YES

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

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [c](x0) = x0 + 1,
   
   [b](x0) = x0 + 12,
   
   [d](x0) = 3x0,
   
   [a](x0) = x0 + 4
  orientation:
   d(a(x1)) = 3x1 + 12 >= 3x1 + 12 = b(d(x1))
   
   b(x1) = x1 + 12 >= x1 + 12 = a(a(a(x1)))
   
   c(d(c(x1))) = 3x1 + 4 >= 3x1 + 4 = a(d(x1))
   
   b(d(d(x1))) = 9x1 + 12 >= 9x1 + 11 = c(c(d(d(c(x1)))))
  problem:
   d(a(x1)) -> b(d(x1))
   b(x1) -> a(a(a(x1)))
   c(d(c(x1))) -> a(d(x1))
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
              [0 0]  
    [c](x0) = [2 2]x0,
    
              [0  0 ]  
    [b](x0) = [-& 3 ]x0,
    
              [0  3 ]  
    [d](x0) = [-& -&]x0,
    
              [0  -&]  
    [a](x0) = [-& 1 ]x0
   orientation:
               [0  4 ]      [0  3 ]             
    d(a(x1)) = [-& -&]x1 >= [-& -&]x1 = b(d(x1))
    
            [0  0 ]      [0  -&]                
    b(x1) = [-& 3 ]x1 >= [-& 3 ]x1 = a(a(a(x1)))
    
                  [5 5]      [0  3 ]             
    c(d(c(x1))) = [7 7]x1 >= [-& -&]x1 = a(d(x1))
   problem:
    d(a(x1)) -> b(d(x1))
    b(x1) -> a(a(a(x1)))
   String Reversal Processor:
    a(d(x1)) -> d(b(x1))
    b(x1) -> a(a(a(x1)))
    Matrix Interpretation Processor: dim=2
     
     interpretation:
                [1 3]     [1]
      [b](x0) = [0 1]x0 + [0],
      
                [1 0]     [0]
      [d](x0) = [0 3]x0 + [1],
      
                [1 1]  
      [a](x0) = [0 1]x0
     orientation:
                 [1 3]     [1]    [1 3]     [1]           
      a(d(x1)) = [0 3]x1 + [1] >= [0 3]x1 + [1] = d(b(x1))
      
              [1 3]     [1]    [1 3]                
      b(x1) = [0 1]x1 + [0] >= [0 1]x1 = a(a(a(x1)))
     problem:
      a(d(x1)) -> d(b(x1))
     Arctic Interpretation Processor:
      dimension: 3
      interpretation:
                 [0  2  -&]  
       [b](x0) = [0  2  1 ]x0
                 [1  1  1 ]  ,
       
                 [0  0  0 ]  
       [d](x0) = [0  0  -&]x0
                 [-& 0  0 ]  ,
       
                 [2  -& 3 ]  
       [a](x0) = [1  1  3 ]x0
                 [-& 2  3 ]  
      orientation:
                  [2 3 3]      [1 2 1]             
       a(d(x1)) = [1 3 3]x1 >= [0 2 1]x1 = d(b(x1))
                  [2 3 3]      [1 2 1]             
      problem:
       
      Qed