# Proving a DFA accepts a Language using Induction

I have the following $$DFA$$ that recognizes the language containing either a $$101$$ substring or a $$010$$ substring.

I need to prove that it accepts exactly the aforementioned language using induction.

This is what I have done so far:

For the base case, an empty string does no contain either of the substrings so the $$DFA$$ correctly rejects the empty string.

For the induction step, I assume that the $$DFA$$ is valid for strings of size $$n-1$$, so assuming that the letter at index $$n$$ is a $$1$$ then it rejects accordingly since the string prior may not contain lead to the required substrings, otherwise it gets to state $$q5$$ and accepts the string. Same thing if the letter at index $$n$$ is a $$0$$.

But I feel that this is a bit too simple and I need to split the proof up into multiple parts.

If anyone knows how I should approach this problem, I would greatly appreciate it!

$$q_0$$ means that nothing has been read yet.
$$q_1$$ means that the last character read was $$1$$ and it is not the case that the last two characters (in order) were $$01$$. Furthermore, neither the substring $$101$$ nor $$010$$ has been seen so far.
$$q_2$$ means that the last two characters read were $$10$$ (in that order) and neither of the two distinguished substrings has been seen so far.
The meanings of $$q_3$$ and $$q_4$$ are similar, and of course, $$q_5$$ means that at least one of the distinguished substrings has been seen.
This is easily proved by induction, and it means that the machine will be in $$q_5$$ if and only if the strings contains one of the two distinguished substrings.