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)))
  Matrix Interpretation Processor: dim=3
   
   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) = [0 1 0]x0 + [0 1 0]x1,
    
              [0]
    [b](x0) = [0]
              [0],
    
                  [0 1 0]     [0 0 1]     [0]
    [p](x0, x1) = [0 0 1]x0 + [0 1 0]x1 + [1]
                  [0 1 0]     [0 1 0]     [0],
    
              [1 1 1]     [1]
    [a](x0) = [1 0 0]x0 + [0]
              [0 1 1]     [0]
   orientation:
    p#(a(a(x0)),p(x1,p(a(x2),x3))) = [1 1 1]x0 + [0 0 1]x1 + [0 1 1]x2 + [0 1 0]x3 + [3] >= [1 1 1]x0 + [0 1 0]x3 + [1] = p#(a(a(x0)),x3)
    
    p#(a(a(x0)),p(x1,p(a(x2),x3))) = [1 1 1]x0 + [0 0 1]x1 + [0 1 1]x2 + [0 1 0]x3 + [3] >= [1 1 1]x0 + [0 1 0]x3 + [2] = p#(a(a(b(x1))),p(a(a(x0)),x3))
    
    p#(a(a(x0)),p(x1,p(a(x2),x3))) = [1 1 1]x0 + [0 0 1]x1 + [0 1 1]x2 + [0 1 0]x3 + [3] >= [1 1 1]x0 + [0 1 0]x2 + [0 1 0]x3 + [2] = p#(x2,p(a(a(b(x1))),p(a(a(x0)),x3)))
    
                                    [1 1 1]     [0 1 0]     [0 1 1]     [0 1 0]     [2]    [1 1 1]     [0 1 0]     [0 1 0]     [2]                                      
    p(a(a(x0)),p(x1,p(a(x2),x3))) = [1 1 1]x0 + [0 0 1]x1 + [0 1 1]x2 + [0 1 0]x3 + [3] >= [1 1 1]x0 + [0 0 1]x2 + [0 1 0]x3 + [3] = p(x2,p(a(a(b(x1))),p(a(a(x0)),x3)))
                                    [1 1 1]     [0 0 1]     [0 1 1]     [0 1 0]     [3]    [1 1 1]     [0 1 0]     [0 1 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