YES

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

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [b](x0) = x0 + 9,
   
   [c](x0) = x0 + 6,
   
   [a](x0) = x0 + 14,
   
   [d](x0) = x0 + 4
  orientation:
   a(a(x1)) = x1 + 28 >= x1 + 27 = b(b(b(x1)))
   
   a(x1) = x1 + 14 >= x1 + 14 = d(c(d(x1)))
   
   b(b(x1)) = x1 + 18 >= x1 + 18 = c(c(c(x1)))
   
   c(c(x1)) = x1 + 12 >= x1 + 12 = d(d(d(x1)))
   
   c(d(d(x1))) = x1 + 14 >= x1 + 14 = a(x1)
  problem:
   a(x1) -> d(c(d(x1)))
   b(b(x1)) -> c(c(c(x1)))
   c(c(x1)) -> d(d(d(x1)))
   c(d(d(x1))) -> a(x1)
  WPO Processor:
   algebra: Sum
   weight function:
    w0 = 0
    w(a) = 4
    w(b) = 3
    w(c) = 2
    w(d) = 1
   status function:
    st(c) = st(d) = st(b) = st(a) = [0]
   precedence:
    b > c > a > d
   problem:
    
   Qed