I'm looking at a proof that says that: If $M_1=(Q_1, \Sigma , q_1, A_1, \delta)$ and $M_2=(Q_2, \Sigma , q_2, A_2, \delta)$ are two finite automata(FA) then $M=M_1 \cup M_2$ is also an FA. We define $M=(Q, \Sigma , q, A, \delta)$ where $Q=Q_1 \times Q_2$, $q_0=(q_1,q_2)$ and $\delta((p,q),\sigma)=(\delta(p,\sigma),\delta(q,\sigma))$.
I'm just having one problem. We have to prove that $\delta^*((p,q),x)=(\delta^*(p,x),\delta^*(q,x))$. (x is a string in $\Sigma^*$) We do it by induction. I've already proven the base case, and I'm now trying to prove it holds for $x'=x\sigma$. This is what I've gotten so far:
I don't really know how to proceed from here. Can someone help?