$P(A_1 \cap A_2 |B)=P(A_1|B)P(A_2|B)$ implies $P(A_1|A_2 \cap B)=P(A_1|B)$

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i want to prove this:

$P(A_1 \cap A_2 /B)=P(A_1/B)P(A_2/B) $$\Rightarrow $$P(A_1/A_2 \cap B)=P(A_1/B)$

i did for $A_2 \cap B \neq \emptyset$ .but i am stuck in case $A_2 \cap B = \emptyset$

i want to see why $P(A_1 \cap A_2 /B)=P(A_1/B)P(A_2/B) $means that $A_1$ and $A_2$ are conditionally independent given $B$.

can you help me?