YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 2
  interpretation:
             [2 0]  
   [a](x0) = [2 0]x0,
   
             [0 0]  
   [b](x0) = [2 1]x0
  orientation:
                 [3 2]      [2 0]          
   b(b(b(x1))) = [4 3]x1 >= [2 0]x1 = a(x1)
   
              [4 2]      [4 2]                
   a(a(x1)) = [4 2]x1 >= [4 2]x1 = a(b(a(x1)))
   
                 [6 4]      [4 2]                
   a(a(a(x1))) = [6 4]x1 >= [6 4]x1 = b(a(a(x1)))
  problem:
   a(a(x1)) -> a(b(a(x1)))
   a(a(a(x1))) -> b(a(a(x1)))
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
              [0 1]  
    [a](x0) = [3 2]x0,
    
              [0  -&]  
    [b](x0) = [0  0 ]x0
   orientation:
               [4 3]      [4 3]                
    a(a(x1)) = [5 4]x1 >= [5 4]x1 = a(b(a(x1)))
    
                  [6 5]      [4 3]                
    a(a(a(x1))) = [7 6]x1 >= [5 4]x1 = b(a(a(x1)))
   problem:
    a(a(x1)) -> a(b(a(x1)))
   Arctic Interpretation Processor:
    dimension: 3
    interpretation:
               [0  -& 0 ]  
     [a](x0) = [-& 0  2 ]x0
               [1  0  1 ]  ,
     
               [0  -& -&]  
     [b](x0) = [0  -& -&]x0
               [0  -& -&]  
    orientation:
                [1 0 1]      [0  -& 0 ]                
     a(a(x1)) = [3 2 3]x1 >= [2  -& 2 ]x1 = a(b(a(x1)))
                [2 1 2]      [1  -& 1 ]                
    problem:
     
    Qed