YES

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

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
                  [1]
   [f](x0) = x0 + [0]
                  [0],
   
             [1 0 0]  
   [c](x0) = [0 0 0]x0
             [0 0 0]  ,
   
             [1 0 0]     [0]
   [b](x0) = [0 0 0]x0 + [1]
             [0 1 1]     [0],
   
             [1 0 0]     [1]
   [a](x0) = [0 0 0]x0 + [1]
             [0 1 1]     [3],
   
             [1 1 2]     [0]
   [d](x0) = [0 0 0]x0 + [1]
             [1 3 3]     [0]
  orientation:
              [1 1 2]     [1]    [1 1 2]     [1]                    
   a(d(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = d(b(c(b(d(x1)))))
              [1 3 3]     [4]    [1 1 2]     [3]                    
   
           [1 0 0]     [1]    [1 0 0]     [1]                    
   a(x1) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = b(b(f(b(b(x1)))))
           [0 1 1]     [3]    [0 1 1]     [3]                    
   
                 [1 2 2]     [1]    [1 1 2]     [1]           
   b(d(b(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = a(d(x1))
                 [1 3 3]     [4]    [1 3 3]     [4]           
   
              [1 1 2]     [1]    [1 1 2]     [0]           
   d(f(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = b(d(x1))
              [1 3 3]     [1]    [1 3 3]     [1]           
  problem:
   a(d(x1)) -> d(b(c(b(d(x1)))))
   a(x1) -> b(b(f(b(b(x1)))))
   b(d(b(x1))) -> a(d(x1))
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
              [0  -&]  
    [f](x0) = [-& -&]x0,
    
              [0  -&]  
    [c](x0) = [-& -&]x0,
    
              [0 0]  
    [b](x0) = [0 1]x0,
    
              [0 1]  
    [a](x0) = [1 2]x0,
    
              [0 0]  
    [d](x0) = [0 1]x0
   orientation:
               [1 2]      [0 1]                      
    a(d(x1)) = [2 3]x1 >= [1 2]x1 = d(b(c(b(d(x1)))))
    
            [0 1]      [0 1]                      
    a(x1) = [1 2]x1 >= [1 2]x1 = b(b(f(b(b(x1)))))
    
                  [1 2]      [1 2]             
    b(d(b(x1))) = [2 3]x1 >= [2 3]x1 = a(d(x1))
   problem:
    a(x1) -> b(b(f(b(b(x1)))))
    b(d(b(x1))) -> a(d(x1))
   String Reversal Processor:
    a(x1) -> b(b(f(b(b(x1)))))
    b(d(b(x1))) -> d(a(x1))
    Bounds Processor:
     bound: 4
     enrichment: match
     automaton:
      final states: {5}
      transitions:
       b3(65) -> 66*
       b3(62) -> 63*
       b3(64) -> 65*
       b3(61) -> 62*
       f3(63) -> 64*
       d3(68) -> 69*
       a3(67) -> 68*
       a3(83) -> 84*
       b4(85) -> 86*
       b4(80) -> 81*
       b4(77) -> 78*
       b4(81) -> 82*
       b4(78) -> 79*
       f4(79) -> 80*
       a0(5) -> 5*
       b0(5) -> 5*
       f0(5) -> 5*
       d0(5) -> 5*
       d1(19) -> 20*
       a1(31) -> 32*
       a1(18) -> 19*
       b1(15) -> 16*
       b1(12) -> 13*
       b1(16) -> 17*
       b1(13) -> 14*
       f1(14) -> 15*
       b2(25) -> 26*
       b2(29) -> 30*
       b2(26) -> 27*
       b2(38) -> 39*
       b2(28) -> 29*
       f2(27) -> 28*
       d2(45) -> 46*
       a2(44) -> 45*
       5 -> 18,12
       16 -> 31*
       17 -> 5*
       18 -> 25*
       20 -> 26,13,5
       29 -> 44*
       30 -> 32,19
       31 -> 38*
       32 -> 19*
       39 -> 26*
       44 -> 61*
       46 -> 27,26,14,5,13
       65 -> 67*
       66 -> 45*
       67 -> 77*
       69 -> 26,13,5,18,12,25,14,27
       81 -> 83*
       82 -> 84,68
       83 -> 85*
       84 -> 68*
       86 -> 78*
     problem:
      
     Qed