YES

1 decompositions
 #1 -----------
   1: a(s(x1)) -> s(a(x1))
   2: b(a(b(s(x2)))) -> a(b(s(a(x2))))
   3: b(a(b(b(x3)))) -> c(s(x3))
   4: c(s(x4)) -> a(b(a(b(x4))))
   5: a(b(a(a(x5)))) -> b(a(b(a(x5))))

@Rule Labeling
--- R
   1: a(s(x1)) -> s(a(x1))
   2: b(a(b(s(x2)))) -> a(b(s(a(x2))))
   3: b(a(b(b(x3)))) -> c(s(x3))
   4: c(s(x4)) -> a(b(a(b(x4))))
   5: a(b(a(a(x5)))) -> b(a(b(a(x5))))

--- S
   1: a(s(x1)) -> s(a(x1))
   2: b(a(b(s(x2)))) -> a(b(s(a(x2))))
   3: b(a(b(b(x3)))) -> c(s(x3))
   4: c(s(x4)) -> a(b(a(b(x4))))
   5: a(b(a(a(x5)))) -> b(a(b(a(x5))))