YES

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

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