MAYBE

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

Proof:
 DP Processor:
  DPs:
   p#(a(x0),p(a(a(a(x1))),x2)) -> p#(a(a(b(x0))),x2)
   p#(a(x0),p(a(a(a(x1))),x2)) -> p#(a(x2),p(a(a(b(x0))),x2))
  TRS:
   p(a(x0),p(a(a(a(x1))),x2)) -> p(a(x2),p(a(a(b(x0))),x2))
  Open