Intersection of Lower Closure of Element with Ideal equals Meet of Element and Ideal

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

Let $I$ be an ideal in $\left({S, \preceq}\right)$

Let $x \in S$.

Then
 * $\left({x^\preceq}\right) \cap I = \left\{ {x \wedge i: i \in I}\right\}$

where $x^\preceq$ denotes the lower closure of $x$.

First Inclusion
Let $z \in \left({x^\preceq}\right) \cap I$

By definition of intersection:
 * $z \in x^\preceq$ and $z \in I$

By definition of lower closure of element:
 * $z \preceq x$

By Preceding iff Meet equals Less Operand:
 * $x \wedge z = z$

Thus
 * $z \in \left\{ {x \wedge i: i \in I}\right\}$

Second Inclusion
Let $z \in \left\{ {x \wedge i: i \in I}\right\}$

Then
 * $\exists i \in I: z = x \wedge i$

By Meet Precedes Operands:
 * $z \preceq x$ and $z \preceq i$

By definition of ideal in ordered set:
 * $I$ is a lower set.

By definitions of lower set and lower closure of element:
 * $z \in I$ and $z \in x^\preceq$

Thus by definition of intersection:
 * $z \in \left({x^\preceq}\right) \cap I$

Thus the result by definition of set equality.