YES

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

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