YES

1 decompositions
 #0 -----------
   1: a(x) -> x
   2: a(b(x)) -> c(b(b(a(a(x)))))
   3: b(x) -> c(x)
   4: c(c(x)) -> x

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

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