YES

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

Proof:
 WPO Processor:
  algebra: Sum
  weight function:
   w0 = 0
   w(a) = 2
   w(b) = 1
   w(c) = 0
  status function:
   st(c) = st(b) = st(a) = [0]
  precedence:
   c > a > b
  problem:
   
  Qed