# How do you prove that $P(E_1 \cup E_2^C |E_3 \cap E_4) = P(E_1 \cup E_2^C)$

The question was to prove that $$P(E_1 \cup E_2^C |E_3 \cap E_4) = P(E_1 \cup E_2^C)$$ when $$E_1,E_2,E_3,E_4$$ are independent events. What I know and can prove is that $$P(E_1 \cap E_2) = P(E_1) P(E_2)$$, $$P(E_1 \cap E_2^c) = P(E_1) P(E_2^c)$$ and $$P(E_1 | E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)} = \frac{P(E_1)P(E_2)}{P(E_2)} = P(E_1)$$

From here, my attempt was that at first I used De Morgan's law to transform $$P(E_1 \cup E_2^C) = P(E_1^c \cap E_2)^c = (P(E_1^c)P(E_2))^c$$

From here I am kind of lost how to proceed.

Alternatively,

$$P(E_1 \cup E_2^C |E_3 \cap E_4) = \frac{P(E_1 \cup E_2^C \cap E_3 \cap E_4)}{P(E_3 \cap E_4)} = \frac{P(E_1 \cup E_2^C)P(E_3)P(E_4)}{P(E_3)P(E_4)} = P(E_1 \cup E_2^C)$$

But I thought this would be too simple.. and also even though it is independent, I don't know whether $$P(E_1 \cup E_2^C)$$ are independent with $$P(E_3 \cap E_4)$$. I would appreciate some help on this.

Hint: $$P((E_1\cup E_2^{c}) \cap E_3\cap E_4)$$ $$=P((E_1\cap E_3 \cap E_4) \cup ( E_2^{c} \cap E_3 \cap E_4))$$ $$=P(E_1\cap E_3 \cap E_4) + P( E_2^{c} \cap E_3 \cap E_4)-P(E_1\cap E_3 \cap E_4 \cap E_2^{c}).$$ Note that $$E_1, E_3,E_4 , E_2^{c}$$ are independent too. So all these terms can be computed by applying definition of independence. I hope you can finish the proof using this computation.