YES

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

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