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)))
  Matrix Interpretation Processor:
   dimension: 1
   interpretation:
    [a#](x0) = x0,
    
    [b](x0) = x0,
    
    [a](x0) = x0 + 1
   orientation:
    a#(b(a(x))) = x + 1 >= x = a#(b(x))
    
    a(b(a(x))) = x + 2 >= x + 1 = b(a(b(x)))
   problem:
    DPs:
     
    TRS:
     a(b(a(x))) -> b(a(b(x)))
   Qed