YES

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

Proof:
 String Reversal Processor:
  c(b(x1)) -> b(c(x1))
  b(c(x1)) -> a(a(a(x1)))
  a(a(a(a(x1)))) -> c(b(x1))
  WPO Processor:
   algebra: Sum
   weight function:
    w0 = 0
    w(b) = w(c) = 2
    w(a) = 1
   status function:
    st(a) = st(b) = st(c) = [0]
   precedence:
    a > c > b
   problem:
    
   Qed