Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 120 / 3

A prime filter of a distributive lattice $P$ is a proper lattice filter: it contains the top element, is upward closed, and is closed under finite meets. Primality means

The Priestley dual space $\hat{L}$ of a distributive lattice $L$ has all prime filters of $L$ as its points. Its order is inclusion. For each $a\in L$, putThe topology is generated by the sets $a^{\ast }$ and their complements. Each $a^{\ast }$ is therefore a clopen up-set, and these sets separate points and order. With this topology and order, $\hat{L}$ is a compact totally order-disconnected ordered space.

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

Let $L$ be any distributive lattice. Its Stone map of a distributive latticeis an injective lattice homomorphism from $L$ into the lattice of clopen up-sets of its Priestley dual space. In particular these images are open subsets of the underlying topological space, and the map preserves $\perp$, $\top$, finite meets and finite joins.

Now assume that an implication-free formula $\varphi$ is valid under every lattice valuation in every topological space. Given any valuation of its variables in any distributive lattice $L$, compose it with the Stone map. Topological validity says that the resulting value of $\varphi$ is the whole Priestley space. Injectivity of the Stone map then says that the original value of $\varphi$ was $1_L$. Hence $\varphi$ is valid in every distributive lattice.

By completeness of implication-free intuitionistic propositional logic for distributive lattices, $\varphi$ is provable in the implication-free fragment of intuitionistic propositional logic.