YES

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

Proof:
 KBO Processor:
  weight function:
   w0 = 1
   w(b) = w(a) = 1
   w(c) = 0
  precedence:
   c > a > b
  problem:
   
  Qed