MAYBE

Problem:
 p(p(b(a(x0)),x1),p(x2,x3)) -> p(p(b(x2),a(a(b(x1)))),p(x3,x0))

Proof:
 DP Processor:
  DPs:
   p#(p(b(a(x0)),x1),p(x2,x3)) -> p#(x3,x0)
   p#(p(b(a(x0)),x1),p(x2,x3)) -> p#(b(x2),a(a(b(x1))))
   p#(p(b(a(x0)),x1),p(x2,x3)) -> p#(p(b(x2),a(a(b(x1)))),p(x3,x0))
  TRS:
   p(p(b(a(x0)),x1),p(x2,x3)) -> p(p(b(x2),a(a(b(x1)))),p(x3,x0))
  Restore Modifier:
   DPs:
    p#(p(b(a(x0)),x1),p(x2,x3)) -> p#(x3,x0)
    p#(p(b(a(x0)),x1),p(x2,x3)) -> p#(b(x2),a(a(b(x1))))
    p#(p(b(a(x0)),x1),p(x2,x3)) -> p#(p(b(x2),a(a(b(x1)))),p(x3,x0))
   TRS:
    p(p(b(a(x0)),x1),p(x2,x3)) -> p(p(b(x2),a(a(b(x1)))),p(x3,x0))
   SCC Processor:
    #sccs: 1
    #rules: 3
    #arcs: 9/9
    DPs:
     p#(p(b(a(x0)),x1),p(x2,x3)) -> p#(x3,x0)
     p#(p(b(a(x0)),x1),p(x2,x3)) -> p#(b(x2),a(a(b(x1))))
     p#(p(b(a(x0)),x1),p(x2,x3)) -> p#(p(b(x2),a(a(b(x1)))),p(x3,x0))
    TRS:
     p(p(b(a(x0)),x1),p(x2,x3)) -> p(p(b(x2),a(a(b(x1)))),p(x3,x0))
    Open