YES

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

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

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