NO

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

Proof:
 String Reversal Processor:
  a(x1) -> x1
  b(a(x1)) -> a(a(c(a(a(b(x1))))))
  c(c(x1)) -> b(x1)
  Unfolding Processor:
   loop length: 7
   terms:
    b(a(a(x5241)))
    a(a(c(a(a(b(a(x5241)))))))
    a(a(c(a(b(a(x5241))))))
    a(a(c(b(a(x5241)))))
    a(a(c(a(a(c(a(a(b(x5241)))))))))
    a(a(c(a(c(a(a(b(x5241))))))))
    a(a(c(c(a(a(b(x5241)))))))
   context: a(a([]))
   substitution:
    x5241 -> b(x5241)
   Qed