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)))
  Subterm Criterion Processor:
   simple projection:
    pi(p) = [1,1]
    pi(p#) = [1,1]
   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
    
    usable rules:
     p(a(a(x0)),p(x1,p(a(x2),x3))) -> p(x2,p(a(a(b(x1))),p(a(a(x0)),x3)))
    interpretation:
     [p#](x0, x1) = [1 1 1 0]x0 + [0 1 0 0]x1 + [1],
     
               [1 0 0 0]  
               [0 0 0 0]  
     [b](x0) = [0 0 0 0]x0
               [0 0 0 0]  ,
     
                   [1 0 1 0]     [0 0 0 1]  
                   [1 0 1 0]     [0 0 0 1]  
     [p](x0, x1) = [0 0 0 0]x0 + [0 0 0 0]x1
                   [1 0 1 0]     [0 0 0 1]  ,
     
               [0 1 0 0]     [0]
               [0 0 0 1]     [0]
     [a](x0) = [1 1 1 0]x0 + [0]
               [0 0 0 0]     [1]
    orientation:
     p#(a(a(x0)),p(x1,p(a(x2),x3))) = [1 2 1 2]x0 + [1 0 1 0]x1 + [1 2 1 0]x2 + [0 0 0 1]x3 + [2] >= [1 2 1 2]x0 + [1 0 0 0]x1 + [1 1 1 0]x2 + [0 0 0 1]x3 + [1] = p#(x2,p(a(a(b(x1))),p(a(a(x0)),x3)))
     
                                     [1 2 1 2]     [1 0 1 0]     [1 2 1 0]     [0 0 0 1]      [1 2 1 2]     [1 0 0 0]     [1 0 1 0]     [0 0 0 1]                                        
                                     [1 2 1 2]     [1 0 1 0]     [1 2 1 0]     [0 0 0 1]      [1 2 1 2]     [1 0 0 0]     [1 0 1 0]     [0 0 0 1]                                        
     p(a(a(x0)),p(x1,p(a(x2),x3))) = [0 0 0 0]x0 + [0 0 0 0]x1 + [0 0 0 0]x2 + [0 0 0 0]x3 >= [0 0 0 0]x0 + [0 0 0 0]x1 + [0 0 0 0]x2 + [0 0 0 0]x3 = p(x2,p(a(a(b(x1))),p(a(a(x0)),x3)))
                                     [1 2 1 2]     [1 0 1 0]     [1 2 1 0]     [0 0 0 1]      [1 2 1 2]     [1 0 0 0]     [1 0 1 0]     [0 0 0 1]                                        
    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