Finite Subset Bounds Element of Finite Infima Set and Upper Closure

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

Let $F$ be filter in $L$.

Let $X$ be non empty finite subset of $S$.

Let $x \in S$ such that
 * $x \in \left({\operatorname{fininfs}\left({F \cup X}\right)}\right)^\succeq$

where
 * $\operatorname{fininfs}$ denotes the finite infima set,
 * $X^\succeq$ denotes the upper closure of $X$.

Then there exists $x \in S$: $a \in F \land x \succeq a \wedge \inf X$

Proof
By definition of upper closure of subset:
 * $\exists u \in \operatorname{fininfs}\left({F \cup X}\right): u \preceq x$

By definition of finite infima set:
 * there exists finite subset $Y$ of $F \cup X$:
 * $Y$ admits an infimum and $u = \inf Y$

We will prove that
 * $Y \setminus X \subseteq F$

Let $a \in Y \setminus X$.

By definition of defference:
 * $a \in Y$ and $a \notin X$

By definition of subset:
 * $a \in F \cup X$

Thus by definition of union:
 * $a \in F$

Define $Z := Y \setminus X$.

Case $Z = \varnothing$:

By $F$ is non-empty:
 * $\exists b: b \in F$

By Meet Precedes Operands:
 * $b \wedge \inf X \preceq \inf X$

By Set Difference with Superset is Empty Set:
 * $Y \subseteq X$