YES

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

Proof:
 String Reversal Processor:
  b(a(x1)) -> d(b(x1))
  c(a(x1)) -> d(d(d(x1)))
  d(b(x1)) -> b(c(a(x1)))
  f(c(x1)) -> c(d(d(x1)))
  d(d(x1)) -> f(x1)
  f(f(x1)) -> a(x1)
  Matrix Interpretation Processor: dim=2
   
   interpretation:
              [1 2]     [2]
    [f](x0) = [0 1]x0 + [0],
    
                
    [c](x0) = x0,
    
              [1 1]     [1]
    [d](x0) = [0 1]x0 + [0],
    
              [1 3]     [3]
    [a](x0) = [0 1]x0 + [0],
    
              [1 0]     [0]
    [b](x0) = [0 3]x0 + [2]
   orientation:
               [1 3]     [3]    [1 3]     [3]           
    b(a(x1)) = [0 3]x1 + [2] >= [0 3]x1 + [2] = d(b(x1))
    
               [1 3]     [3]    [1 3]     [3]              
    c(a(x1)) = [0 1]x1 + [0] >= [0 1]x1 + [0] = d(d(d(x1)))
    
               [1 3]     [3]    [1 3]     [3]              
    d(b(x1)) = [0 3]x1 + [2] >= [0 3]x1 + [2] = b(c(a(x1)))
    
               [1 2]     [2]    [1 2]     [2]              
    f(c(x1)) = [0 1]x1 + [0] >= [0 1]x1 + [0] = c(d(d(x1)))
    
               [1 2]     [2]    [1 2]     [2]        
    d(d(x1)) = [0 1]x1 + [0] >= [0 1]x1 + [0] = f(x1)
    
               [1 4]     [4]    [1 3]     [3]        
    f(f(x1)) = [0 1]x1 + [0] >= [0 1]x1 + [0] = a(x1)
   problem:
    b(a(x1)) -> d(b(x1))
    c(a(x1)) -> d(d(d(x1)))
    d(b(x1)) -> b(c(a(x1)))
    f(c(x1)) -> c(d(d(x1)))
    d(d(x1)) -> f(x1)
   Arctic Interpretation Processor:
    dimension: 2
    interpretation:
                 
     [f](x0) = x0,
     
               [0  -&]  
     [c](x0) = [0  -&]x0,
     
               [0  -&]  
     [d](x0) = [0  0 ]x0,
     
               [0 0]  
     [a](x0) = [1 2]x0,
     
               [0 0]  
     [b](x0) = [0 0]x0
    orientation:
                [1 2]      [0 0]             
     b(a(x1)) = [1 2]x1 >= [0 0]x1 = d(b(x1))
     
                [0 0]      [0  -&]                
     c(a(x1)) = [0 0]x1 >= [0  0 ]x1 = d(d(d(x1)))
     
                [0 0]      [0 0]                
     d(b(x1)) = [0 0]x1 >= [0 0]x1 = b(c(a(x1)))
     
                [0  -&]      [0  -&]                
     f(c(x1)) = [0  -&]x1 >= [0  -&]x1 = c(d(d(x1)))
     
                [0  -&]                
     d(d(x1)) = [0  0 ]x1 >= x1 = f(x1)
    problem:
     c(a(x1)) -> d(d(d(x1)))
     d(b(x1)) -> b(c(a(x1)))
     f(c(x1)) -> c(d(d(x1)))
     d(d(x1)) -> f(x1)
    String Reversal Processor:
     a(c(x1)) -> d(d(d(x1)))
     b(d(x1)) -> a(c(b(x1)))
     c(f(x1)) -> d(d(c(x1)))
     d(d(x1)) -> f(x1)
     Matrix Interpretation Processor: dim=1
      
      interpretation:
       [f](x0) = x0 + 4,
       
       [c](x0) = x0,
       
       [d](x0) = x0 + 2,
       
       [a](x0) = x0 + 7,
       
       [b](x0) = 8x0 + 9
      orientation:
       a(c(x1)) = x1 + 7 >= x1 + 6 = d(d(d(x1)))
       
       b(d(x1)) = 8x1 + 25 >= 8x1 + 16 = a(c(b(x1)))
       
       c(f(x1)) = x1 + 4 >= x1 + 4 = d(d(c(x1)))
       
       d(d(x1)) = x1 + 4 >= x1 + 4 = f(x1)
      problem:
       c(f(x1)) -> d(d(c(x1)))
       d(d(x1)) -> f(x1)
      String Reversal Processor:
       f(c(x1)) -> c(d(d(x1)))
       d(d(x1)) -> f(x1)
       Bounds Processor:
        bound: 1
        enrichment: match
        automaton:
         final states: {5,1}
         transitions:
          d0(2) -> 3*
          d0(3) -> 4*
          f0(2) -> 5*
          f1(6) -> 7*
          f50() -> 2*
          c0(4) -> 1*
          1 -> 7,5,4
          2 -> 6*
          5 -> 3*
          7 -> 4*
        problem:
         
        Qed