YES

1 decompositions
 #1 -----------
   1: a(a(x1)) -> a(b(a(x1)))
   2: b(a(b(x2))) -> a(c(a(x2)))

@Rule Labeling
--- R
   1: a(a(x1)) -> a(b(a(x1)))
   2: b(a(b(x2))) -> a(c(a(x2)))

--- S
   1: a(a(x1)) -> a(b(a(x1)))
   2: b(a(b(x2))) -> a(c(a(x2)))