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))
    Matrix Interpretation Processor: dim=3
     
     interpretation:
                     [1]
      [A](x0) = x0 + [0]
                     [1],
      
                [1 1 1]     [1]
      [a](x0) = [0 1 0]x0 + [0]
                [0 1 1]     [1],
      
                [1 1 0]     [0]
      [p](x0) = [0 1 0]x0 + [1]
                [0 0 1]     [1]
     orientation:
                 [1 2 1]     [3]    [1 2 1]     [3]              
      a(p(x1)) = [0 1 0]x1 + [1] >= [0 1 0]x1 + [1] = p(a(A(x1)))
                 [0 1 1]     [3]    [0 1 1]     [3]              
      
                 [1 1 1]     [3]    [1 1 1]     [2]           
      a(A(x1)) = [0 1 0]x1 + [0] >= [0 1 0]x1 + [0] = A(a(x1))
                 [0 1 1]     [2]    [0 1 1]     [2]           
     problem:
      a(p(x1)) -> p(a(A(x1)))
     Arctic Interpretation Processor:
      dimension: 3
      interpretation:
                 [0  -& -&]  
       [A](x0) = [0  -& 0 ]x0
                 [-& -& -&]  ,
       
                 [0  -& 1 ]  
       [a](x0) = [-& 1  1 ]x0
                 [0  0  1 ]  ,
       
                 [0  -& 0 ]  
       [p](x0) = [-& -& -&]x0
                 [0  -& 0 ]  
      orientation:
                  [1  -& 1 ]      [0  -& 0 ]                
       a(p(x1)) = [1  -& 1 ]x1 >= [-& -& -&]x1 = p(a(A(x1)))
                  [1  -& 1 ]      [0  -& 0 ]                
      problem:
       
      Qed