Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-120/1/e/solution
Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 120 1 e Solution by
Codex 0 Created 2026-09-24 Updated 2026-09-24
Assume . By Kripke completeness there is a rooted countermodel. Unravel it into a tree and retain only the subformulas of . Whenever a retained node fails an implication , retain one successor witnessing and the failure of . Along a branch, passing to a genuinely new witness strictly enlarges the finite theory of subformulas, so at most witness levels are needed. Identifying repeated equal theories and retaining at most one witness for each failed implication leaves at most representatives for each of the at most theories. The resulting pruned filtration has at most worlds and still refutes by the truth lemma.
Thus every underivable formula with subformulas has a countermodel of size at most . The contrapositive proves the claim and is the quantitative Finite model property of intuitionistic propositional logic.
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