YES

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

Proof:
 String Reversal Processor:
  s(a(x1)) -> a(s(x1))
  s(b(a(b(x1)))) -> a(s(b(a(x1))))
  b(b(a(b(x1)))) -> s(c(x1))
  s(c(x1)) -> b(a(b(a(x1))))
  a(a(b(a(x1)))) -> a(b(a(b(x1))))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [c](x0) = 8x0 + 15,
    
    [b](x0) = 2x0 + 2,
    
    [a](x0) = 2x0 + 2,
    
    [s](x0) = 2x0
   orientation:
    s(a(x1)) = 4x1 + 4 >= 4x1 + 2 = a(s(x1))
    
    s(b(a(b(x1)))) = 16x1 + 28 >= 16x1 + 26 = a(s(b(a(x1))))
    
    b(b(a(b(x1)))) = 16x1 + 30 >= 16x1 + 30 = s(c(x1))
    
    s(c(x1)) = 16x1 + 30 >= 16x1 + 30 = b(a(b(a(x1))))
    
    a(a(b(a(x1)))) = 16x1 + 30 >= 16x1 + 30 = a(b(a(b(x1))))
   problem:
    b(b(a(b(x1)))) -> s(c(x1))
    s(c(x1)) -> b(a(b(a(x1))))
    a(a(b(a(x1)))) -> a(b(a(b(x1))))
   Matrix Interpretation Processor: dim=4
    
    interpretation:
               [1 0 0 1]     [1]
               [0 1 0 0]     [1]
     [c](x0) = [0 0 1 1]x0 + [1]
               [0 0 1 0]     [0],
     
               [1 1 0 0]     [0]
               [0 1 1 1]     [0]
     [b](x0) = [0 0 0 0]x0 + [1]
               [0 0 0 0]     [0],
     
               [1 0 0 1]     [0]
               [0 0 0 0]     [1]
     [a](x0) = [0 0 1 0]x0 + [0]
               [0 1 0 1]     [0],
     
               [1 0 0 1]     [1]
               [0 1 1 0]     [1]
     [s](x0) = [0 0 0 0]x0 + [1]
               [0 0 0 0]     [0]
    orientation:
                      [1 2 1 1]     [3]    [1 0 1 1]     [2]           
                      [0 1 1 1]     [3]    [0 1 1 1]     [3]           
     b(b(a(b(x1)))) = [0 0 0 0]x1 + [1] >= [0 0 0 0]x1 + [1] = s(c(x1))
                      [0 0 0 0]     [0]    [0 0 0 0]     [0]           
     
                [1 0 1 1]     [2]    [1 0 0 1]     [2]                 
                [0 1 1 1]     [3]    [0 1 1 1]     [3]                 
     s(c(x1)) = [0 0 0 0]x1 + [1] >= [0 0 0 0]x1 + [1] = b(a(b(a(x1))))
                [0 0 0 0]     [0]    [0 0 0 0]     [0]                 
     
                      [1 1 1 2]     [2]    [1 1 0 0]     [1]                 
                      [0 0 0 0]     [1]    [0 0 0 0]     [1]                 
     a(a(b(a(x1)))) = [0 0 0 0]x1 + [1] >= [0 0 0 0]x1 + [1] = a(b(a(b(x1))))
                      [0 1 1 1]     [2]    [0 1 1 1]     [2]                 
    problem:
     s(c(x1)) -> b(a(b(a(x1))))
    Arctic Interpretation Processor:
     dimension: 2
     interpretation:
                [0 0]  
      [c](x0) = [3 2]x0,
      
                [0  -&]  
      [b](x0) = [-& -&]x0,
      
                [2 0]  
      [a](x0) = [3 3]x0,
      
                [0  2 ]  
      [s](x0) = [-& -&]x0
     orientation:
                 [5  4 ]      [4  2 ]                   
      s(c(x1)) = [-& -&]x1 >= [-& -&]x1 = b(a(b(a(x1))))
     problem:
      
     Qed