Past exam of the mathematics course of the University of Cambridge 2023 iii Paper 120 1 d Solution 2026-09-28
Both sequents follow directly from the introduction and elimination rules of the implication-free fragment of intuitionistic propositional logic. From a proof of , eliminate the conjunction to obtain and . Eliminate the disjunction: in the branch introduce and then the left disjunct; in the branch introduce and then the right disjunct. This yieldsConversely, eliminate the outer disjunction. From , obtain and introduce the left side of ; from , obtain and introduce its right side. In either branch, conjunction introduction produces . HenceThis is the proof-theoretic form of the distributive law for lattices.
Past exam of the mathematics course of the University of Cambridge 2023 iii Paper 120 3 e Solution 2026-09-28
Let be any distributive lattice. Its Stone map of a distributive latticeis an injective lattice homomorphism from 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 , , finite meets and finite joins.
Now assume that an implication-free formula is valid under every lattice valuation in every topological space. Given any valuation of its variables in any distributive lattice , compose it with the Stone map. Topological validity says that the resulting value of is the whole Priestley space. Injectivity of the Stone map then says that the original value of was . Hence is valid in every distributive lattice.