I studied automaton theory and came across the pumping lemma, which I understand as:
To let $ mathcal {A} = langle Sigma, Q, q_0, F, delta rangle $ to be an NFA. Accept $ x in L ( mathcal {A}) $ s.t. $ | x | geq | Q | $, Then can be divided into x $ u, v, w, | v | geq 1 $s.t. $ x = uvw $ and for everyone $ i geq 0 $. $ uv ^ iw in L ( mathcal {A}) $
I had to consider if the following statement is correct:
To let $ L $ be a language. If for every word $ x in L $, there is $ u, v, w $s.t. $ uvw = x $ and for everyone $ i geq 0 $. $ uv ^ iw in L $, then $ L $ is regular
My idea is that no matter what $ u, v, w $ we decide to partition $ x $We can always construct NFAs that accept $ u, v, w $, respectively. And we can also construct an NFA that accepts $ v ^ i $ to the $ i geq 0 $, So if we take over the Union of NFAs, we'll get it $ A_x = bigcup ^ {} _ {uvw s.t uvw = x} A_ {uvw} $, Where $ A_ {uvw} $ is the NFA that has accepted the words of the form $ uv ^ iw $, to the $ i geq 0 $, if $ uvw = x $, If we take the Union of these NFAs, we get $ A_x $This is the NFA that takes care of all the possible words that are generated by $ x $, The final NFA is then $ A = bigcup {} _ {x in L} A_x $, in order to $ L $ is regular.
But I'm really new to automata theory, so I'm not sure if my idea is right. Every answer is grateful, thanks in advance.
