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)))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
    [p#](x0, x1) = [1 0 0]x0 + [0 1 0]x1,
    
                  [0 0 1]     [0 0 0]  
    [p](x0, x1) = [2 1 0]x0 + [1 2 1]x1
                  [0 0 0]     [0 0 0]  ,
    
              [0 0 0]  
    [b](x0) = [0 0 0]x0
              [0 2 0]  ,
    
              [0 0 0]     [0]
    [a](x0) = [0 1 0]x0 + [1]
              [0 0 0]     [0]
   orientation:
    p#(p(b(a(x0)),x1),p(x2,x3)) = [0 2 0]x0 + [2 1 0]x2 + [1 2 1]x3 + [2] >= [0] = p#(b(a(x1)),b(x0))
    
    p#(p(b(a(x0)),x1),p(x2,x3)) = [0 2 0]x0 + [2 1 0]x2 + [1 2 1]x3 + [2] >= [0 1 0]x2 + [1 0 0]x3 + [1] = p#(x3,a(x2))
    
    p#(p(b(a(x0)),x1),p(x2,x3)) = [0 2 0]x0 + [2 1 0]x2 + [1 2 1]x3 + [2] >= [0 2 0]x0 + [0 0 1]x3 = p#(p(x3,a(x2)),p(b(a(x1)),b(x0)))
    
                                 [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]                                   
    p(p(b(a(x0)),x1),p(x2,x3)) = [0 4 0]x0 + [1 2 1]x1 + [4 2 1]x2 + [2 4 2]x3 + [4] >= [0 4 0]x0 + [0 2 0]x1 + [0 2 0]x2 + [2 1 2]x3 + [4] = p(p(x3,a(x2)),p(b(a(x1)),b(x0)))
                                 [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]                                   
   problem:
    DPs:
     
    TRS:
     p(p(b(a(x0)),x1),p(x2,x3)) -> p(p(x3,a(x2)),p(b(a(x1)),b(x0)))
   Qed