CompltIntuiLogic3[1].20.00

CompltIntuiLogic3[1].20.00 - COMPLETENESS OF INTUITIONISTIC...

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COMPLETENESS OF INTUITIONISTIC PROPOSITIONAL CALCULUS by Harvey Friedman March 20, 2000 This is detailed proof of results we obtained in the 1980’s. We give the uniform interpretation of HPC. We let HPC’, HQC’ be obtained by dropping negation and absurdity. Define the type structure S( j ) for every formula j of HPC’, using Cartesian powers, products, and disjoint unions. S(A) for atoms A is taken to be N. An assignment is a function f that assigns subsets of N to some atoms. Then f is extended to f* which sends every formula A of HPC to a subset of S(A). We will prove that for any formula j in HPC, j is provable in HPC if and only if the intersection of f*( j ) is nonempty. We begin by constructing a particular Kripke model that is complete for formulas j in HPC’. We say that a set X of formulas in HPC’ without absurdity is saturated if and only if it is nonempty, logically closed, and any given disjunction lies in X if and only if at least one of the disjuncts lies in X. Let K be the set of all saturated sets of formulas in HPC’. The particular Kripke model M has frame (P(K), ). We define forcing for atoms as follows. Let p K. 1) any two elements of p are comparable under inclusion. Then p forces an atom if and only if the atom lies in the union of Y; 2) otherwise. Then p forces every atom. LEMMA 1. For any nonempty p in M and j in HPC’, p forces j if and only if j lies in an element of p or p has incomparable elements. Proof: By induction on j . 1. j is an atom. By construction.
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2. j y . Suppose p forces j y . Then j lies in an element of p or p has incomparable elements, and y lies in an element of p or p has incomparable elements. If any two elements of p are comparable, then j , y lie in an element of p. Hence j , y lie in some common element of p, and so j y lies in some element of p.
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CompltIntuiLogic3[1].20.00 - COMPLETENESS OF INTUITIONISTIC...

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