YES

Problem:
 a(a(x1)) -> b(c(c(c(x1))))
 b(c(x1)) -> d(d(d(d(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:
   [c](x0) = x0 + 3,
   
   [d](x0) = x0 + 2,
   
   [a](x0) = x0 + 7,
   
   [b](x0) = x0 + 5
  orientation:
   a(a(x1)) = x1 + 14 >= x1 + 14 = b(c(c(c(x1))))
   
   b(c(x1)) = x1 + 8 >= x1 + 8 = d(d(d(d(x1))))
   
   a(x1) = x1 + 7 >= x1 + 7 = d(c(d(x1)))
   
   b(b(x1)) = x1 + 10 >= x1 + 9 = c(c(c(x1)))
   
   c(c(x1)) = x1 + 6 >= x1 + 6 = d(d(d(x1)))
   
   c(d(d(x1))) = x1 + 7 >= x1 + 7 = a(x1)
  problem:
   a(a(x1)) -> b(c(c(c(x1))))
   b(c(x1)) -> d(d(d(d(x1))))
   a(x1) -> d(c(d(x1)))
   c(c(x1)) -> d(d(d(x1)))
   c(d(d(x1))) -> a(x1)
  String Reversal Processor:
   a(a(x1)) -> c(c(c(b(x1))))
   c(b(x1)) -> d(d(d(d(x1))))
   a(x1) -> d(c(d(x1)))
   c(c(x1)) -> d(d(d(x1)))
   d(d(c(x1))) -> a(x1)
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [c](x0) = x0 + 2,
     
     [d](x0) = x0 + 1,
     
     [a](x0) = x0 + 4,
     
     [b](x0) = x0 + 2
    orientation:
     a(a(x1)) = x1 + 8 >= x1 + 8 = c(c(c(b(x1))))
     
     c(b(x1)) = x1 + 4 >= x1 + 4 = d(d(d(d(x1))))
     
     a(x1) = x1 + 4 >= x1 + 4 = d(c(d(x1)))
     
     c(c(x1)) = x1 + 4 >= x1 + 3 = d(d(d(x1)))
     
     d(d(c(x1))) = x1 + 4 >= x1 + 4 = a(x1)
    problem:
     a(a(x1)) -> c(c(c(b(x1))))
     c(b(x1)) -> d(d(d(d(x1))))
     a(x1) -> d(c(d(x1)))
     d(d(c(x1))) -> a(x1)
    String Reversal Processor:
     a(a(x1)) -> b(c(c(c(x1))))
     b(c(x1)) -> d(d(d(d(x1))))
     a(x1) -> d(c(d(x1)))
     c(d(d(x1))) -> a(x1)
     WPO Processor:
      algebra: Sum
      weight function:
       w0 = 0
       w(b) = 4
       w(a) = 2
       w(d) = 1
       w(c) = 0
      status function:
       st(d) = st(b) = st(c) = st(a) = [0]
      precedence:
       c > a > b > d
      problem:
       
      Qed