YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 2
  interpretation:
             [0 0]  
   [d](x0) = [2 2]x0,
   
             [0  0 ]  
   [b](x0) = [0  -&]x0,
   
             [0  -&]  
   [c](x0) = [0  -&]x0,
   
             [0  -&]  
   [a](x0) = [0  -&]x0
  orientation:
                 [0  -&]      [0  -&]                
   a(c(a(x1))) = [0  -&]x1 >= [0  -&]x1 = c(a(c(x1)))
   
                 [0 0]      [0 0]                
   a(a(b(x1))) = [0 0]x1 >= [0 0]x1 = a(d(b(x1)))
   
              [0 0]      [0  -&]                
   a(b(x1)) = [0 0]x1 >= [0  -&]x1 = b(a(a(x1)))
   
              [2 2]      [0 0]                
   d(d(x1)) = [4 4]x1 >= [0 0]x1 = a(d(b(x1)))
   
              [0 0]      [0  -&]             
   b(b(x1)) = [0 0]x1 >= [0  -&]x1 = b(c(x1))
   
                 [0  -&]      [0  -&]             
   a(d(c(x1))) = [0  -&]x1 >= [0  -&]x1 = c(a(x1))
   
              [0  -&]      [0  -&]                
   b(c(x1)) = [0  -&]x1 >= [0  -&]x1 = a(a(a(x1)))
  problem:
   a(c(a(x1))) -> c(a(c(x1)))
   a(a(b(x1))) -> a(d(b(x1)))
   a(b(x1)) -> b(a(a(x1)))
   b(b(x1)) -> b(c(x1))
   a(d(c(x1))) -> c(a(x1))
   b(c(x1)) -> a(a(a(x1)))
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [d](x0) = x0,
    
    [b](x0) = 2x0,
    
    [c](x0) = x0,
    
    [a](x0) = x0
   orientation:
    a(c(a(x1))) = x1 >= x1 = c(a(c(x1)))
    
    a(a(b(x1))) = 2x1 >= 2x1 = a(d(b(x1)))
    
    a(b(x1)) = 2x1 >= 2x1 = b(a(a(x1)))
    
    b(b(x1)) = 4x1 >= 2x1 = b(c(x1))
    
    a(d(c(x1))) = x1 >= x1 = c(a(x1))
    
    b(c(x1)) = 2x1 >= x1 = a(a(a(x1)))
   problem:
    a(c(a(x1))) -> c(a(c(x1)))
    a(a(b(x1))) -> a(d(b(x1)))
    a(b(x1)) -> b(a(a(x1)))
    a(d(c(x1))) -> c(a(x1))
   String Reversal Processor:
    a(c(a(x1))) -> c(a(c(x1)))
    b(a(a(x1))) -> b(d(a(x1)))
    b(a(x1)) -> a(a(b(x1)))
    c(d(a(x1))) -> a(c(x1))
    Bounds Processor:
     bound: 1
     enrichment: match
     automaton:
      final states: {4,8,5,1}
      transitions:
       c1(15) -> 16*
       c1(17) -> 18*
       c1(21) -> 22*
       a1(16) -> 17*
       f40() -> 2*
       c0(2) -> 3*
       c0(4) -> 1*
       a0(10) -> 8*
       a0(2) -> 6*
       a0(9) -> 10*
       a0(3) -> 4*
       b0(7) -> 5*
       b0(2) -> 9*
       d0(6) -> 7*
       1 -> 6,4
       3 -> 15*
       4 -> 3*
       5 -> 9*
       8 -> 9*
       16 -> 21*
       18 -> 17,4,3
       22 -> 16*
     problem:
      
     Qed