YES

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

Proof:
 DP Processor:
  DPs:
   a#(b(a(x1))) -> b#(b(a(x1)))
   a#(b(a(x1))) -> a#(b(b(a(x1))))
  TRS:
   a(b(a(x1))) -> a(b(b(a(x1))))
   b(b(b(x1))) -> b(b(x1))
  Matrix Interpretation Processor: dim=4
   
   interpretation:
    [b#](x0) = [0],
    
    [a#](x0) = [1 0 1 0]x0,
    
              [0 0 0 1]  
              [0 0 1 0]  
    [b](x0) = [0 0 1 0]x0
              [0 0 0 0]  ,
    
              [0 0 1 0]     [0]
              [0 1 0 0]     [0]
    [a](x0) = [1 0 0 1]x0 + [0]
              [0 0 0 0]     [1]
   orientation:
    a#(b(a(x1))) = [1 0 0 1]x1 + [1] >= [0] = b#(b(a(x1)))
    
    a#(b(a(x1))) = [1 0 0 1]x1 + [1] >= [1 0 0 1]x1 = a#(b(b(a(x1))))
    
                  [1 0 0 1]     [0]    [1 0 0 1]     [0]                 
                  [1 0 0 1]     [0]    [1 0 0 1]     [0]                 
    a(b(a(x1))) = [0 0 0 0]x1 + [1] >= [0 0 0 0]x1 + [0] = a(b(b(a(x1))))
                  [0 0 0 0]     [1]    [0 0 0 0]     [1]                 
    
                  [0 0 0 0]      [0 0 0 0]             
                  [0 0 1 0]      [0 0 1 0]             
    b(b(b(x1))) = [0 0 1 0]x1 >= [0 0 1 0]x1 = b(b(x1))
                  [0 0 0 0]      [0 0 0 0]             
   problem:
    DPs:
     
    TRS:
     a(b(a(x1))) -> a(b(b(a(x1))))
     b(b(b(x1))) -> b(b(x1))
   Qed