YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 1
  interpretation:
   [b](x0) = 2x0,
   
   [d](x0) = 8x0,
   
   [c](x0) = 5x0,
   
   [a](x0) = 1x0
  orientation:
   c(c(c(a(x1)))) = 16x1 >= 16x1 = d(d(x1))
   
   d(b(x1)) = 10x1 >= 10x1 = c(c(x1))
   
   c(x1) = 5x1 >= 4x1 = a(a(a(a(x1))))
   
   d(x1) = 8x1 >= 8x1 = b(b(b(b(x1))))
   
   b(d(x1)) = 10x1 >= 10x1 = c(c(x1))
   
   a(c(c(c(x1)))) = 16x1 >= 16x1 = d(d(x1))
  problem:
   c(c(c(a(x1)))) -> d(d(x1))
   d(b(x1)) -> c(c(x1))
   d(x1) -> b(b(b(b(x1))))
   b(d(x1)) -> c(c(x1))
   a(c(c(c(x1)))) -> d(d(x1))
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [b](x0) = x0,
    
    [d](x0) = x0,
    
    [c](x0) = x0,
    
    [a](x0) = 9x0
   orientation:
    c(c(c(a(x1)))) = 9x1 >= x1 = d(d(x1))
    
    d(b(x1)) = x1 >= x1 = c(c(x1))
    
    d(x1) = x1 >= x1 = b(b(b(b(x1))))
    
    b(d(x1)) = x1 >= x1 = c(c(x1))
    
    a(c(c(c(x1)))) = 9x1 >= x1 = d(d(x1))
   problem:
    d(b(x1)) -> c(c(x1))
    d(x1) -> b(b(b(b(x1))))
    b(d(x1)) -> c(c(x1))
   Arctic Interpretation Processor:
    dimension: 2
    interpretation:
               [0 0]  
     [b](x0) = [0 0]x0,
     
               [0 3]  
     [d](x0) = [2 0]x0,
     
               [1 1]  
     [c](x0) = [0 0]x0
    orientation:
                [3 3]      [2 2]             
     d(b(x1)) = [2 2]x1 >= [1 1]x1 = c(c(x1))
     
             [0 3]      [0 0]                   
     d(x1) = [2 0]x1 >= [0 0]x1 = b(b(b(b(x1))))
     
                [2 3]      [2 2]             
     b(d(x1)) = [2 3]x1 >= [1 1]x1 = c(c(x1))
    problem:
     d(x1) -> b(b(b(b(x1))))
     b(d(x1)) -> c(c(x1))
    Arctic Interpretation Processor:
     dimension: 2
     interpretation:
                [0 0]  
      [b](x0) = [0 0]x0,
      
                [2 1]  
      [d](x0) = [3 3]x0,
      
                [0 0]  
      [c](x0) = [2 1]x0
     orientation:
              [2 1]      [0 0]                   
      d(x1) = [3 3]x1 >= [0 0]x1 = b(b(b(b(x1))))
      
                 [3 3]      [2 1]             
      b(d(x1)) = [3 3]x1 >= [3 2]x1 = c(c(x1))
     problem:
      b(d(x1)) -> c(c(x1))
     Arctic Interpretation Processor:
      dimension: 3
      interpretation:
                 [0  -& 0 ]  
       [b](x0) = [0  0  0 ]x0
                 [0  1  -&]  ,
       
                 [0  0  3 ]  
       [d](x0) = [-& 0  1 ]x0
                 [1  3  0 ]  ,
       
                 [0  0  0 ]  
       [c](x0) = [-& 0  2 ]x0
                 [-& 0  0 ]  
      orientation:
                  [1 3 3]      [0  0  2 ]             
       b(d(x1)) = [1 3 3]x1 >= [-& 2  2 ]x1 = c(c(x1))
                  [0 1 3]      [-& 0  2 ]             
      problem:
       
      Qed