MAYBE

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

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