YES

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

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