YES

Problem:
 strict:
  b(c(a(x1))) -> d(d(x1))
  b(x1) -> c(c(x1))
  a(a(x1)) -> 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(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 1]x0
              [0 0 0]  ,
    
              [1 2 0]     [1]
    [d](x0) = [0 0 0]x0 + [0]
              [0 0 0]     [2],
    
              [1 2 0]     [1]
    [a](x0) = [0 0 0]x0 + [0]
              [0 0 0]     [3],
    
              [1 2 0]     [0]
    [b](x0) = [0 2 0]x0 + [0]
              [0 0 0]     [2]
   orientation:
                  [1 2 0]     [5]    [1 2 0]     [2]           
    a(c(b(x1))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = d(d(x1))
                  [0 0 0]     [3]    [0 0 0]     [2]           
    
            [1 2 0]     [0]    [1 0 0]             
    b(x1) = [0 2 0]x1 + [0] >= [0 0 0]x1 = c(c(x1))
            [0 0 0]     [2]    [0 0 0]             
    
               [1 2 0]     [2]    [1 2 0]     [1]        
    a(a(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = a(x1)
               [0 0 0]     [3]    [0 0 0]     [3]        
    
               [1 2 0]     [1]    [1 2 0]     [1]        
    b(a(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = d(x1)
               [0 0 0]     [2]    [0 0 0]     [2]        
    
            [1 2 0]     [1]    [1 2 0]     [1]           
    d(x1) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = b(a(x1))
            [0 0 0]     [2]    [0 0 0]     [2]           
   problem:
    strict:
     b(x1) -> c(c(x1))
    weak:
     b(a(x1)) -> d(x1)
     d(x1) -> b(a(x1))
   String Reversal Processor:
    strict:
     b(x1) -> c(c(x1))
    weak:
     a(b(x1)) -> d(x1)
     d(x1) -> a(b(x1))
    Arctic Interpretation Processor:
     dimension: 1
     interpretation:
      [c](x0) = x0,
      
      [d](x0) = 8x0,
      
      [a](x0) = x0,
      
      [b](x0) = 8x0
     orientation:
      b(x1) = 8x1 >= x1 = c(c(x1))
      
      a(b(x1)) = 8x1 >= 8x1 = d(x1)
      
      d(x1) = 8x1 >= 8x1 = a(b(x1))
     problem:
      strict:
       
      weak:
       a(b(x1)) -> d(x1)
       d(x1) -> a(b(x1))
     Qed