YES

1 decompositions
 #0 -----------
   1: a(s(x)) -> s(a(x))
   2: b(a(b(s(x)))) -> a(b(s(a(x))))
   3: b(a(b(b(x)))) -> a(b(a(b(x))))
   4: a(b(a(a(x)))) -> b(a(b(a(x))))

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

--- S
   1: a(s(x)) -> s(a(x))
   2: b(a(b(s(x)))) -> a(b(s(a(x))))
   3: b(a(b(b(x)))) -> a(b(a(b(x))))
   4: a(b(a(a(x)))) -> b(a(b(a(x))))