YES

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

Proof:
 String Reversal Processor:
  strict:
   a(c(b(x1))) -> d(d(x1))
   b(x1) -> c(c(x1))
   a(a(x1)) -> a(b(c(a(x1))))
  weak:
   b(a(x1)) -> d(x1)
   d(x1) -> b(a(x1))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 0 0]  
    [c](x0) = [0 0 2]x0
              [0 0 0]  ,
    
              [2 2 0]     [1]
    [d](x0) = [0 0 0]x0 + [1]
              [2 2 0]     [3],
    
              [2 2 0]     [0]
    [a](x0) = [0 0 0]x0 + [2]
              [0 2 0]     [3],
    
              [1 0 0]     [1]
    [b](x0) = [0 0 0]x0 + [1]
              [1 1 0]     [1]
   orientation:
                  [6 4 0]     [6]    [4 4 0]     [5]           
    a(c(b(x1))) = [0 0 0]x1 + [2] >= [0 0 0]x1 + [1] = d(d(x1))
                  [4 4 0]     [7]    [4 4 0]     [7]           
    
            [1 0 0]     [1]    [1 0 0]             
    b(x1) = [0 0 0]x1 + [1] >= [0 0 0]x1 = c(c(x1))
            [1 1 0]     [1]    [0 0 0]             
    
               [4 4 0]     [4]    [4 4 0]     [4]                 
    a(a(x1)) = [0 0 0]x1 + [2] >= [0 0 0]x1 + [2] = a(b(c(a(x1))))
               [0 0 0]     [7]    [0 0 0]     [5]                 
    
               [2 2 0]     [1]    [2 2 0]     [1]        
    b(a(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = d(x1)
               [2 2 0]     [3]    [2 2 0]     [3]        
    
            [2 2 0]     [1]    [2 2 0]     [1]           
    d(x1) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = b(a(x1))
            [2 2 0]     [3]    [2 2 0]     [3]           
   problem:
    strict:
     a(a(x1)) -> a(b(c(a(x1))))
    weak:
     b(a(x1)) -> d(x1)
     d(x1) -> b(a(x1))
   Matrix Interpretation Processor: dim=2
    
    interpretation:
                    [1]
     [c](x0) = x0 + [0],
     
               [1 1]  
     [d](x0) = [0 0]x0,
     
               [1 1]     [0]
     [a](x0) = [0 0]x0 + [2],
     
               [1 0]  
     [b](x0) = [0 0]x0
    orientation:
                [1 1]     [2]    [1 1]     [1]                 
     a(a(x1)) = [0 0]x1 + [2] >= [0 0]x1 + [2] = a(b(c(a(x1))))
     
                [1 1]      [1 1]          
     b(a(x1)) = [0 0]x1 >= [0 0]x1 = d(x1)
     
             [1 1]      [1 1]             
     d(x1) = [0 0]x1 >= [0 0]x1 = b(a(x1))
    problem:
     strict:
      
     weak:
      b(a(x1)) -> d(x1)
      d(x1) -> b(a(x1))
    Qed