Meet in Inclusion Ordered Set

Theorem
Let $P = \left({X, \subseteq}\right)$ be an inclusion ordered set.

Let $A, B \in X$ such that
 * $A \cap B \in X$

Then $A \wedge B = A \cap B$

Proof
By Intersection is Subset:
 * $A \cap B \subseteq A$ and $A \cap B \subseteq B$

By definition:
 * $A \cap B$ is lower bound for $\left\{ {A, B}\right\}$

We will prove that
 * $\forall C \in X: C$ is lower bound for $\left\{ {A, B}\right\} \implies C \subseteq A \cap B$

Let $C \in X$ such that
 * $C$ is lower bound for $\left\{ {A, B}\right\}$.

By definition of lower bound:
 * $C \subseteq A$ and $C \subseteq B$

Thus by Intersection is Largest Subset:
 * $C \subseteq A \cap B$

By definition of infimum:
 * $\inf \left\{ {A, B}\right\} = A \cap B$

Thus by definition of meet:
 * $A \wedge B = A \cap B$