YES

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

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [c](x0) = 2x0,
   
   [b](x0) = x0 + 1,
   
   [a](x0) = x0 + 4,
   
   [d](x0) = 2x0
  orientation:
   a(a(d(d(x1)))) = 4x1 + 8 >= 4x1 + 8 = d(d(b(b(x1))))
   
   a(a(x1)) = x1 + 8 >= x1 + 6 = b(b(b(b(b(b(x1))))))
   
   b(b(d(d(b(b(x1)))))) = 4x1 + 10 >= 4x1 + 8 = a(a(c(c(x1))))
   
   c(c(x1)) = 4x1 >= 4x1 = d(d(x1))
  problem:
   a(a(d(d(x1)))) -> d(d(b(b(x1))))
   c(c(x1)) -> d(d(x1))
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
              [0 2]  
    [c](x0) = [0 2]x0,
    
              [0 0]  
    [b](x0) = [0 1]x0,
    
              [1 1]  
    [a](x0) = [1 2]x0,
    
              [0  0 ]  
    [d](x0) = [-& 1 ]x0
   orientation:
                     [2 5]      [2 3]                   
    a(a(d(d(x1)))) = [3 6]x1 >= [3 4]x1 = d(d(b(b(x1))))
    
               [2 4]      [0  1 ]             
    c(c(x1)) = [2 4]x1 >= [-& 2 ]x1 = d(d(x1))
   problem:
    a(a(d(d(x1)))) -> d(d(b(b(x1))))
   Arctic Interpretation Processor:
    dimension: 2
    interpretation:
               [0  0 ]  
     [b](x0) = [-& -&]x0,
     
               [2  -&]  
     [a](x0) = [-& 1 ]x0,
     
               [0 0]  
     [d](x0) = [1 0]x0
    orientation:
                      [5 4]      [1 1]                   
     a(a(d(d(x1)))) = [3 3]x1 >= [1 1]x1 = d(d(b(b(x1))))
    problem:
     
    Qed