YES

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

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

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