The proof is pretty clear from the following picture:
Since the [math]A_i
[/math] are disjoint, we have that the sets [math]B \href{/cs2800/wiki/index.php/%E2%88%A9}{∩} A_i
[/math] are disjoint; and since every element of [math]S
[/math] is in one of the [math]A_i
[/math], we have that every element of [math]B
[/math] is in one of the [math]B \href{/cs2800/wiki/index.php/%E2%88%A9}{∩} A_i
[/math].
Therefore, we can apply the third Kolmogorov axiom to conclude
[math]\begin{aligned}
\href{/cs2800/wiki/index.php/Pr}{Pr}(B)
&= \href{/cs2800/wiki/index.php/Pr}{Pr}((B \href{/cs2800/wiki/index.php/%E2%88%A9}{∩} A_1) \href{/cs2800/wiki/index.php/%E2%88%AA}{∪} \cdots \href{/cs2800/wiki/index.php/%E2%88%AA}{∪} (B \href{/cs2800/wiki/index.php/%E2%88%A9}{∩} A_n)) && \text{as argued above} \\
&= \href{/cs2800/wiki/index.php/Pr}{Pr}(B \href{/cs2800/wiki/index.php/%E2%88%A9}{∩} A_1) + \cdots + \href{/cs2800/wiki/index.php/Pr}{Pr}(B \href{/cs2800/wiki/index.php/%E2%88%A9}{∩} A_n) && \href{/cs2800/wiki/index.php/Kolmogorov_axiom}{\text{Kologorov's third axiom}} \\
&= \href{/cs2800/wiki/index.php/Pr}{Pr}(B \href{/cs2800/wiki/index.php/%5Cmid}{\mid} A_1)\href{/cs2800/wiki/index.php/Pr}{Pr}(A_1) + \cdots + \href{/cs2800/wiki/index.php/Pr}{Pr}(B \href{/cs2800/wiki/index.php/%5Cmid}{\mid} A_n)\href{/cs2800/wiki/index.php/Pr}{Pr}(A_n) && \href{/cs2800/wiki/index.php/Conditional_probability}{\text{by definition of }Pr(B \mid A)}
\end{aligned}
[/math]
as required.