YES

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

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

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