# Show that if $(A\cap C)\subseteq (B\cap C)$ and $(A\cap \overline C)\subseteq (B\cap \overline C)$, then $A\subseteq B$

Show that if $(A\cap C)\subseteq (B\cap C)$ and $(A\cap \overline C)\subseteq (B\cap \overline C)$, then $A\subseteq B$.

Let $a \in A$ be arbitrary. We will show that $a \in B$. Suppose $a \in C$. Then $a \in A \cap C$, hence $a \in B \cap C$, which implies $a \in B$. Similarly, if $a \in \overline{C}$ it follows that $a \in B \cap \overline{C}$, hence $a \in B$. Since $a \in B$ for all $a \in A$, we conclude that $A$ is a subset of $B$ or $A \subseteq B$.
Assuming that $\overline{C}$ is the complement of $C$:
$$A = A \cap (C \cup \overline{C}) = (A \cap C) \cup (A \cap \overline{C}) \subseteq (B \cap C) \cup (B \cap \overline{C}) = B \cap (C \cup \overline{C}) = B$$