YES

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

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

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