Prove: $A \cap(B\cup C^*)=(A\cap B)\cup C^*$ How do you prove this mathematically, when $C^*$ is the complement of C? I know from drawing a Venn diagram that this equation should hold. 
$A \cap(B\cup C^*)=(A\cap B)\cup C^*$
Thanks!
 A: Hints:
$A \cap(B\cup C^*)=(A\cap B)\cup (A\cap C^*)$
So if $A \cap C=\emptyset$, then $A \subset C^*$ and the equality is holds.
A: The "sensible" statement to consider is:

Conjecture. For all sets $X$, the power set of $X$ satisfies the
  following identity.
$$A∩(B∪C)=(A∩B)∪C$$

This is false.
E.g. take $A=\emptyset_X$ and $C=X$. Then $A \cap (B \cup C) = \emptyset_X$, while $(A \cap B) \cup C = X$.
However, it can be salvaged by adjoining the condition that $C \subseteq A,$ which isn't altogether surprising since this rules out the case $A=\emptyset_X$ and $C=X,$ as well as all similar cases.

Proposition. For all sets $X$, the power set of $X$ satisfies the following quasi-identity.
$$C \subseteq A \rightarrow A∩(B∪C)=(A∩B)∪C$$

Remark. In the language of lattice theory, this just says that the power set of $X$ is always a modular lattice.
Proof. Consider a set $X$ and subsets $A,B,C$. Assume $C \subseteq A$. Then $A \cap C = C.$ So we may argue as follows:
$$A∩(B∪C)=(A \cap B) \cup (A \cap C) = (A \cap B) \cup C$$
Of course, this is just the proof that every distributive lattice is modular.
