YES

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

Proof:
 DP Processor:
  DPs:
   p#(p(b(a(x0)),x1),p(x2,x3)) -> p#(b(a(x1)),b(x0))
   p#(p(b(a(x0)),x1),p(x2,x3)) -> p#(x3,a(x2))
   p#(p(b(a(x0)),x1),p(x2,x3)) -> p#(p(x3,a(x2)),p(b(a(x1)),b(x0)))
  TRS:
   p(p(b(a(x0)),x1),p(x2,x3)) -> p(p(x3,a(x2)),p(b(a(x1)),b(x0)))
  EDG Processor:
   DPs:
    p#(p(b(a(x0)),x1),p(x2,x3)) -> p#(b(a(x1)),b(x0))
    p#(p(b(a(x0)),x1),p(x2,x3)) -> p#(x3,a(x2))
    p#(p(b(a(x0)),x1),p(x2,x3)) -> p#(p(x3,a(x2)),p(b(a(x1)),b(x0)))
   TRS:
    p(p(b(a(x0)),x1),p(x2,x3)) -> p(p(x3,a(x2)),p(b(a(x1)),b(x0)))
   graph:
    p#(p(b(a(x0)),x1),p(x2,x3)) -> p#(p(x3,a(x2)),p(b(a(x1)),b(x0))) ->
    p#(p(b(a(x0)),x1),p(x2,x3)) -> p#(b(a(x1)),b(x0))
    p#(p(b(a(x0)),x1),p(x2,x3)) -> p#(p(x3,a(x2)),p(b(a(x1)),b(x0))) ->
    p#(p(b(a(x0)),x1),p(x2,x3)) -> p#(x3,a(x2))
    p#(p(b(a(x0)),x1),p(x2,x3)) -> p#(p(x3,a(x2)),p(b(a(x1)),b(x0))) ->
    p#(p(b(a(x0)),x1),p(x2,x3)) -> p#(p(x3,a(x2)),p(b(a(x1)),b(x0)))
   SCC Processor:
    #sccs: 1
    #rules: 1
    #arcs: 3/9
    DPs:
     p#(p(b(a(x0)),x1),p(x2,x3)) -> p#(p(x3,a(x2)),p(b(a(x1)),b(x0)))
    TRS:
     p(p(b(a(x0)),x1),p(x2,x3)) -> p(p(x3,a(x2)),p(b(a(x1)),b(x0)))
    Bounds Processor:
     bound: 0
     enrichment: match
     automaton:
      final states: {9,1}
      transitions:
       p{#,0}(8,6) -> 1*
       a0(5) -> 15*
       a0(15) -> 4*
       a0(7) -> 4*
       a0(2) -> 7,4
       p0(8,6) -> 9*
       p0(3,15) -> 8*
       p0(2,7) -> 8*
       p0(5,3) -> 6*
       f40() -> 2*
       b0(4) -> 5*
       b0(2) -> 3*
     problem:
      DPs:
       
      TRS:
       p(p(b(a(x0)),x1),p(x2,x3)) -> p(p(x3,a(x2)),p(b(a(x1)),b(x0)))
     Qed