YES

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

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