   4: c(x4) -> x4
   5: c(b(x5)) -> b(a(c(x5)))

@Rule Labeling
--- R
   1: a(x1) -> x1
   2: a(a(x2)) -> b(c(x2))
   3: b(x3) -> x3
   4: c(x4) -> x4
   5: c(b(x5)) -> b(a(c(x5)))

--- S
   1: a(x1) -> x1
   2: a(a(x2)) -> b(c(x2))
   3: b(x3) -> x3
   4: c(x4) -> x4
   5: c(b(x5)) -> b(a(c(x5)))