YES

Problem:
 R(x1) -> r(x1)
 r(p(x1)) -> p(p(r(P(x1))))
 r(r(x1)) -> x1
 r(P(P(x1))) -> P(P(r(x1)))
 p(P(x1)) -> x1
 P(p(x1)) -> x1
 r(R(x1)) -> x1
 R(r(x1)) -> x1

Proof:
 Arctic Interpretation Processor:
  dimension: 2
  interpretation:
             [0 0]  
   [P](x0) = [0 0]x0,
   
             [0  0 ]  
   [p](x0) = [0  -&]x0,
   
             [1 0]  
   [r](x0) = [1 0]x0,
   
             [2 2]  
   [R](x0) = [3 1]x0
  orientation:
           [2 2]      [1 0]          
   R(x1) = [3 1]x1 >= [1 0]x1 = r(x1)
   
              [1 1]      [1 1]                   
   r(p(x1)) = [1 1]x1 >= [1 1]x1 = p(p(r(P(x1))))
   
              [2 1]             
   r(r(x1)) = [2 1]x1 >= x1 = x1
   
                 [1 1]      [1 0]                
   r(P(P(x1))) = [1 1]x1 >= [1 0]x1 = P(P(r(x1)))
   
              [0 0]             
   p(P(x1)) = [0 0]x1 >= x1 = x1
   
              [0 0]             
   P(p(x1)) = [0 0]x1 >= x1 = x1
   
              [3 3]             
   r(R(x1)) = [3 3]x1 >= x1 = x1
   
              [3 2]             
   R(r(x1)) = [4 3]x1 >= x1 = x1
  problem:
   r(p(x1)) -> p(p(r(P(x1))))
   r(P(P(x1))) -> P(P(r(x1)))
   p(P(x1)) -> x1
   P(p(x1)) -> x1
  String Reversal Processor:
   p(r(x1)) -> P(r(p(p(x1))))
   P(P(r(x1))) -> r(P(P(x1)))
   P(p(x1)) -> x1
   p(P(x1)) -> x1
   Bounds Processor:
    bound: 0
    enrichment: match
    automaton:
     final states: {2,6,1}
     transitions:
      p0(2) -> 3*
      p0(3) -> 4*
      f40() -> 2*
      P0(5) -> 1*
      P0(7) -> 8*
      P0(2) -> 7*
      r0(4) -> 5*
      r0(8) -> 6*
      1 -> 4,3
      2 -> 4,8,3,7
      5 -> 4*
      6 -> 7,8
    problem:
     
    Qed