YES

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

Proof:
 DP Processor:
  DPs:
   a#(a(x)) -> a#(b(a(x)))
  TRS:
   a(a(x)) -> a(b(a(x)))
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [a#](x0) = 12x0 + 0,
    
    [b](x0) = 0,
    
    [a](x0) = 1x0 + 3
   orientation:
    a#(a(x)) = 13x + 15 >= 12 = a#(b(a(x)))
    
    a(a(x)) = 2x + 4 >= 3 = a(b(a(x)))
   problem:
    DPs:
     
    TRS:
     a(a(x)) -> a(b(a(x)))
   Qed