In the book "Probability and statistics" by Morris H. DeGroot and Mark J. Schervish, on page 80, the conditional version of Bayes' theorem is given with no explanation:
$$\Pr(B_i\mid A \cap C) = \dfrac{\Pr(B_i\mid C) \cdot \Pr(A\mid B_i \cap C)}{\sum (\Pr(B_i\mid C)\cdot \Pr(A\mid B_i \cap C))}$$
What is the intuition or mathematical proof behind this?
It's directly equivalent to: $$\begin{align} \Pr(B_i\mid A) & = \frac{\Pr(B_i \cap A)}{\Pr(A)} \\[1ex] &= \frac{\Pr(B_i)\cdot \Pr(A\mid B_i)}{\sum_k \Pr(B_k)\cdot\Pr(A\mid B_k)} \end{align}$$ But with conditioning when given $C$: $$\begin{align} \Pr(B_i\mid A\cap C) & = \frac{\Pr(B_i \cap A\mid C)}{\Pr(A\mid C)} \\[1ex] &= \frac{\Pr(B_i \mid C)\cdot\Pr(A\mid B_i\cap C)}{\sum_k \Pr(B_k\mid C)\cdot\Pr(A\mid B_k\cap C)} \end{align}$$
How do you know equation 3 is true?
$\begin{align} \Pr(B_i\mid A\cap C) & = \frac{\Pr(B_i\cap A \cap C)}{\Pr(A\cap C)} & \text{by definition} \\[1ex] & = \frac{\Pr(B_i\cap A\mid C)\cdot \Pr(C)}{\Pr(A\mid C)\cdot\Pr(C)} & \text{by the same} \\[1ex] & = \frac{\Pr(B_i\cap A\mid C)}{\Pr(A\mid C)} & \text{by cancelation} \end{align}$
Also, can you explain how you derived the denominator in equation 4?
By the Law of Total Probability. If the set of $\{B_k: k\in \{1..n\}\}$ partitions the sample space then the measure of event $A$ is the sum of the products of the measure of the partition and the measure of the event conditional on the partition.
$$\begin{align}\Pr(A) \quad & = \Pr(B_1)\cdot\Pr(A\mid B_1) + \cdots + \Pr(B_k)\cdot\Pr(A\mid B_k) + \cdots \Pr(B_n)\cdot\Pr(A\mid B_n) \\ & = \sum_{k=1}^n \Pr(B_k)\cdot\Pr(A\mid B_k) \end{align}$$
I understand everything you said except the actual steps transforming equation three to equation four. Are you treating Pr(A|C) as sum(Pr(A|C|Bk)Pr(Bk)).
Yes. Thuswise: $$\begin{align} \Pr(A\mid C) \quad & = \frac{\Pr(A\cap C)}{\Pr(C)} & \text{by conditional probability} \\[1ex] & = \frac{\sum_k \Pr(A\cap C\cap B_k)}{\Pr(C)} & \text{by total probability} \\[1ex] & = \frac{\sum_k \Pr(B_k\cap C)\cdot\Pr(A\mid C\cap B_k)}{\Pr(C)} & \text{by conditional probability} \\[1ex] & = \sum_k \frac{\Pr(B_k\cap C)}{\Pr(C)}\cdot\Pr(A\mid C\cap B_k) & \text{by rearrangement} \\[1ex] & = \sum_k \Pr(B_k\mid C)\cdot\Pr(A\mid C\cap B_k) & \text{by conditional probability} \end{align}$$
