YES

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

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

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