YES

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

Proof:
 WPO Processor:
  algebra: Pol
  weight function:
   w0 = 0
   w(a) = 4
   w(b) = 1
   w(c) = 0
  status function:
   st(b) = st(a) = st(c) = [0]
  subterm coefficient function:
   sc(c, 0) = 4
   sc(b, 0) = sc(a, 0) = 1
  precedence:
   a > b ~ c
  problem:
   
  Qed