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    By intuitionistic logic and the decidability of quantifie... — Carmelics
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    Challenges→The rule 'If ∀x(A ∨ B(x)) is a theorem of HA, then A ∨ ∀xB(x) is a theorem of HA' is not admissible for HA, assuming HA is consistent.

    By intuitionistic logic and the decidability of quantifier-free arithmetical formulas, HA proves ∀x(∃yG(y) ∨ ¬G(x)).

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    By the disjunction property of HA, HA would then prove either ∃yG(y) or ∀x¬G(x).Gödel numbering provides a quantifier-free formula G(x) that numeralwise express...HA cannot prove ∀x¬G(x) because that formula expresses the consistency of HA, an...

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    HA cannot prove ∃yG(y) because that would assert HA proves (0=1), contradicting ...
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    If the rule were admissible, HA would prove ∃yG(y) ∨ ∀x¬G(x).The rule 'If ∀x(A ∨ B(x)) is a theorem of HA, then A ∨ ∀xB(x) is a theorem of HA...

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    If A contains no disjunction (∨) or existential quantifier (∃), then ¬...82%For every formula A, g(A) is provable intuitionistically if and only i...80%Prawitz's conjecture — that intuitionistic logic is complete in the se...80%If B ∧ ¬B were classically provable for some formula B, then g(B) ∧ ¬g...80%

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    Here is a proof that the rule “If \(\forall x (A \vee B(x))\) is a theorem, so is \(A \vee \forall x B(x)\)” (where \(x\) is not free in \(A)\) is not admissible for \(\mathbf{HA},\) if \(\mathbf{HA}\) is consistent. Gödel numbering provides a quantifier-free formula \(G(x)\) which (numeralwise) expresses the predicate “\(x\) is the code of a proof in \(\mathbf{HA}\) of \((0 = 1).\)” By intuitionistic logic with the decidability of quantifier-free arithmetical formulas, \(\mathbf{HA}\) proves \(

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