YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 1
  interpretation:
   [d](x0) = 6x0,
   
   [b](x0) = 8x0,
   
   [c](x0) = 9x0,
   
   [a](x0) = 9x0
  orientation:
   a(a(x1)) = 18x1 >= 17x1 = b(c(x1))
   
   b(b(x1)) = 16x1 >= 15x1 = c(d(x1))
   
   c(c(x1)) = 18x1 >= 18x1 = d(d(d(x1)))
   
   d(d(d(x1))) = 18x1 >= 18x1 = a(c(x1))
  problem:
   c(c(x1)) -> d(d(d(x1)))
   d(d(d(x1))) -> a(c(x1))
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
              [1  -&]  
    [d](x0) = [0  2 ]x0,
    
              [2  -&]  
    [c](x0) = [2  3 ]x0,
    
              [0  -&]  
    [a](x0) = [1  0 ]x0
   orientation:
               [4  -&]      [3  -&]                
    c(c(x1)) = [5  6 ]x1 >= [4  6 ]x1 = d(d(d(x1)))
    
                  [3  -&]      [2  -&]             
    d(d(d(x1))) = [4  6 ]x1 >= [3  3 ]x1 = a(c(x1))
   problem:
    c(c(x1)) -> d(d(d(x1)))
   Arctic Interpretation Processor:
    dimension: 2
    interpretation:
               [0  -&]  
     [d](x0) = [0  -&]x0,
     
               [1 0]  
     [c](x0) = [0 0]x0
    orientation:
                [2 1]      [0  -&]                
     c(c(x1)) = [1 0]x1 >= [0  -&]x1 = d(d(d(x1)))
    problem:
     
    Qed