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: 1
     enrichment: match-dp
     automaton:
      final states: {5}
      transitions:
       p{#,0}(11,9) -> 5*
       p0(3,1) -> 3*
       p0(3,3) -> 3*
       p0(4,10) -> 11*
       p0(1,2) -> 3*
       p0(2,1) -> 3*
       p0(2,3) -> 3*
       p0(3,2) -> 3*
       p0(8,6) -> 9*
       p0(1,1) -> 3*
       p0(1,3) -> 3*
       p0(2,2) -> 3*
       b0(7) -> 8*
       b0(2) -> 2*
       b0(4) -> 6*
       b0(1) -> 2*
       b0(3) -> 2*
       a0(2) -> 1*
       a0(4) -> 10,7
       a0(1) -> 1*
       a0(3) -> 1*
       p{#,1}(23,21) -> 5*
       p1(6,22) -> 23*
       p1(18,24) -> 23*
       p1(20,18) -> 21*
       a1(20) -> 24*
       a1(10) -> 19*
       a1(22) -> 19*
       a1(24) -> 19*
       a1(8) -> 22*
       b1(2) -> 18*
       b1(19) -> 20*
       b1(1) -> 18*
       b1(3) -> 18*
       1 -> 4*
       2 -> 4*
       3 -> 4*
     problem:
      DPs:
       
      TRS:
       p(p(b(a(x0)),x1),p(x2,x3)) -> p(p(x3,a(x2)),p(b(a(x1)),b(x0)))
     Qed