# Show that 0^n1^n is decidable

I would like to know how would one go about and answer this question. I know that a language is decidable if a turing machine exists that accepts the strings in that language, and rejects otherwise. Would I need to create a turing machine to show this or is it enough to create a PDA and claim a TM can be created from a PDA such that it accepts and halts all strings in it and rejects otherwise. Also, if anyone can refer a good post to learn how to create a PDA from CFL would be greatly appreciated. Thanks!

• @Boris_s What do you mean by "computatonal complexity of CFG"? There is a simple $O(n^3)$ algorithm deciding if a string of length $n$ belongs to $L(G)$. – fade2black Nov 26 '17 at 22:29