YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 1
  interpretation:
   [c](x0) = 1x0,
   
   [d](x0) = x0,
   
   [b](x0) = 2x0,
   
   [a](x0) = 3x0
  orientation:
   a(a(x1)) = 6x1 >= 6x1 = b(b(b(x1)))
   
   b(x1) = 2x1 >= 2x1 = c(c(d(x1)))
   
   c(x1) = 1x1 >= x1 = d(d(d(x1)))
   
   b(c(x1)) = 3x1 >= 3x1 = c(b(x1))
   
   b(c(d(x1))) = 3x1 >= 3x1 = a(x1)
  problem:
   a(a(x1)) -> b(b(b(x1)))
   b(x1) -> c(c(d(x1)))
   b(c(x1)) -> c(b(x1))
   b(c(d(x1))) -> a(x1)
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
                
    [c](x0) = x0,
    
              [1  0 ]  
    [d](x0) = [-& 1 ]x0,
    
              [1  1 ]  
    [b](x0) = [-& 1 ]x0,
    
              [2  2 ]  
    [a](x0) = [-& 2 ]x0
   orientation:
               [4  4 ]      [3  3 ]                
    a(a(x1)) = [-& 4 ]x1 >= [-& 3 ]x1 = b(b(b(x1)))
    
            [1  1 ]      [1  0 ]                
    b(x1) = [-& 1 ]x1 >= [-& 1 ]x1 = c(c(d(x1)))
    
               [1  1 ]      [1  1 ]             
    b(c(x1)) = [-& 1 ]x1 >= [-& 1 ]x1 = c(b(x1))
    
                  [2  2 ]      [2  2 ]          
    b(c(d(x1))) = [-& 2 ]x1 >= [-& 2 ]x1 = a(x1)
   problem:
    b(x1) -> c(c(d(x1)))
    b(c(x1)) -> c(b(x1))
    b(c(d(x1))) -> a(x1)
   Arctic Interpretation Processor:
    dimension: 2
    interpretation:
               [0 1]  
     [c](x0) = [0 0]x0,
     
               [1  -&]  
     [d](x0) = [0  -&]x0,
     
               [3  -&]  
     [b](x0) = [2  3 ]x0,
     
               [0  -&]  
     [a](x0) = [0  -&]x0
    orientation:
             [3  -&]      [2  -&]                
     b(x1) = [2  3 ]x1 >= [1  -&]x1 = c(c(d(x1)))
     
                [3 4]      [3 4]             
     b(c(x1)) = [3 3]x1 >= [3 3]x1 = c(b(x1))
     
                   [4  -&]      [0  -&]          
     b(c(d(x1))) = [4  -&]x1 >= [0  -&]x1 = a(x1)
    problem:
     b(c(x1)) -> c(b(x1))
    KBO Processor:
     weight function:
      w0 = 1
      w(c) = w(b) = 1
     precedence:
      b > c
     problem:
      
     Qed