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In a strong enough meta-logic such as ZFC or CC, for any $L_{\Omega}$-theory $T$, is $\Omega \models T$ equivalent to that $T$ is $\omega$-consistent?

Here, $L_{\Omega}$ is a first-order language whose signature is $\left\langle + , \cdot , S , 0 , < \right\rangle$, and $\Omega$ is the standard model of $L_{\Omega}$.

Kijeong Lim
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    What is $L_\Omega$, and what is $\Omega$? Please edit your question to give more context. – Alex Kruckman Nov 09 '21 at 15:09
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    @AlexKruckman $L_{\Omega}$ is a first-order language whose signature is $\left\langle + , \cdot , S , 0 , < \right\rangle$, and $\Omega$ is the standard model of $L_{\Omega}$. – Kijeong Lim Nov 09 '21 at 15:13
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    Ok - Please edit your question to include these important details. I think this notation is very unusual: I've never seen the standard model ($\omega$ or $\mathbb{N}$) referred to as $\Omega$. – Alex Kruckman Nov 09 '21 at 15:15
  • @AlexKruckman Jobs done. I meant $\Omega = \left\langle \mathbb{N} , +{\mathbb{N}} , \times{\mathbb{N}} , S_{\mathbb{N}} , 0_{\mathbb{N}} , <_{\mathbb{N}} \right\rangle$, i.e., $\left| \Omega \right| = \omega$. – Kijeong Lim Nov 09 '21 at 15:24
  • See $\omega$ ordinal and $\omega$-consistency – Mauro ALLEGRANZA Nov 09 '21 at 15:33
  • As you see from the linked def, the notation originates with $\omega$-inconsistency that is a sort of "inconsistency at infinity", that needs the capability of the theory to "represent" natural numbers. – Mauro ALLEGRANZA Nov 10 '21 at 11:44

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No. This is explained (but not very clearly, I think) on the Wikipedia page for $\omega$-consistent theory.

Let $L$ be a language containing $L_A = \{+,\cdot,S,0,<\}$, the language of arithmetic. Let $T$ be an $L$-theory.

  • We say $T$ is $\omega$-inconsistent if there is an $L$-formula $\varphi(x)$ in one free variable $x$ such that $T\vdash \varphi(n)$ for every term $n = S(S(\dots S(0)\dots ))$ representing a natural number, but $T\vdash \lnot \forall x\,\varphi(x)$. Otherwise, we say $T$ is $\omega$-consistent.
  • $\omega$-logic is first-order logic extended with the infinitary rule with hypotheses $\varphi(0)$, $\varphi(1)$, $\varphi(2)$, $\dots$ and conclusion $\forall x\, \varphi(x)$. We say $T$ is consistent in $\omega$-logic if $T$ does not prove a contradiction in $\omega$-logic.
  • We say $T$ has an $\omega$-model if there is a model $M\models T$ such that the underlying set of $M$ is $\omega$, and the symbols of $L_A$ are interpreted in the standard way on $\omega$. In the setting of your question, where $T$ is an $L_A$-theory, the only $\omega$ model is $\omega$ itself (with the symbols of $L_A$ interpreted in the standard way). So $T$ has an $\omega$-model if and only if $\omega\models T$.

Now $T$ has an $\omega$-model if and only if $T$ is consistent in $\omega$-logic, and these conditions imply that $T$ is $\omega$-consistent. But the converse is not true: there are $\omega$-consistent theories $T$ which are not consistent in $\omega$-logic, and hence have no $\omega$-model. An example is given in the section "Arithmetically unsound, $\omega$-consistent theories" on Wikipedia: $\mathsf{PA}+\lnot \omega\mathsf{-ConPA}$, where $\omega\mathsf{-ConPA}$ is the sentence asserting that $\mathsf{PA}$ is $\omega$-consistent.

See also the discussion on this question and this question.

Alex Kruckman
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  • Thank you! My question started by wondering why the term '$\omega$-consistency' refers to $\omega$-consistency. Now, I conclude: as consistency is to the guarantee of a model, $\omega$-consistency is to the guarantee of an $\omega$-model. – Kijeong Lim Nov 09 '21 at 16:08