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    If Δ has an infinite model, then σ has an infinite model — Carmelics
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    Challenges→Any sentence σ equivalent to a conjunction of first-order sentences that has models of all finite cardinalities must also have an infinite model, contradicting the assumption that σ characterizes only finite structures.

    If Δ has an infinite model, then σ has an infinite model

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    Any sentence σ equivalent to a conjunction of first-order sentences that has mod...If Δ has models of arbitrarily large finite cardinalities, the compactness theor...The compactness theorem for sets of first-order sentences states that if every f...σ (and hence Δ) has models of all finite cardinalities
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    σ is equivalent to the conjunction of the set Δ = {σi : i ∈ I}, where each σi is...

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    σ (and hence Δ) has models of all finite cardinalities83%Any sentence σ equivalent to a conjunction of first-order sentences th...83%Non-standard models of F must contain 'infinite' non-natural numbers b...76%A static Newtonian model involves an infinite container with an infini...76%

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    where each Qk is ∀ or ∃ and φi is a (possibly infinitary) conjunction or disjunction of formulas of the form xk = xl or xk ≠ xl. Since each σi is a sentence, there are only finitely many variables in each φi, and it is easy to see that each φi is then equivalent to a first-order formula. Accordingly each σi may be taken to be a first-order sentence. Since σ is assumed to be equivalent to the conjunction of the σi, it follows that σ and the set Δ = {σi : i ∈ I} have the same models. But obviously

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