YES

Problem:
 p(a(x0),p(a(b(x1)),x2)) -> p(a(b(a(x2))),p(a(a(x1)),x2))

Proof:
 Reflection Processor:
  p(p(x2,a(b(x1))),a(x0)) -> p(p(x2,a(a(x1))),a(b(a(x2))))
  LPO Processor:
   precedence:
    p > b > a
   problem:
    
   Qed