Finite Subset Bounds Element of Finite Infima Set and Upper Closure

Theorem
Let $L = \struct {S, \wedge, \preceq}$ be meet semilattice.

Let $F$ be filter in $L$.

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

Let $x \in S$ such that
 * $x \in \paren {\map {\operatorname{fininfs} } {F \cup X} }^\succeq$

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

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

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

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 difference:
 * $a \in Y$ 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 $Z = \O$:

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:
 * $Y \subseteq X$

By Existence of Non-Empty Finite Infima in Meet Semilattice:
 * $X$ admits an infimum.

By Infimum of Subset:
 * $\inf X \preceq u$

Thus by definition of trantisivity:
 * $b \wedge \inf X \preceq x$

Case $Z \ne \O$:

By Set Difference is Subset:
 * $Z \subseteq Y$

By Subset of Finite Set is Finite:
 * $Z$ is finite.

By Union of Finite Sets is Finite:
 * $Z \cup X$ is finite.

By definitions of non-empty set and union:
 * $Z \cup X \ne \O$

By Existence of Non-Empty Finite Infima in Meet Semilattice:
 * $X$ admits an infimum

and
 * $Z \cup X$ admits an infimum.

By Set is Subset of Union:
 * $Y \subseteq Y \cup X$

By Set Difference Union Second Set is Union:
 * $Y \subseteq Z \cup X$

By Infimum of Subset:
 * $\map \inf {Z \cup X} \preceq u$

By Existence of Non-Empty Finite Infima in Meet Semilattice:
 * $Z$ admits an infimum

By Infimum of Infima:
 * $\map \inf {Z \cup X} = \paren {\inf Z} \wedge \inf X$

By Filtered in Meet Semilattice with Finite Infima:
 * $\inf Z \in F$

By definition of transitivity:
 * $\paren {\inf Z} \wedge \inf X \preceq x$

Thus
 * $\exists a \in S: a \in F \land x \succeq a \wedge \inf X$