YES

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

Proof:
 WPO Processor:
  algebra: Sum
  weight function:
   w0 = 0
   w(a) = 3
   w(b) = 2
   w(c) = 1
  status function:
   st(a) = st(b) = st(c) = [0]
  precedence:
   b > a > c
  problem:
   
  Qed