NO

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

Proof:
 String Reversal Processor:
  a(x1) -> x1
  b(a(x1)) -> a(a(b(c(a(b(x1))))))
  c(b(c(x1))) -> x1
  Unfolding Processor:
   loop length: 7
   terms:
    b(a(a(a(x4731))))
    a(a(b(c(a(b(a(a(x4731))))))))
    a(a(b(c(b(a(a(x4731)))))))
    a(a(b(c(a(a(b(c(a(b(a(x4731)))))))))))
    a(a(b(c(a(b(c(a(b(a(x4731))))))))))
    a(a(b(c(b(c(a(b(a(x4731)))))))))
    a(a(b(a(b(a(x4731))))))
   context: a(a([]))
   substitution:
    x4731 -> b(c(a(b(x4731))))
   Qed