YES

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

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