YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 1
  interpretation:
   [c](x0) = 13x0,
   
   [d](x0) = 14x0,
   
   [b](x0) = 9x0,
   
   [a](x0) = 6x0
  orientation:
   a(a(a(x1))) = 18x1 >= 18x1 = b(b(x1))
   
   b(b(b(x1))) = 27x1 >= 27x1 = c(d(x1))
   
   c(x1) = 13x1 >= 12x1 = a(a(x1))
   
   d(x1) = 14x1 >= 13x1 = c(x1)
  problem:
   a(a(a(x1))) -> b(b(x1))
   b(b(b(x1))) -> c(d(x1))
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
              [0  0 ]  
    [c](x0) = [-& 0 ]x0,
    
              [0  -&]  
    [d](x0) = [0  -&]x0,
    
              [0  -&]  
    [b](x0) = [0  1 ]x0,
    
              [2  0 ]  
    [a](x0) = [1  -&]x0
   orientation:
                  [6 4]      [0  -&]             
    a(a(a(x1))) = [5 3]x1 >= [1  2 ]x1 = b(b(x1))
    
                  [0  -&]      [0  -&]             
    b(b(b(x1))) = [2  3 ]x1 >= [0  -&]x1 = c(d(x1))
   problem:
    b(b(b(x1))) -> c(d(x1))
   Arctic Interpretation Processor:
    dimension: 3
    interpretation:
               [0  0  0 ]  
     [c](x0) = [0  0  0 ]x0
               [-& -& 0 ]  ,
     
               [0 0 0]  
     [d](x0) = [0 0 0]x0
               [1 0 0]  ,
     
               [0  0  -&]  
     [b](x0) = [-& -& 1 ]x0
               [1  0  1 ]  
    orientation:
                   [2 1 2]      [1 0 0]             
     b(b(b(x1))) = [3 2 3]x1 >= [1 0 0]x1 = c(d(x1))
                   [3 2 3]      [1 0 0]             
    problem:
     
    Qed