YES

Problem:
 a(p(x1)) -> p(a(A(x1)))
 a(A(x1)) -> A(a(x1))
 p(A(A(x1))) -> a(p(x1))

Proof:
 String Reversal Processor:
  p(a(x1)) -> A(a(p(x1)))
  A(a(x1)) -> a(A(x1))
  A(A(p(x1))) -> p(a(x1))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 0 0]     [1]
    [A](x0) = [0 1 0]x0 + [1]
              [0 1 0]     [0],
    
                   [0]
    [a](x0) = x0 + [1]
                   [1],
    
              [1 1 0]     [0]
    [p](x0) = [0 1 1]x0 + [1]
              [0 1 1]     [0]
   orientation:
               [1 1 0]     [1]    [1 1 0]     [1]              
    p(a(x1)) = [0 1 1]x1 + [3] >= [0 1 1]x1 + [3] = A(a(p(x1)))
               [0 1 1]     [2]    [0 1 1]     [2]              
    
               [1 0 0]     [1]    [1 0 0]     [1]           
    A(a(x1)) = [0 1 0]x1 + [2] >= [0 1 0]x1 + [2] = a(A(x1))
               [0 1 0]     [1]    [0 1 0]     [1]           
    
                  [1 1 0]     [2]    [1 1 0]     [1]           
    A(A(p(x1))) = [0 1 1]x1 + [3] >= [0 1 1]x1 + [3] = p(a(x1))
                  [0 1 1]     [2]    [0 1 1]     [2]           
   problem:
    p(a(x1)) -> A(a(p(x1)))
    A(a(x1)) -> a(A(x1))
   String Reversal Processor:
    a(p(x1)) -> p(a(A(x1)))
    a(A(x1)) -> A(a(x1))
    Bounds Processor:
     bound: 1
     enrichment: match
     automaton:
      final states: {5,1}
      transitions:
       A1(8) -> 9*
       a1(7) -> 8*
       f30() -> 2*
       p0(4) -> 1*
       a0(2) -> 6*
       a0(3) -> 4*
       A0(2) -> 3*
       A0(6) -> 5*
       1 -> 8,6
       2 -> 7*
       5 -> 8,6
       9 -> 4*
     problem:
      
     Qed