NO

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

Proof:
 Unfolding Processor:
  loop length: 7
  terms:
   a(b(c(b(x16562))))
   b(a(a(c(b(a(c(b(x16562))))))))
   b(a(a(c(a(c(b(x16562)))))))
   b(a(a(c(c(b(x16562))))))
   b(a(a(b(x16562))))
   b(a(b(a(a(c(b(a(x16562))))))))
   b(a(b(a(c(b(a(x16562)))))))
  context: b([])
  substitution:
   x16562 -> a(x16562)
  Qed