YES

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

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