YES

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

Proof:
 DP Processor:
  DPs:
   b#(b(b(x1))) -> a#(b(x1))
   b#(b(b(x1))) -> b#(a(b(x1)))
   a#(a(x1)) -> b#(x1)
   a#(a(x1)) -> b#(b(x1))
   a#(a(x1)) -> b#(b(b(x1)))
  TRS:
   a(b(a(x1))) -> b(a(x1))
   b(b(b(x1))) -> b(a(b(x1)))
   a(a(x1)) -> b(b(b(x1)))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
    [b#](x0) = [0 0 1]x0,
    
    [a#](x0) = [1 1 0]x0 + [1],
    
              [0 0 0]     [0]
    [b](x0) = [0 0 1]x0 + [0]
              [0 0 1]     [1],
    
              [0 0 1]     [1]
    [a](x0) = [1 0 1]x0 + [1]
              [1 1 0]     [1]
   orientation:
    b#(b(b(x1))) = [0 0 1]x1 + [2] >= [0 0 1]x1 + [1] = a#(b(x1))
    
    b#(b(b(x1))) = [0 0 1]x1 + [2] >= [0 0 1]x1 + [1] = b#(a(b(x1)))
    
    a#(a(x1)) = [1 0 2]x1 + [3] >= [0 0 1]x1 = b#(x1)
    
    a#(a(x1)) = [1 0 2]x1 + [3] >= [0 0 1]x1 + [1] = b#(b(x1))
    
    a#(a(x1)) = [1 0 2]x1 + [3] >= [0 0 1]x1 + [2] = b#(b(b(x1)))
    
                  [1 1 0]     [3]    [0 0 0]     [0]           
    a(b(a(x1))) = [1 1 0]x1 + [3] >= [1 1 0]x1 + [1] = b(a(x1))
                  [1 1 0]     [2]    [1 1 0]     [2]           
    
                  [0 0 0]     [0]    [0 0 0]     [0]              
    b(b(b(x1))) = [0 0 1]x1 + [2] >= [0 0 1]x1 + [1] = b(a(b(x1)))
                  [0 0 1]     [3]    [0 0 1]     [2]              
    
               [1 1 0]     [2]    [0 0 0]     [0]              
    a(a(x1)) = [1 1 1]x1 + [3] >= [0 0 1]x1 + [2] = b(b(b(x1)))
               [1 0 2]     [3]    [0 0 1]     [3]              
   problem:
    DPs:
     
    TRS:
     a(b(a(x1))) -> b(a(x1))
     b(b(b(x1))) -> b(a(b(x1)))
     a(a(x1)) -> b(b(b(x1)))
   Qed