YES

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

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