NO

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

Proof:
 String Reversal Processor:
  a(x1) -> b(x1)
  c(a(x1)) -> x1
  b(b(c(x1))) -> c(c(a(a(a(x1)))))
  Unfolding Processor:
   loop length: 12
   terms:
    a(b(c(b(c(c(x198979))))))
    b(b(c(b(c(c(x198979))))))
    c(c(a(a(a(b(c(c(x198979))))))))
    c(c(b(a(a(b(c(c(x198979))))))))
    c(c(b(b(a(b(c(c(x198979))))))))
    c(c(b(b(b(b(c(c(x198979))))))))
    c(c(b(b(c(c(a(a(a(c(x198979))))))))))
    c(c(c(c(a(a(a(c(a(a(a(c(x198979))))))))))))
    c(c(c(c(a(a(b(c(a(a(a(c(x198979))))))))))))
    c(c(c(c(a(a(b(c(b(a(a(c(x198979))))))))))))
    c(c(c(c(a(a(b(c(b(b(a(c(x198979))))))))))))
    c(c(c(c(a(a(b(c(b(b(b(c(x198979))))))))))))
   context: c(c(c(c(a([])))))
   substitution:
    x198979 -> a(a(a(x198979)))
   Qed