YES

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

Proof:
 String Reversal Processor:
  b(a(x1)) -> a(c(b(x1)))
  c(b(x1)) -> b(b(c(x1)))
  c(a(x1)) -> b(a(c(x1)))
  WPO Processor:
   algebra: Sum
   weight function:
    w0 = 0
    w(a) = 2
    w(c) = w(b) = 0
   status function:
    st(c) = st(a) = st(b) = [0]
   precedence:
    c > b > a
   problem:
    
   Qed