YES

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

Proof:
 DP Processor:
  DPs:
   p#(a(a(x0)),p(x1,p(a(x2),x3))) -> p#(a(a(x0)),x3)
   p#(a(a(x0)),p(x1,p(a(x2),x3))) -> p#(a(a(b(x1))),p(a(a(x0)),x3))
   p#(a(a(x0)),p(x1,p(a(x2),x3))) -> p#(x2,p(a(a(b(x1))),p(a(a(x0)),x3)))
  TRS:
   p(a(a(x0)),p(x1,p(a(x2),x3))) -> p(x2,p(a(a(b(x1))),p(a(a(x0)),x3)))
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [p#](x0, x1) = 2x1 + 0,
    
    [b](x0) = 2x0,
    
    [p](x0, x1) = 1x1 + 1,
    
    [a](x0) = x0 + 0
   orientation:
    p#(a(a(x0)),p(x1,p(a(x2),x3))) = 4x3 + 4 >= 2x3 + 0 = p#(a(a(x0)),x3)
    
    p#(a(a(x0)),p(x1,p(a(x2),x3))) = 4x3 + 4 >= 3x3 + 3 = p#(a(a(b(x1))),p(a(a(x0)),x3))
    
    p#(a(a(x0)),p(x1,p(a(x2),x3))) = 4x3 + 4 >= 4x3 + 4 = p#(x2,p(a(a(b(x1))),p(a(a(x0)),x3)))
    
    p(a(a(x0)),p(x1,p(a(x2),x3))) = 3x3 + 3 >= 3x3 + 3 = p(x2,p(a(a(b(x1))),p(a(a(x0)),x3)))
   problem:
    DPs:
     p#(a(a(x0)),p(x1,p(a(x2),x3))) -> p#(a(a(x0)),x3)
     p#(a(a(x0)),p(x1,p(a(x2),x3))) -> p#(x2,p(a(a(b(x1))),p(a(a(x0)),x3)))
    TRS:
     p(a(a(x0)),p(x1,p(a(x2),x3))) -> p(x2,p(a(a(b(x1))),p(a(a(x0)),x3)))
   KBO Processor:
    argument filtering:
     pi(a) = 0
     pi(p) = [1]
     pi(b) = 0
     pi(p#) = [1]
    weight function:
     w0 = 1
     w(p#) = w(b) = w(a) = 1
     w(p) = 0
    precedence:
     p# ~ b ~ p ~ a
    problem:
     DPs:
      p#(a(a(x0)),p(x1,p(a(x2),x3))) -> p#(x2,p(a(a(b(x1))),p(a(a(x0)),x3)))
     TRS:
      p(a(a(x0)),p(x1,p(a(x2),x3))) -> p(x2,p(a(a(b(x1))),p(a(a(x0)),x3)))
    Matrix Interpretation Processor: dim=4
     
     interpretation:
      [p#](x0, x1) = [0 0 0 1]x0 + [0 1 0 0]x1,
      
                [0]
                [0]
      [b](x0) = [0]
                [0],
      
                    [0 0 0 0]     [0 0 0 0]     [0]
                    [1 0 0 0]     [0 1 0 0]     [0]
      [p](x0, x1) = [0 0 0 1]x0 + [0 1 0 0]x1 + [0]
                    [0 0 0 1]     [0 0 1 0]     [1],
      
                [1 0 0 1]     [0]
                [1 1 0 1]     [1]
      [a](x0) = [1 0 0 0]x0 + [0]
                [0 1 0 0]     [0]
     orientation:
      p#(a(a(x0)),p(x1,p(a(x2),x3))) = [1 1 0 1]x0 + [1 0 0 0]x1 + [1 0 0 1]x2 + [0 1 0 0]x3 + [1] >= [1 1 0 1]x0 + [0 0 0 1]x2 + [0 1 0 0]x3 = p#(x2,p(a(a(b(x1))),p(a(a(x0)),x3)))
      
                                      [0 0 0 0]     [0 0 0 0]     [0 0 0 0]     [0 0 0 0]     [0]    [0 0 0 0]     [0 0 0 0]     [0 0 0 0]     [0]                                      
                                      [1 1 0 1]     [1 0 0 0]     [1 0 0 1]     [0 1 0 0]     [0]    [1 1 0 1]     [1 0 0 0]     [0 1 0 0]     [0]                                      
      p(a(a(x0)),p(x1,p(a(x2),x3))) = [1 1 0 1]x0 + [1 0 0 0]x1 + [1 0 0 1]x2 + [0 1 0 0]x3 + [1] >= [1 1 0 1]x0 + [0 0 0 1]x2 + [0 1 0 0]x3 + [0] = p(x2,p(a(a(b(x1))),p(a(a(x0)),x3)))
                                      [1 1 0 1]     [0 0 0 1]     [1 0 0 1]     [0 1 0 0]     [2]    [1 1 0 1]     [0 0 0 1]     [0 1 0 0]     [2]                                      
     problem:
      DPs:
       
      TRS:
       p(a(a(x0)),p(x1,p(a(x2),x3))) -> p(x2,p(a(a(b(x1))),p(a(a(x0)),x3)))
     Qed