NO

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

Proof:
 Unfolding Processor:
  loop length: 7
  terms:
   a(b(b(x10285)))
   a(c(b(b(a(a(b(x10285)))))))
   a(c(b(a(a(b(x10285))))))
   a(c(a(a(b(x10285)))))
   a(c(a(b(x10285))))
   a(c(a(c(b(b(a(a(x10285))))))))
   a(c(c(b(b(a(a(x10285)))))))
  context: []
  substitution:
   x10285 -> a(a(x10285))
  Qed