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:
 String Reversal Processor:
  R(x1) -> r(x1)
  p(r(x1)) -> P(r(p(p(x1))))
  r(r(x1)) -> x1
  P(P(r(x1))) -> r(P(P(x1)))
  P(p(x1)) -> x1
  p(P(x1)) -> x1
  R(r(x1)) -> x1
  r(R(x1)) -> x1
  WPO Processor:
   algebra: Sum
   weight function:
    w0 = 0
    w(R) = 4
    w(r) = 1
    w(P) = w(p) = 0
   status function:
    st(P) = st(p) = st(r) = st(R) = [0]
   precedence:
    p > P > r ~ R
   problem:
    
   Qed