Let $F_1, \dots, F_n$ be a partition of $\Omega$ the sample space. Let $E$ be an event. The law of total probability tells us: \begin{align} P(E) = \sum_{i=1}^n P(E\vert F_i) P(F_i) \end{align}
Let's take now an other arbitrary event $B$, intuitively conditioning on this new event, we should have: \begin{align} P(E\vert B) = \sum_{i=1}^n P(E\vert F_i, B) P(F_i \vert B) \end{align}
Here is my proof:
On the left hand side we have: $$ P(E\vert B) = \frac{P(E \cap B)}{P(B)}$$ On the right hand side we have: \begin{align} \sum_{i=1}^n P(E\vert F_i, B) P(F_i \vert B) &= \sum_{i=1}^n \frac{P(E\cap F_i \cap B) }{P(F_i \cap B)} \frac{P(F_i \cap B)}{ P(B)} \\ &= \sum_{i=1}^n \frac{P((E\cap B) \cap F_i)}{P(B)} \\ &= \frac{1}{P(B) }\sum_{i=1}^n P((E\cap B) \cap F_i) \\ &= \frac{P(E\cap B)}{P(B)} \qquad \text{by the law of total probability} \end{align}
Is my demonstration correct?
Yes. It is.