YES

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

Proof:
 DP Processor:
  DPs:
   a#(b(a(b(x)))) -> a#(a(b(x)))
   a#(b(a(b(x)))) -> a#(b(a(a(b(x)))))
  TRS:
   a(b(a(b(x)))) -> b(a(b(a(a(b(x))))))
  EDG Processor:
   DPs:
    a#(b(a(b(x)))) -> a#(a(b(x)))
    a#(b(a(b(x)))) -> a#(b(a(a(b(x)))))
   TRS:
    a(b(a(b(x)))) -> b(a(b(a(a(b(x))))))
   graph:
    
   Qed