NO

Problem:
 a(x1) -> x1
 a(b(x1)) -> b(a(c(b(a(a(x1))))))
 c(b(c(x1))) -> x1

Proof:
 Unfolding Processor:
  loop length: 7
  terms:
   a(b(b(c(b(x14133)))))
   b(a(c(b(a(a(b(c(b(x14133)))))))))
   b(a(c(b(a(b(a(c(b(a(a(c(b(x14133)))))))))))))
   b(a(c(b(a(b(a(c(b(a(c(b(x14133))))))))))))
   b(a(c(b(a(b(a(c(b(c(b(x14133)))))))))))
   b(a(c(b(a(b(a(b(x14133))))))))
   b(a(c(b(a(b(b(a(c(b(a(a(x14133))))))))))))
  context: b(a(c(b([]))))
  substitution:
   x14133 -> a(a(x14133))
  Qed