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))
  WPO Processor:
   algebra: Pol
   weight function:
    w0 = 0
    w(A) = 2
    w(a) = 1
    w(p) = 0
   status function:
    st(A) = st(a) = st(p) = [0]
   subterm coefficient function:
    sc(p, 0) = 3
    sc(A, 0) = sc(a, 0) = 1
   precedence:
    p > A > a
   problem:
    
   Qed