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:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [d](x0) = x0 + 4,
   
   [b](x0) = x0 + 9,
   
   [c](x0) = x0 + 6,
   
   [f](x0) = x0 + 14,
   
   [a](x0) = x0
  orientation:
   f(f(x1)) = x1 + 28 >= x1 + 27 = b(b(b(x1)))
   
   a(f(x1)) = x1 + 14 >= x1 + 14 = f(a(a(x1)))
   
   b(b(x1)) = x1 + 18 >= x1 + 18 = c(c(a(c(x1))))
   
   d(b(x1)) = x1 + 13 >= x1 + 13 = d(a(b(x1)))
   
   c(c(x1)) = x1 + 12 >= x1 + 12 = d(d(d(x1)))
   
   b(d(x1)) = x1 + 13 >= x1 + 13 = d(b(x1))
   
   c(d(d(x1))) = x1 + 14 >= x1 + 14 = 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)
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [d](x0) = x0 + 1,
    
    [b](x0) = 4x0 + 2,
    
    [c](x0) = 2x0 + 1,
    
    [f](x0) = 2x0 + 4,
    
    [a](x0) = x0
   orientation:
    a(f(x1)) = 2x1 + 4 >= 2x1 + 4 = f(a(a(x1)))
    
    b(b(x1)) = 16x1 + 10 >= 8x1 + 7 = c(c(a(c(x1))))
    
    d(b(x1)) = 4x1 + 3 >= 4x1 + 3 = d(a(b(x1)))
    
    c(c(x1)) = 4x1 + 3 >= x1 + 3 = d(d(d(x1)))
    
    b(d(x1)) = 4x1 + 6 >= 4x1 + 3 = d(b(x1))
    
    c(d(d(x1))) = 2x1 + 5 >= 2x1 + 4 = f(x1)
   problem:
    a(f(x1)) -> f(a(a(x1)))
    d(b(x1)) -> d(a(b(x1)))
    c(c(x1)) -> d(d(d(x1)))
   String Reversal Processor:
    f(a(x1)) -> a(a(f(x1)))
    b(d(x1)) -> b(a(d(x1)))
    c(c(x1)) -> d(d(d(x1)))
    Bounds Processor:
     bound: 0
     enrichment: match
     automaton:
      final states: {8,5,1}
      transitions:
       f50() -> 2*
       f0(2) -> 3*
       a0(6) -> 7*
       a0(4) -> 1*
       a0(3) -> 4*
       b0(7) -> 5*
       d0(2) -> 6*
       d0(6) -> 9*
       d0(9) -> 8*
       1 -> 3*
     problem:
      
     Qed