Extended transition function of a DFA - a proof I would like to write a proof of the following statement
$$
\delta^+(q,PQ) = \delta^+(\delta^+(q,P),Q)
$$
$\delta^+$ - Extended transition function
I have to do it by induction. However, I'm not sure what I should do. At first, check it for $P = \epsilon$ and $Q = \epsilon$, right? What next?

You don’t have to worry about $P$: the induction is on the length of $Q$. Your base case is $Q=\epsilon$: check that $$\delta^+(q,P\epsilon)=\delta^+(\delta^+(q,P),\epsilon)\;.$$ Your induction step will then be to show that if $$\delta^+(q,PQ)=\delta^+(\delta^+(q,P),Q)\;,$$ then $$\delta^+(q,P(Qx))=\delta^+(\delta^+(q,P),Qx)\;,$$ where $x$ is any symbol in the input alphabet.
Assuming that $\delta^+$ has been defined by $\delta^+(q,Px) = \delta(\delta^+(q,P),x)$, this shouldn’t be too hard to do.
