YES

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

Proof:
 String Reversal Processor:
  b(a(x)) -> a(b(x))
  c(a(x)) -> x
  KBO Processor:
   weight function:
    w0 = 1
    w(c) = w(a) = 1
    w(b) = 0
   precedence:
    b > a > c
   problem:
    
   Qed