YES

Problem:
 a(a(x1)) -> b(b(b(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:
   [c](x0) = 6x0,
   
   [d](x0) = 4x0,
   
   [b](x0) = 9x0,
   
   [a](x0) = 14x0
  orientation:
   a(a(x1)) = 28x1 >= 27x1 = b(b(b(x1)))
   
   a(x1) = 14x1 >= 14x1 = d(c(d(x1)))
   
   b(b(x1)) = 18x1 >= 18x1 = c(c(c(x1)))
   
   c(c(x1)) = 12x1 >= 12x1 = d(d(d(x1)))
   
   c(d(d(x1))) = 14x1 >= 14x1 = a(x1)
  problem:
   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)
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
              [0  0 ]  
    [c](x0) = [-& 0 ]x0,
    
              [0  0 ]  
    [d](x0) = [-& 0 ]x0,
    
              [0 1]  
    [b](x0) = [0 0]x0,
    
              [0  0 ]  
    [a](x0) = [-& 0 ]x0
   orientation:
            [0  0 ]      [0  0 ]                
    a(x1) = [-& 0 ]x1 >= [-& 0 ]x1 = d(c(d(x1)))
    
               [1 1]      [0  0 ]                
    b(b(x1)) = [0 1]x1 >= [-& 0 ]x1 = c(c(c(x1)))
    
               [0  0 ]      [0  0 ]                
    c(c(x1)) = [-& 0 ]x1 >= [-& 0 ]x1 = d(d(d(x1)))
    
                  [0  0 ]      [0  0 ]          
    c(d(d(x1))) = [-& 0 ]x1 >= [-& 0 ]x1 = a(x1)
   problem:
    a(x1) -> d(c(d(x1)))
    c(c(x1)) -> d(d(d(x1)))
    c(d(d(x1))) -> a(x1)
   Arctic Interpretation Processor:
    dimension: 2
    interpretation:
               [2 2]  
     [c](x0) = [0 0]x0,
     
               [0  0 ]  
     [d](x0) = [-& 0 ]x0,
     
               [2 2]  
     [a](x0) = [0 0]x0
    orientation:
             [2 2]      [2 2]                
     a(x1) = [0 0]x1 >= [0 0]x1 = d(c(d(x1)))
     
                [4 4]      [0  0 ]                
     c(c(x1)) = [2 2]x1 >= [-& 0 ]x1 = d(d(d(x1)))
     
                   [2 2]      [2 2]          
     c(d(d(x1))) = [0 0]x1 >= [0 0]x1 = a(x1)
    problem:
     a(x1) -> d(c(d(x1)))
     c(d(d(x1))) -> a(x1)
    String Reversal Processor:
     a(x1) -> d(c(d(x1)))
     d(d(c(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 -> 7,3
        9 -> 5*
      problem:
       
      Qed