YES

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

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