YES

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

Proof:
 WPO Processor:
  algebra: Pol
  weight function:
   w0 = 0
   w(b) = 3
   w(a) = 2
   w(c) = 1
  status function:
   st(c) = st(b) = st(a) = [0]
  subterm coefficient function:
   sc(c, 0) = 3
   sc(b, 0) = sc(a, 0) = 1
  precedence:
   b > c ~ a
  problem:
   
  Qed