YES

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

Proof:
 DP Processor:
  DPs:
   a#(b(a(b(x1)))) -> a#(x1)
   a#(b(a(b(x1)))) -> a#(a(x1))
   a#(b(a(b(x1)))) -> a#(b(b(a(a(x1)))))
  TRS:
   a(a(x1)) -> x1
   a(b(a(b(x1)))) -> b(a(b(b(a(a(x1))))))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
    [a#](x0) = [0 0 2]x0 + [1],
    
              [0 1 1]     [1]
    [b](x0) = [0 0 0]x0 + [0]
              [0 2 0]     [0],
    
              [0 1 1]  
    [a](x0) = [1 0 0]x0
              [0 2 2]  
   orientation:
    a#(b(a(b(x1)))) = [0 4 4]x1 + [5] >= [0 0 2]x1 + [1] = a#(x1)
    
    a#(b(a(b(x1)))) = [0 4 4]x1 + [5] >= [0 4 4]x1 + [1] = a#(a(x1))
    
    a#(b(a(b(x1)))) = [0 4 4]x1 + [5] >= [1] = a#(b(b(a(a(x1)))))
    
               [1 2 2]             
    a(a(x1)) = [0 1 1]x1 >= x1 = x1
               [2 4 4]             
    
                     [0 2 2]     [2]    [0 2 2]     [2]                       
    a(b(a(b(x1)))) = [0 5 1]x1 + [2] >= [0 0 0]x1 + [0] = b(a(b(b(a(a(x1))))))
                     [0 4 4]     [4]    [0 4 4]     [2]                       
   problem:
    DPs:
     
    TRS:
     a(a(x1)) -> x1
     a(b(a(b(x1)))) -> b(a(b(b(a(a(x1))))))
   Qed