YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 2
  interpretation:
             [0 1]  
   [b](x0) = [1 0]x0,
   
             [1  0 ]  
   [a](x0) = [0  -&]x0
  orientation:
              [2 1]      [2 1]                
   a(a(x1)) = [1 0]x1 >= [1 0]x1 = a(b(a(x1)))
   
              [2 1]      [2 1]             
   b(b(x1)) = [1 2]x1 >= [1 0]x1 = a(a(x1))
   
                    [4 3]             
   a(b(b(a(x1)))) = [3 2]x1 >= x1 = x1
  problem:
   a(a(x1)) -> a(b(a(x1)))
   b(b(x1)) -> a(a(x1))
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
              [0 0]  
    [b](x0) = [2 3]x0,
    
              [0  -&]  
    [a](x0) = [0  -&]x0
   orientation:
               [0  -&]      [0  -&]                
    a(a(x1)) = [0  -&]x1 >= [0  -&]x1 = a(b(a(x1)))
    
               [2 3]      [0  -&]             
    b(b(x1)) = [5 6]x1 >= [0  -&]x1 = a(a(x1))
   problem:
    a(a(x1)) -> a(b(a(x1)))
   Arctic Interpretation Processor:
    dimension: 3
    interpretation:
               [0  -& -&]  
     [b](x0) = [-& 0  -&]x0
               [0  0  -&]  ,
     
               [0  0  0 ]  
     [a](x0) = [-& 0  0 ]x0
               [1  1  1 ]  
    orientation:
                [1 1 1]      [0 0 0]                
     a(a(x1)) = [1 1 1]x1 >= [0 0 0]x1 = a(b(a(x1)))
                [2 2 2]      [1 1 1]                
    problem:
     
    Qed