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:
 DP Processor:
  DPs:
   b#(b(b(x1))) -> a#(b(x1))
   a#(a(x1)) -> b#(a(x1))
   a#(a(x1)) -> a#(b(a(x1)))
   a#(a(a(x1))) -> b#(x1)
   a#(a(a(x1))) -> b#(b(x1))
   a#(a(a(x1))) -> a#(b(b(x1)))
  TRS:
   b(b(b(x1))) -> a(b(x1))
   a(a(x1)) -> a(b(a(x1)))
   a(a(a(x1))) -> a(b(b(x1)))
  Arctic Interpretation Processor:
   dimension: 2
   usable rules:
    b(b(b(x1))) -> a(b(x1))
    a(a(x1)) -> a(b(a(x1)))
    a(a(a(x1))) -> a(b(b(x1)))
   interpretation:
    [a#](x0) = [0  -2]x0 + [0],
    
    [b#](x0) = [-& 0 ]x0 + [0],
    
              [2  0 ]     [2 ]
    [a](x0) = [-2 -4]x0 + [-4],
    
              [-1 2 ]     [0]
    [b](x0) = [1  -&]x0 + [1]
   orientation:
    b#(b(b(x1))) = [0 3]x1 + [1] >= [-1 2 ]x1 + [0] = a#(b(x1))
    
    a#(a(x1)) = [2 0]x1 + [2] >= [-2 -4]x1 + [0] = b#(a(x1))
    
    a#(a(x1)) = [2 0]x1 + [2] >= [1  -1]x1 + [1] = a#(b(a(x1)))
    
    a#(a(a(x1))) = [4 2]x1 + [4] >= [-& 0 ]x1 + [0] = b#(x1)
    
    a#(a(a(x1))) = [4 2]x1 + [4] >= [1  -&]x1 + [1] = b#(b(x1))
    
    a#(a(a(x1))) = [4 2]x1 + [4] >= [3 1]x1 + [3] = a#(b(b(x1)))
    
                  [2 5]     [3]    [1  4 ]     [2 ]           
    b(b(b(x1))) = [4 2]x1 + [4] >= [-3 0 ]x1 + [-2] = a(b(x1))
    
               [4  2 ]     [4]    [3  1 ]     [3 ]              
    a(a(x1)) = [0  -2]x1 + [0] >= [-1 -3]x1 + [-1] = a(b(a(x1)))
    
                  [6 4]     [6]    [5  3 ]     [5]              
    a(a(a(x1))) = [2 0]x1 + [2] >= [1  -1]x1 + [1] = a(b(b(x1)))
   problem:
    DPs:
     
    TRS:
     b(b(b(x1))) -> a(b(x1))
     a(a(x1)) -> a(b(a(x1)))
     a(a(a(x1))) -> a(b(b(x1)))
   Qed