YES

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

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [c](x0) = x0 + 5,
   
   [b](x0) = x0 + 2,
   
   [a](x0) = x0 + 1,
   
   [d](x0) = x0 + 8
  orientation:
   c(c(c(a(x1)))) = x1 + 16 >= x1 + 16 = d(d(x1))
   
   d(b(x1)) = x1 + 10 >= x1 + 10 = c(c(x1))
   
   c(x1) = x1 + 5 >= x1 + 4 = a(a(a(a(x1))))
   
   d(x1) = x1 + 8 >= x1 + 8 = b(b(b(b(x1))))
   
   b(d(x1)) = x1 + 10 >= x1 + 10 = c(c(x1))
   
   a(c(c(c(x1)))) = x1 + 16 >= x1 + 16 = d(d(x1))
  problem:
   c(c(c(a(x1)))) -> d(d(x1))
   d(b(x1)) -> c(c(x1))
   d(x1) -> b(b(b(b(x1))))
   b(d(x1)) -> c(c(x1))
   a(c(c(c(x1)))) -> d(d(x1))
  String Reversal Processor:
   a(c(c(c(x1)))) -> d(d(x1))
   b(d(x1)) -> c(c(x1))
   d(x1) -> b(b(b(b(x1))))
   d(b(x1)) -> c(c(x1))
   c(c(c(a(x1)))) -> d(d(x1))
   WPO Processor:
    algebra: Sum
    weight function:
     w0 = 0
     w(a) = 1
     w(b) = w(d) = w(c) = 0
    status function:
     st(b) = st(d) = st(c) = st(a) = [0]
    precedence:
     d > b > c ~ a
    problem:
     
    Qed