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When given one grammar, we can formally prove that it can recognize one language using QA_1

Since Kleene's Theorem gives the equivalence between the regular grammar and the NFA, we can also use QA_1 to prove that one NFA can recognize one language by transfroming the NFA to the grammar firstly.

Then how to prove that one Turing Machine recognizes one specific language? This QA_2 answer uses contradiction to prove. But it seems informal for me.

Take $\{0^n 1^n:n\ge 1\}$ for example. I followed this lecture.

Since there is no non-terminal/terminal symbols for the Turing Machine, it seems inconvenient to construct $M_i$ when following QA_1.

So I followed QA_3 answer. It is trivial to prove the Turing Machine in p4 of the above lecture can accept $\{0^n 1^n:n\ge 1\}$, but how to prove that it can't accept others?

Both the reference QA_3 answer and QA_4 answer gives one strategy to construct states, but they didn't give one formal proof.

Q:

How to prove the Turing Machine in the above reference lecture recognizes $\{0^n 1^n:n\ge 1\}$? Is it impossible to give one formal proof since the halting problem is an unsolvable decision problem?

An5Drama
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1 Answers1

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The halting problem doesn't really tell you anything about whether you can prove that a TM recognises a given language.

We can prove that the TM recognises $\{0^n1^m : n \geq 1\}$ using the definition of $L(M)$ and $\mapsto_M$. One thing to note is that the machine is deterministic, so for each state and character there's at most one transition. Therefore for each input $w$ there's only one sequence of IDs (one computation) we need to consider to decide if $w \in L(M)$.

These proofs can get very long since there are a lot of cases to consider, so I'll just give a rough sketch.

For simplicity we'll write $\mapsto$ instead of $\mapsto_M$, note the following:

Lemma 1: $X^kq_00^nY^k1^m \mapsto^* X^{k + 1}q_00^{n - 1}Y^{k + 1}1^{m - 1}$ for $k, n, m \in \mathbb{N}; n,m \geq 1$ using only states $q_0, q_1, q_2 \notin F$.

This can be proven by applying the definitions of $\mapsto$ and $M$. A corollary of Lemma 1 is that for $n \geq m$

$$X^kq_00^nY^k1^m \mapsto^* X^{k + m}q_00^{n - m}Y^{k + m}$$

and for $m \geq n$

$$X^kq_00^nY^k1^m \mapsto^* X^{k + n}q_0Y^{k + n}1^{m - n}$$

using only transitions through $q_0, q_1$ and $q_2$. From this it follows that for all $n \in \mathbb{N}; n \geq 1$

$$\underbrace{q_00^n1^n}_{=\ X^0q_00^nY^01^n} \mapsto^* X^nq_0Y^n \mapsto X^nYq_3Y^{n - 1} \mapsto^* X^nY^nq_3B \mapsto X^nY^nBq_4B$$

with $q_4 \in F$, so by definition $0^n1^n \in L(M)$.

Now assume that $w \in \Sigma^*; w \notin \{0^n1^n : n \geq 1\}$, then one of the following cases must be true:

  • $w = \varepsilon$ or $w = 1x$ for $x \in \Sigma^*$, then $M$ on $w$ directly halts in $q_0 \notin F \Rightarrow w \notin L(M)$.
  • $w = 0^n$ for $n \in \mathbb{N}; n \geq 1$, then $$q_0w \mapsto Xq_10^{n - 1} \mapsto^* X0^{n - 1}q_1B$$ so the machine halts in $q_1$ without ever entering a final state. Thus $w \notin L(M)$.
  • $w = 0^n1^m$ for $n, m \geq 1; n \neq m$, if $n > m$ then by the corollary of Lemma 1 $$q_0w \mapsto^* X^mq_00^{n - m}Y^m \mapsto^* X^{m + 1}0^{n - m - 1}Y^mq_1B$$ so $M$ halts in $q_1$ without having entered a final state. If $m > n$ then again by Lemma 1 $$q_0w \mapsto^* X^nq_0Y^n1^{m - n} \mapsto^* X^nY^nq_31^{m - n}$$ so $M$ halts in $q_3$ without having entered a final state. Thus $w \notin L(M)$.
  • $w = 0^n1^m0x$ for $n, m \geq 1$ and $x \in \Sigma^*$, $w \notin L(M)$ can be proven similar to the cases above by considering $n = m$, $n > m$ and $m > n$.

So $w \in \{0^n1^n : n \geq 1\} \iff w \in L(M)$, thus $L(M) = \{0^n1^n : n \geq 1\}$.

Knogger
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