How to demonstrate the Generalization of Bayes' Theorem?

Question

The theorem

Let $(Ω,\mathscr{C},P)$ be a probability space. If the events $A_1,A_2,...,A_n∈\mathscr{C}$ are incompatible two to two, with $P(A_i)> 0, ∀i=1,2,...,n$ and such that $\bigcup_{i=1}^{i=n}A_{i}=Ω$ , then, for any event $B∈C$, such that $P(B)>0$, it follows that:

$$P(A_k|B)=\frac{P(B|A_k)\cdot P(A_k)}{\sum_{i=1}^{i=n}P(B|A_i)\cdot P(A_i)}\quad\text{for all }k=1,2,...,n.$$

My demonstration

By definition of Conditional Probability:

1. $\displaystyle P(A_k|B)=\frac{P(A_k∩B)}{P(B)}$
2. $\displaystyle P(B|A_k)=\frac{P(A_k∩B)}{P(A_k)}$
3. $\displaystyle P(A_k∩B)=P(B|A_k)\cdot P(A_k)$
4. $\displaystyle P(A_k|B)=\frac{P(B|A_k)\cdot P(A_k)}{P(B)}$

By the Total Probability Theorem, concludes that:

5. $\displaystyle P(B) = \sum_i P(B\cap A_i) = \sum_i P(B|A_i)\cdot P(A_i)$
6. $\displaystyle P(A_k|B)=\frac{P(B|A_k)\cdot P(A_k)}{\sum_{i=1}^{i=n}P(B|A_i)\cdot P(A_i)}$

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Is it well demonstrated that way?