YES

Problem:
 a(b(x1)) -> b(b(a(x1)))
 b(c(x1)) -> c(b(b(x1)))
 c(a(x1)) -> a(c(c(x1)))

Proof:
 String Reversal Processor:
  b(a(x1)) -> a(b(b(x1)))
  c(b(x1)) -> b(b(c(x1)))
  a(c(x1)) -> c(c(a(x1)))
  WPO Processor:
   algebra: Pol
   weight function:
    w0 = 0
    w(a) = 4
    w(c) = 1
    w(b) = 0
   status function:
    st(c) = st(a) = st(b) = [0]
   subterm coefficient function:
    sc(a, 0) = 3
    sc(c, 0) = sc(b, 0) = 1
   precedence:
    c > b > a
   problem:
    
   Qed