Past exam of the mathematics course of the University of Cambridge 2024 ii Paper 1 16I iii Created 2026-09-23 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2024 ii Paper 1 16I ii Created 2026-09-23 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2024 ii Paper 1 16I i Created 2026-09-23 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2024 ii Paper 1 16I Solution Created 2026-09-23 Updated 2026-09-25
The soundness theorem for propositional logic says that if , then every valuation satisfying every member of also satisfies .
The proposed function need not respect the connectives. For example, if a primitive proposition is independent of , then neither nor is provable from , so the definition gives
Order the consistent supersets of by inclusion. The union of a chain is consistent, since a finite proof of a contradiction would already use assumptions from one member of the chain. Zorn lemma therefore gives a maximal consistent extension . It is deductively closed: if , adjoining preserves consistency, so maximality forces . Moreover, for every , exactly one of and belongs to . They cannot both belong by consistency; if neither belonged, the inconsistency of both proper extensions would give and , again a contradiction. The usual induction on formulae now shows thatdefines a valuation satisfying , and hence .
Now suppose every finite subset of has a model. By soundness every finite subset is consistent. Any proof of a contradiction from uses only finitely many assumptions, so itself is consistent. Applying the preceding maximal-consistent-extension construction gives a model of . This proves the propositional compactness theorem.
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