在陈述语言 L = {0^(n)1^(n) | 的证明中发现错误 n>0} 是一个正则表达式

Question

Proof: Let M be the following NFA:

Automata for L Now, if x is in L, then x = 0^(n)1^(n). Thus while processing x, M will start in state q0, loop in state q0 n times, then transition to state q1 on the rst 1, and follow the loop in state q1 a total of n 􀀀 1 times. Since it ends in state q1, x will be accepted. Thus M recognizes every string in L, so L is NFA-recognizable (and by Kleene's theorem, is thus regular).

Answer 1

You've shown that M accepts every string of L. But that doesn't mean that M recognizes L: for that to be true, you have to also show that M accepts exactly the strings of L and no others (or equivalently: it doesn't accept any string not in L). Unfortunately in your case, M does accept some non-L strings: for example, "011" .

Answer 2

Your proof is indeed wrong. To prove that the language L is regular you need to prove that there exists a finite state automaton M that recognizes the language L. Hence, there are three things that you need to prove about the automaton M:

1. The automaton M exists.
2. The automaton M recognizes the language L (nothing more, nothing less).
3. The automaton M has a finite number of states.

The problem with your proof is that the automaton you posted does not recognize the language L:

This automaton recognizes the language 0*1+ which is not the same as the language L ( 0^n1^n ). Yes, every string in L is recognized by this automaton. However, this automaton also recognizes strings like 1 which is clearly not in L. Hence, using this automaton in your proof is wrong.

So, how do you prove that the automaton M exists? Well, the only way to prove that the automaton M exists is by showing that M exists. The only problem is that L is not a regular language. Hence, there is no finite state automaton M that recognizes L.

The opposite way is to prove that L is not a regular language. To prove that L is not a regular language you need to prove that there is no finite state automaton M that recognizes L. This is an easier proof because you don't have to find the automaton M. You can just assume that it exists.

Here's what you need to do to prove that L is not a regular language:

1. Assume that an automaton M exists that recognizes the language L.
2. Show that M must necessarily have an infinite number of states.

The standard way of proving that an automaton M cannot have a finite number of states is by using the Pumping Lemma . I'll leave it as an exercise for you to figure that out because it's not in the scope of your question.