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The Stone prime filter theorem says that if a lattice filter $F$ and a lattice ideal $I$ of a distributive lattice are disjoint, then there is a prime filter of a distributive lattice $P$ such thatEquivalently, whenever $a\nleqq b$, there is a prime filter containing $a$ and omitting $b$.

For a prime filter $P\in \hat{H}$, first suppose $a\Rightarrow b\in P$. If $Q\supseteq P$ and $a\in Q$, then $a\Rightarrow b\in Q$ andso $b\in Q$. Thus no prime filter above $P$ belongs to $a^{\ast }\setminus b^{\ast }$, and

Conversely, suppose $a\Rightarrow b\notin P$. The lattice filter generated by $P\cup \{a\}$ is disjoint from the principal lattice ideal $\downarrow b$. Indeed, an intersection would give some $p\in P$ with $p\wedge a\le b$, whence $p\le a\Rightarrow b$ and then $a\Rightarrow b\in P$, a contradiction. The Stone prime filter theorem therefore extends this filter to a prime filter $Q\supseteq P$ that omits $b$. Then $Q\in a^{\ast }\setminus b^{\ast }$, so $P\in \uparrow (a^{\ast }\setminus b^{\ast })$.

We have proved, for every $P$,which is the required identity

The Stone map sends an element $a$ of a distributive lattice to the clopen up-set $a^{\ast }=\{P:a\in P\}$ of its Priestley dual space. The Stone prime filter theorem makes this lattice homomorphism injective.