YES

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

Proof:
 DP Processor:
  DPs:
   a#(b(b(x1))) -> a#(x1)
   a#(a(x1)) -> b#(x1)
   a#(a(x1)) -> b#(b(x1))
   a#(a(x1)) -> b#(b(b(x1)))
   b#(b(a(x1))) -> a#(b(a(x1)))
  TRS:
   a(b(b(x1))) -> a(x1)
   a(a(x1)) -> b(b(b(x1)))
   b(b(a(x1))) -> a(b(a(x1)))
  Arctic Interpretation Processor:
   dimension: 2
   usable rules:
    a(b(b(x1))) -> a(x1)
    a(a(x1)) -> b(b(b(x1)))
    b(b(a(x1))) -> a(b(a(x1)))
   interpretation:
    [b#](x0) = [-3 1 ]x0 + [2],
    
    [a#](x0) = [3 0]x0 + [-4],
    
              [3  0 ]     [3 ]
    [a](x0) = [-1 -&]x0 + [-&],
    
              [-& 1 ]     [0]
    [b](x0) = [0  -&]x0 + [0]
   orientation:
    a#(b(b(x1))) = [4 1]x1 + [4] >= [3 0]x1 + [-4] = a#(x1)
    
    a#(a(x1)) = [6 3]x1 + [6] >= [-3 1 ]x1 + [2] = b#(x1)
    
    a#(a(x1)) = [6 3]x1 + [6] >= [1  -2]x1 + [2] = b#(b(x1))
    
    a#(a(x1)) = [6 3]x1 + [6] >= [-2 2 ]x1 + [2] = b#(b(b(x1)))
    
    b#(b(a(x1))) = [4 1]x1 + [4] >= [3 0]x1 + [3] = a#(b(a(x1)))
    
                  [4  1 ]     [4]    [3  0 ]     [3 ]        
    a(b(b(x1))) = [0  -&]x1 + [0] >= [-1 -&]x1 + [-&] = a(x1)
    
               [6  3 ]     [6]    [-& 2 ]     [1]              
    a(a(x1)) = [2  -1]x1 + [2] >= [1  -&]x1 + [1] = b(b(b(x1)))
    
                  [4  1 ]     [4]    [3  0 ]     [3 ]              
    b(b(a(x1))) = [0  -&]x1 + [0] >= [-1 -&]x1 + [-1] = a(b(a(x1)))
   problem:
    DPs:
     
    TRS:
     a(b(b(x1))) -> a(x1)
     a(a(x1)) -> b(b(b(x1)))
     b(b(a(x1))) -> a(b(a(x1)))
   Qed