YES

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

Proof:
 DP Processor:
  DPs:
   a#(b(a(x))) -> a#(b(x))
  TRS:
   a(b(a(x))) -> b(a(b(x)))
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [a#](x0) = 10x0 + 0,
    
    [b](x0) = 1x0 + 0,
    
    [a](x0) = 2x0 + 5
   orientation:
    a#(b(a(x))) = 13x + 16 >= 11x + 10 = a#(b(x))
    
    a(b(a(x))) = 5x + 8 >= 4x + 6 = b(a(b(x)))
   problem:
    DPs:
     
    TRS:
     a(b(a(x))) -> b(a(b(x)))
   Qed