YES

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

Proof:
 String Reversal Processor:
  d(a(x1)) -> b(d(x1))
  a(x1) -> b(b(b(x1)))
  b(d(b(x1))) -> c(a(x1))
  c(x1) -> d(x1)
  Matrix Interpretation Processor: dim=2
   
   interpretation:
              [1 0]     [0]
    [c](x0) = [0 2]x0 + [1],
    
              [1 1]  
    [b](x0) = [0 1]x0,
    
              [1 3]     [1]
    [a](x0) = [0 1]x0 + [0],
    
              [1 0]     [0]
    [d](x0) = [0 2]x0 + [1]
   orientation:
               [1 3]     [1]    [1 2]     [1]           
    d(a(x1)) = [0 2]x1 + [1] >= [0 2]x1 + [1] = b(d(x1))
    
            [1 3]     [1]    [1 3]                
    a(x1) = [0 1]x1 + [0] >= [0 1]x1 = b(b(b(x1)))
    
                  [1 3]     [1]    [1 3]     [1]           
    b(d(b(x1))) = [0 2]x1 + [1] >= [0 2]x1 + [1] = c(a(x1))
    
            [1 0]     [0]    [1 0]     [0]        
    c(x1) = [0 2]x1 + [1] >= [0 2]x1 + [1] = d(x1)
   problem:
    d(a(x1)) -> b(d(x1))
    b(d(b(x1))) -> c(a(x1))
    c(x1) -> d(x1)
   Arctic Interpretation Processor:
    dimension: 2
    interpretation:
               [1  0 ]  
     [c](x0) = [3  -&]x0,
     
               [2  1 ]  
     [b](x0) = [-& 2 ]x0,
     
               [3  -&]  
     [a](x0) = [-& -&]x0,
     
               [1  -&]  
     [d](x0) = [3  -&]x0
    orientation:
                [4  -&]      [4  -&]             
     d(a(x1)) = [6  -&]x1 >= [5  -&]x1 = b(d(x1))
     
                   [6 5]      [4  -&]             
     b(d(b(x1))) = [7 6]x1 >= [6  -&]x1 = c(a(x1))
     
             [1  0 ]      [1  -&]          
     c(x1) = [3  -&]x1 >= [3  -&]x1 = d(x1)
    problem:
     d(a(x1)) -> b(d(x1))
     c(x1) -> d(x1)
    Arctic Interpretation Processor:
     dimension: 3
     interpretation:
                [3 2 3]  
      [c](x0) = [2 1 3]x0
                [0 2 3]  ,
      
                [0  0  2 ]  
      [b](x0) = [-& -& -&]x0
                [0  -& 0 ]  ,
      
                [2  0  2 ]  
      [a](x0) = [0  2  2 ]x0
                [1  -& 1 ]  ,
      
                [0  0  0 ]  
      [d](x0) = [1  0  2 ]x0
                [-& 0  0 ]  
     orientation:
                 [2 2 2]      [1  2  2 ]             
      d(a(x1)) = [3 2 3]x1 >= [-& -& -&]x1 = b(d(x1))
                 [1 2 2]      [0  0  0 ]             
      
              [3 2 3]      [0  0  0 ]          
      c(x1) = [2 1 3]x1 >= [1  0  2 ]x1 = d(x1)
              [0 2 3]      [-& 0  0 ]          
     problem:
      d(a(x1)) -> b(d(x1))
     Arctic Interpretation Processor:
      dimension: 3
      interpretation:
                 [0  0  -&]  
       [b](x0) = [0  2  1 ]x0
                 [0  -& 0 ]  ,
       
                 [2 0 2]  
       [a](x0) = [1 2 0]x0
                 [3 3 3]  ,
       
                 [1  0  0 ]  
       [d](x0) = [0  -& 0 ]x0
                 [-& 0  0 ]  
      orientation:
                  [3 3 3]      [1 0 0]             
       d(a(x1)) = [3 3 3]x1 >= [2 1 2]x1 = b(d(x1))
                  [3 3 3]      [1 0 0]             
      problem:
       
      Qed