YES

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

Proof:
 DP Processor:
  DPs:
   p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(b(x1),x3)
   p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(a(a(x0)),p(b(x1),x3))
   p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(x2,p(a(a(x0)),p(b(x1),x3)))
  TRS:
   p(a(x0),p(b(x1),p(a(x2),x3))) -> p(x2,p(a(a(x0)),p(b(x1),x3)))
  EDG Processor:
   DPs:
    p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(b(x1),x3)
    p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(a(a(x0)),p(b(x1),x3))
    p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(x2,p(a(a(x0)),p(b(x1),x3)))
   TRS:
    p(a(x0),p(b(x1),p(a(x2),x3))) -> p(x2,p(a(a(x0)),p(b(x1),x3)))
   graph:
    p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(a(a(x0)),p(b(x1),x3)) ->
    p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(b(x1),x3)
    p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(a(a(x0)),p(b(x1),x3)) ->
    p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(a(a(x0)),p(b(x1),x3))
    p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(a(a(x0)),p(b(x1),x3)) ->
    p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(x2,p(a(a(x0)),p(b(x1),x3)))
    p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(x2,p(a(a(x0)),p(b(x1),x3))) ->
    p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(b(x1),x3)
    p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(x2,p(a(a(x0)),p(b(x1),x3))) ->
    p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(a(a(x0)),p(b(x1),x3))
    p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(x2,p(a(a(x0)),p(b(x1),x3))) ->
    p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(x2,p(a(a(x0)),p(b(x1),x3)))
   SCC Processor:
    #sccs: 1
    #rules: 2
    #arcs: 6/9
    DPs:
     p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(a(a(x0)),p(b(x1),x3))
     p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(x2,p(a(a(x0)),p(b(x1),x3)))
    TRS:
     p(a(x0),p(b(x1),p(a(x2),x3))) -> p(x2,p(a(a(x0)),p(b(x1),x3)))
    KBO Processor:
     argument filtering:
      pi(a) = []
      pi(b) = 0
      pi(p) = [1]
      pi(p#) = 1
     weight function:
      w0 = 1
      w(p#) = w(p) = w(b) = w(a) = 1
     precedence:
      p# ~ p ~ b ~ a
     problem:
      DPs:
       p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(x2,p(a(a(x0)),p(b(x1),x3)))
      TRS:
       p(a(x0),p(b(x1),p(a(x2),x3))) -> p(x2,p(a(a(x0)),p(b(x1),x3)))
     Matrix Interpretation Processor: dim=2
      
      interpretation:
       [p#](x0, x1) = [2 0]x0 + [1 0]x1,
       
                     [2 1]     [0 1]  
       [p](x0, x1) = [2 0]x0 + [0 1]x1,
       
                 [0 2]     [0]
       [b](x0) = [2 1]x0 + [1],
       
                 [1 2]  
       [a](x0) = [0 0]x0
      orientation:
       p#(a(x0),p(b(x1),p(a(x2),x3))) = [2 4]x0 + [2 5]x1 + [2 4]x2 + [0 1]x3 + [1] >= [2 4]x0 + [0 4]x1 + [2 0]x2 + [0 1]x3 = p#(x2,p(a(a(x0)),p(b(x1),x3)))
       
                                       [2 4]     [0 4]     [2 4]     [0 1]      [2 4]     [0 4]     [2 1]     [0 1]                                  
       p(a(x0),p(b(x1),p(a(x2),x3))) = [2 4]x0 + [0 4]x1 + [2 4]x2 + [0 1]x3 >= [2 4]x0 + [0 4]x1 + [2 0]x2 + [0 1]x3 = p(x2,p(a(a(x0)),p(b(x1),x3)))
      problem:
       DPs:
        
       TRS:
        p(a(x0),p(b(x1),p(a(x2),x3))) -> p(x2,p(a(a(x0)),p(b(x1),x3)))
      Qed