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)))
  Restore Modifier:
   DPs:
    a#(a(x)) -> a#(b(a(x)))
   TRS:
    a(a(x)) -> a(b(a(x)))
   SCC Processor:
    #sccs: 1
    #rules: 1
    #arcs: 1/1
    DPs:
     a#(a(x)) -> a#(b(a(x)))
    TRS:
     a(a(x)) -> a(b(a(x)))
    Matrix Interpretation Processor:
     dimension: 1
     interpretation:
      [a#](x0) = x0,
      
      [b](x0) = 0,
      
      [a](x0) = 1
     orientation:
      a#(a(x)) = 1 >= 0 = a#(b(a(x)))
      
      a(a(x)) = 1 >= 1 = a(b(a(x)))
     problem:
      DPs:
       
      TRS:
       a(a(x)) -> a(b(a(x)))
     Qed