YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 1
  interpretation:
   [d](x0) = 4x0,
   
   [c](x0) = 6x0,
   
   [a](x0) = x0,
   
   [b](x0) = 9x0,
   
   [f](x0) = 14x0
  orientation:
   f(f(x1)) = 28x1 >= 27x1 = b(b(b(x1)))
   
   a(f(x1)) = 14x1 >= 14x1 = f(a(a(x1)))
   
   b(b(x1)) = 18x1 >= 18x1 = c(c(a(c(x1))))
   
   d(b(x1)) = 13x1 >= 13x1 = d(a(b(x1)))
   
   c(c(x1)) = 12x1 >= 12x1 = d(d(d(x1)))
   
   b(d(x1)) = 13x1 >= 13x1 = d(b(x1))
   
   c(d(d(x1))) = 14x1 >= 14x1 = f(x1)
  problem:
   a(f(x1)) -> f(a(a(x1)))
   b(b(x1)) -> c(c(a(c(x1))))
   d(b(x1)) -> d(a(b(x1)))
   c(c(x1)) -> d(d(d(x1)))
   b(d(x1)) -> d(b(x1))
   c(d(d(x1))) -> f(x1)
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [d](x0) = x0,
    
    [c](x0) = 8x0,
    
    [a](x0) = x0,
    
    [b](x0) = 13x0,
    
    [f](x0) = x0
   orientation:
    a(f(x1)) = x1 >= x1 = f(a(a(x1)))
    
    b(b(x1)) = 26x1 >= 24x1 = c(c(a(c(x1))))
    
    d(b(x1)) = 13x1 >= 13x1 = d(a(b(x1)))
    
    c(c(x1)) = 16x1 >= x1 = d(d(d(x1)))
    
    b(d(x1)) = 13x1 >= 13x1 = d(b(x1))
    
    c(d(d(x1))) = 8x1 >= x1 = f(x1)
   problem:
    a(f(x1)) -> f(a(a(x1)))
    d(b(x1)) -> d(a(b(x1)))
    b(d(x1)) -> d(b(x1))
   Arctic Interpretation Processor:
    dimension: 2
    interpretation:
               [0 1]  
     [d](x0) = [1 2]x0,
     
               [0  0 ]  
     [a](x0) = [-& -&]x0,
     
               [0 1]  
     [b](x0) = [0 1]x0,
     
               [0  0 ]  
     [f](x0) = [-& 0 ]x0
    orientation:
                [0  0 ]      [0  0 ]                
     a(f(x1)) = [-& -&]x1 >= [-& -&]x1 = f(a(a(x1)))
     
                [1 2]      [0 1]                
     d(b(x1)) = [2 3]x1 >= [1 2]x1 = d(a(b(x1)))
     
                [2 3]      [1 2]             
     b(d(x1)) = [2 3]x1 >= [2 3]x1 = d(b(x1))
    problem:
     a(f(x1)) -> f(a(a(x1)))
     b(d(x1)) -> d(b(x1))
    KBO Processor:
     weight function:
      w0 = 1
      w(d) = w(b) = w(f) = 1
      w(a) = 0
     precedence:
      a > b > d ~ f
     problem:
      
     Qed