YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 1
  interpretation:
   [d](x0) = 2x0,
   
   [b](x0) = 5x0,
   
   [c](x0) = 3x0,
   
   [a](x0) = 7x0
  orientation:
   a(a(x1)) = 14x1 >= 14x1 = b(c(c(c(x1))))
   
   b(c(x1)) = 8x1 >= 8x1 = d(d(d(d(x1))))
   
   a(x1) = 7x1 >= 7x1 = d(c(d(x1)))
   
   b(b(x1)) = 10x1 >= 9x1 = c(c(c(x1)))
   
   c(c(x1)) = 6x1 >= 6x1 = d(d(d(x1)))
   
   c(d(d(x1))) = 7x1 >= 7x1 = a(x1)
  problem:
   a(a(x1)) -> b(c(c(c(x1))))
   b(c(x1)) -> d(d(d(d(x1))))
   a(x1) -> d(c(d(x1)))
   c(c(x1)) -> d(d(d(x1)))
   c(d(d(x1))) -> a(x1)
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [d](x0) = 1x0,
    
    [b](x0) = x0,
    
    [c](x0) = 4x0,
    
    [a](x0) = 6x0
   orientation:
    a(a(x1)) = 12x1 >= 12x1 = b(c(c(c(x1))))
    
    b(c(x1)) = 4x1 >= 4x1 = d(d(d(d(x1))))
    
    a(x1) = 6x1 >= 6x1 = d(c(d(x1)))
    
    c(c(x1)) = 8x1 >= 3x1 = d(d(d(x1)))
    
    c(d(d(x1))) = 6x1 >= 6x1 = a(x1)
   problem:
    a(a(x1)) -> b(c(c(c(x1))))
    b(c(x1)) -> d(d(d(d(x1))))
    a(x1) -> d(c(d(x1)))
    c(d(d(x1))) -> a(x1)
   String Reversal Processor:
    a(a(x1)) -> c(c(c(b(x1))))
    c(b(x1)) -> d(d(d(d(x1))))
    a(x1) -> d(c(d(x1)))
    d(d(c(x1))) -> a(x1)
    Matrix Interpretation Processor: dim=3
     
     interpretation:
                [1 1 0]  
      [d](x0) = [1 0 0]x0
                [0 0 0]  ,
      
                [3 3 0]  
      [b](x0) = [0 2 0]x0
                [1 0 0]  ,
      
                [1 0 2]     [0]
      [c](x0) = [0 1 3]x0 + [1]
                [0 0 0]     [0],
      
                [2 1 1]     [1]
      [a](x0) = [1 1 0]x0 + [1]
                [0 0 0]     [0]
     orientation:
                 [5 3 2]     [4]    [5 3 0]     [0]                 
      a(a(x1)) = [3 2 1]x1 + [3] >= [3 2 0]x1 + [3] = c(c(c(b(x1))))
                 [0 0 0]     [0]    [0 0 0]     [0]                 
      
                 [5 3 0]     [0]    [5 3 0]                   
      c(b(x1)) = [3 2 0]x1 + [1] >= [3 2 0]x1 = d(d(d(d(x1))))
                 [0 0 0]     [0]    [0 0 0]                   
      
              [2 1 1]     [1]    [2 1 0]     [1]              
      a(x1) = [1 1 0]x1 + [1] >= [1 1 0]x1 + [0] = d(c(d(x1)))
              [0 0 0]     [0]    [0 0 0]     [0]              
      
                    [2 1 7]     [1]    [2 1 1]     [1]        
      d(d(c(x1))) = [1 1 5]x1 + [1] >= [1 1 0]x1 + [1] = a(x1)
                    [0 0 0]     [0]    [0 0 0]     [0]        
     problem:
      c(b(x1)) -> d(d(d(d(x1))))
      a(x1) -> d(c(d(x1)))
      d(d(c(x1))) -> a(x1)
     Arctic Interpretation Processor:
      dimension: 2
      interpretation:
                 [0  -&]  
       [d](x0) = [0  0 ]x0,
       
                 [1  0 ]  
       [b](x0) = [0  -&]x0,
       
                 [3 3]  
       [c](x0) = [1 1]x0,
       
                 [3 3]  
       [a](x0) = [3 3]x0
      orientation:
                  [4 3]      [0  -&]                   
       c(b(x1)) = [2 1]x1 >= [0  0 ]x1 = d(d(d(d(x1))))
       
               [3 3]      [3 3]                
       a(x1) = [3 3]x1 >= [3 3]x1 = d(c(d(x1)))
       
                     [3 3]      [3 3]          
       d(d(c(x1))) = [3 3]x1 >= [3 3]x1 = a(x1)
      problem:
       a(x1) -> d(c(d(x1)))
       d(d(c(x1))) -> a(x1)
      String Reversal Processor:
       a(x1) -> d(c(d(x1)))
       c(d(d(x1))) -> a(x1)
       Bounds Processor:
        bound: 1
        enrichment: match
        automaton:
         final states: {5,1}
         transitions:
          a0(2) -> 5*
          d1(6) -> 7*
          d1(8) -> 9*
          c1(7) -> 8*
          f40() -> 2*
          d0(2) -> 3*
          d0(4) -> 1*
          c0(3) -> 4*
          2 -> 6*
          5 -> 8,4
          9 -> 5*
        problem:
         
        Qed