YES

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

Proof:
 DP Processor:
  DPs:
   a#(b(b(x1))) -> a#(x1)
   a#(b(b(x1))) -> a#(a(x1))
  TRS:
   a(b(b(x1))) -> b(b(a(a(x1))))
   a(b(a(x1))) -> b(b(x1))
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
    [a#](x0) = [-& 0 ]x0 + [0],
    
              [0  2 ]     [2]
    [a](x0) = [-& 0 ]x0 + [0],
    
              [-& 2 ]     [2]
    [b](x0) = [3  -&]x0 + [2]
   orientation:
    a#(b(b(x1))) = [-& 5 ]x1 + [5] >= [-& 0 ]x1 + [0] = a#(x1)
    
    a#(b(b(x1))) = [-& 5 ]x1 + [5] >= [-& 0 ]x1 + [0] = a#(a(x1))
    
                  [5  7 ]     [7]    [5  7 ]     [7]                 
    a(b(b(x1))) = [-& 5 ]x1 + [5] >= [-& 5 ]x1 + [5] = b(b(a(a(x1))))
    
                  [5 7]     [7]    [5  -&]     [4]           
    a(b(a(x1))) = [3 5]x1 + [5] >= [-& 5 ]x1 + [5] = b(b(x1))
   problem:
    DPs:
     
    TRS:
     a(b(b(x1))) -> b(b(a(a(x1))))
     a(b(a(x1))) -> b(b(x1))
   Qed