YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 1
  interpretation:
   [f](x0) = 10x0,
   
   [d](x0) = 8x0,
   
   [b](x0) = 10x0,
   
   [c](x0) = 12x0,
   
   [a](x0) = 12x0
  orientation:
   a(a(x1)) = 24x1 >= 22x1 = b(c(x1))
   
   b(b(x1)) = 20x1 >= 20x1 = c(d(x1))
   
   c(c(x1)) = 24x1 >= 24x1 = d(d(d(x1)))
   
   d(c(x1)) = 20x1 >= 20x1 = b(f(x1))
   
   d(d(d(x1))) = 24x1 >= 24x1 = a(c(x1))
   
   f(f(x1)) = 20x1 >= 20x1 = f(b(x1))
  problem:
   b(b(x1)) -> c(d(x1))
   c(c(x1)) -> d(d(d(x1)))
   d(c(x1)) -> b(f(x1))
   d(d(d(x1))) -> a(c(x1))
   f(f(x1)) -> f(b(x1))
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
              [1 0]  
    [f](x0) = [1 0]x0,
    
              [0 0]  
    [d](x0) = [1 0]x0,
    
              [1 0]  
    [b](x0) = [1 1]x0,
    
              [0 0]  
    [c](x0) = [2 1]x0,
    
              [0  -&]  
    [a](x0) = [0  -&]x0
   orientation:
               [2 1]      [1 0]             
    b(b(x1)) = [2 2]x1 >= [2 2]x1 = c(d(x1))
    
               [2 1]      [1 1]                
    c(c(x1)) = [3 2]x1 >= [2 1]x1 = d(d(d(x1)))
    
               [2 1]      [2 1]             
    d(c(x1)) = [2 1]x1 >= [2 1]x1 = b(f(x1))
    
                  [1 1]      [0 0]             
    d(d(d(x1))) = [2 1]x1 >= [0 0]x1 = a(c(x1))
    
               [2 1]      [2 1]             
    f(f(x1)) = [2 1]x1 >= [2 1]x1 = f(b(x1))
   problem:
    b(b(x1)) -> c(d(x1))
    c(c(x1)) -> d(d(d(x1)))
    d(c(x1)) -> b(f(x1))
    f(f(x1)) -> f(b(x1))
   Arctic Interpretation Processor:
    dimension: 2
    interpretation:
               [0 0]  
     [f](x0) = [1 1]x0,
     
               [0  -&]  
     [d](x0) = [0  -&]x0,
     
               [0  -&]  
     [b](x0) = [0  -&]x0,
     
               [0 0]  
     [c](x0) = [0 0]x0
    orientation:
                [0  -&]      [0  -&]             
     b(b(x1)) = [0  -&]x1 >= [0  -&]x1 = c(d(x1))
     
                [0 0]      [0  -&]                
     c(c(x1)) = [0 0]x1 >= [0  -&]x1 = d(d(d(x1)))
     
                [0 0]      [0 0]             
     d(c(x1)) = [0 0]x1 >= [0 0]x1 = b(f(x1))
     
                [1 1]      [0  -&]             
     f(f(x1)) = [2 2]x1 >= [1  -&]x1 = f(b(x1))
    problem:
     b(b(x1)) -> c(d(x1))
     c(c(x1)) -> d(d(d(x1)))
     d(c(x1)) -> b(f(x1))
    Arctic Interpretation Processor:
     dimension: 2
     interpretation:
                [0  -&]  
      [f](x0) = [0  -&]x0,
      
                [0  -&]  
      [d](x0) = [0  0 ]x0,
      
                [0 1]  
      [b](x0) = [0 1]x0,
      
                [1 0]  
      [c](x0) = [0 1]x0
     orientation:
                 [1 2]      [1 0]             
      b(b(x1)) = [1 2]x1 >= [1 1]x1 = c(d(x1))
      
                 [2 1]      [0  -&]                
      c(c(x1)) = [1 2]x1 >= [0  0 ]x1 = d(d(d(x1)))
      
                 [1 0]      [1  -&]             
      d(c(x1)) = [1 1]x1 >= [1  -&]x1 = b(f(x1))
     problem:
      b(b(x1)) -> c(d(x1))
      d(c(x1)) -> b(f(x1))
     Arctic Interpretation Processor:
      dimension: 2
      interpretation:
                 [2 0]  
       [f](x0) = [0 0]x0,
       
                 [2  -&]  
       [d](x0) = [0  3 ]x0,
       
                 [0 1]  
       [b](x0) = [2 3]x0,
       
                 [1 1]  
       [c](x0) = [3 3]x0
      orientation:
                  [3 4]      [3 4]             
       b(b(x1)) = [5 6]x1 >= [5 6]x1 = c(d(x1))
       
                  [3 3]      [2 1]             
       d(c(x1)) = [6 6]x1 >= [4 3]x1 = b(f(x1))
      problem:
       b(b(x1)) -> c(d(x1))
      Arctic Interpretation Processor:
       dimension: 3
       interpretation:
                  [0  -& 0 ]  
        [d](x0) = [0  -& 2 ]x0
                  [0  -& 2 ]  ,
        
                  [0  0  2 ]  
        [b](x0) = [-& 0  2 ]x0
                  [1  -& 1 ]  ,
        
                  [0  0  0 ]  
        [c](x0) = [-& 0  0 ]x0
                  [0  0  0 ]  
       orientation:
                   [3 0 3]      [0  -& 2 ]             
        b(b(x1)) = [3 0 3]x1 >= [0  -& 2 ]x1 = c(d(x1))
                   [2 1 3]      [0  -& 2 ]             
       problem:
        
       Qed