YES

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

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