NO

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

Proof:
 String Reversal Processor:
  a(x1) -> x1
  b(a(x1)) -> a(a(c(a(b(c(b(x1)))))))
  c(c(x1)) -> x1
  Unfolding Processor:
   loop length: 10
   terms:
    b(a(a(a(x183026))))
    a(a(c(a(b(c(b(a(a(x183026)))))))))
    a(a(c(a(b(c(a(a(c(a(b(c(b(a(x183026))))))))))))))
    a(a(c(a(b(c(a(c(a(b(c(b(a(x183026)))))))))))))
    a(a(c(a(b(c(c(a(b(c(b(a(x183026))))))))))))
    a(a(c(a(b(a(b(c(b(a(x183026))))))))))
    a(a(c(a(b(a(b(c(a(a(c(a(b(c(b(x183026)))))))))))))))
    a(a(c(a(b(a(b(c(a(c(a(b(c(b(x183026))))))))))))))
    a(a(c(a(b(a(b(c(c(a(b(c(b(x183026)))))))))))))
    a(a(c(a(b(a(b(a(b(c(b(x183026)))))))))))
   context: a(a(c(a([]))))
   substitution:
    x183026 -> c(a(b(c(b(b(c(b(x183026))))))))
   Qed