YES

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

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