Prime Ideal in Lattice

Theorem
If $(L,\le)$ is a lattice and $I$ is an ideal in $L$, then

$I$ is a prime ideal if and only if for all $a$ and $b$ in $L$, $a \wedge b \in I$ implies that $a \in I$ or $b \in I$.

Proof
Suppose that $I$ is a prime ideal, $a,\,b \in L$, $a \notin I$, and $b \notin I$.

Then $a,\, b \in L \setminus I$

By the definition of prime ideal, $L \setminus I$ is a filter.

By the definition of a filter, $a \wedge b \in L \setminus I$.

Thus $a \wedge b \notin I$.

Contrapositively, if $a \wedge b \in I$, it must hold that $a \in I$ or $b \in I$.

Suppose that for all $a$ and $b$ in $L$, $a \wedge b \in I$ implies that $a \in I$ or $b \in I$.

Let $x,\,y \in L \setminus I$.

Then $x,\, y \notin I$.

By supposition, $x \wedge y \notin I$, so

$x \wedge y \in L \setminus I$.

Let $p \in L \setminus I$, let $q \in L$, and suppose that $p \le q$.

Then if $q \in I$, the definition of an ideal would imply that $p \in I$, a contradiction, so $q \notin I$.

Thus $q \in L \setminus I$.

Therefore, $L \setminus I$ is a filter.