Meet is Intersection in Set of Ideals

Theorem
Let $\mathscr S = \left({S, \wedge, \preceq}\right)$ be a meet semilattice.

Let $\mathit{Ids}\left({\mathscr S}\right)$ be the set of all ideals in $\mathscr S$.

Let $P = \left({\mathit{Ids}\left({\mathscr S}\right), \precsim}\right)$ be an ordered set where $\mathord\precsim = \mathord\subseteq\restriction_{\mathit{Ids}\left({\mathscr S}\right) \times \mathit{Ids}\left({\mathscr S}\right)}$

Let $I_1, I_2$ be ideals in $\mathscr S$.

Then
 * $I_1 \wedge_P I_2 = I_1 \cap I_2$

Proof
By Intersection of Ideals is Ideal:
 * $I_1 \cap I_2 \in \mathit{Ids}\left({\mathscr S}\right)$

We will prove that
 * $I_1 \cap I_2$ is lower bound for $\left\{ {I_1, I_2}\right\}$

Let $x \in \left\{ {I_1, I_2}\right\}$.

Then by definition of unordered tuple:
 * $x = I_1$ or $x = I_2$

By Intersection is Subset
 * $I_1 \cap I_2 \subseteq x$

Thus by definition of $\precsim$:
 * $I_1 \cap I_2 \precsim x$

We will prove that
 * $\forall I \in \mathit{Ids}\left({\mathscr S}\right): I$ is lower bound for $\left\{ {I_1, I_2}\right\} \implies I \precsim I_1 \cap I_2$

Let $I \in \mathit{Ids}\left({\mathscr S}\right)$ such that
 * $I$ is lower bound for $\left\{ {I_1, I_2}\right\}$

By definition of lower bound:
 * $I \precsim I_1$ and $I \precsim I_2$

By definition of $\precsim$:
 * $I \subseteq I_1$ and $I \subseteq I_2$

By Intersection is Largest Subset:
 * $I \subseteq I_1 \cap I_2$

Thus by definition of $\precsim$
 * $I \precsim I_1 \cap I_2$

By definition of infimum:
 * $\inf \left\{ {I_1, I_2}\right\} = I_1 \cap I_2$

Hence by definition of meet:
 * $I_1 \wedge I_2 = I_1 \cap I_2$