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    HA cannot prove ∃yG(y) because that would assert HA prove... — 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.

    HA cannot prove ∃yG(y) because that would assert HA proves (0=1), contradicting the consistency assumption together with the fact that HA proves all true quantifier-free sentences.

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    Modality & PossibilityTruth & Knowledge

    Key Terms

    0=1(as the ultimate contradiction the system would produce)
    A statement that is obviously false in mathematics; it's used as a symbol for logical contradiction or absurdity.
    Consistency assumption(the key assumption being preserved in the argument)
    The assumption that a logical system doesn't contradict itself — that it never proves both a statement and its opposite at the same time.
    G(y)

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    Browse more in Modality & Possibility
    Related propositions within the same area of thought.
    (in the notation ∃yG(y))
    A placeholder for some property or formula that applies to a thing — think of it like a test where you plug in different values to see if they pass.
    HA(Constructive mathematics)
    Heyting Arithmetic — the standard formal system for intuitionistic/constructive arithmetic
    Quantifier-free sentences(the type of sentences HA is known to prove correctly)
    Mathematical statements that don't use 'there exists' or 'for all' — basically, statements you can check by plugging in specific numbers rather than reasoning about infinite collections.
    ∃y (existential quantifier)(as used in formal logic notation)
    A symbol meaning 'there exists at least one'—it says that at least one thing satisfying the condition is out there.

    Related

    By intuitionistic logic and the decidability of quantifier-free arithmetical for...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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    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...

    Similar

    Church's argument, if not blocked by devices like disallowing quantify...82%The consistency statement Cons(F) cannot be provable in F81%The set of valid sentences of most infinite quantifier languages is no...79%The Gödel sentence G_F is true (when F is consistent and the provabili...79%

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    SEP: logic-intuitionistic
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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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