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I have a basic question about the use of judgment-level negation $\nvdash$. Though I meet it in some of my courses on proof theory, I usually treat it as a metalevel expression and I don't know how to handle it if it appears in the object language. The concrete question that I meet is as follows:

Here are two “statements”:

  • (a) $\Gamma\nvdash\neg\exists xA(x)$.
  • (b) for some $x$, $\Gamma\nvdash\neg A(x)$.

Assuming that we are working with classical logic, are (a) and (b) logically equivalent? Thanks in advance!

Taroccoesbrocco
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ferdinand
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1 Answers1

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I assume that $x$ does not occur free in $\Gamma$.

Yes, it is equivalent to say

  1. $\Gamma \not\vdash \lnot \exists x A(x)$
  2. for some $x$, $Γ⊬¬A(x)$

Both of them mean that there is a model of $\Gamma$ and $\exists x A(x)$. Roughly, it means that it is possible to make $\Gamma$ and $\exists x A(x)$ true simultaneously.

Indeed, $\Gamma \not\vdash \lnot \exists x A(x)$ means that $\lnot \exists x A(x)$ is not provable from the hypothesis $\Gamma$, which amounts to say that there is a model of $\Gamma$ and $\exists x A(x)$.

Under the assumption that $x$ is not free in $\Gamma$, $\Gamma \vdash \lnot A(x)$ means that $\lnot A(x)$ is provable from the hypothesis $\Gamma$, for any $x$. It amounts to say that $\Gamma \vdash \forall x \lnot A(x)$.

Therefore, saying that $\Gamma \not\vdash \lnot A(x)$ for some $x$ (i.e. negating that $\Gamma \vdash \lnot A(x)$ for any $x$) means that $\Gamma \not\vdash \forall x \lnot A(x)$, which amounts to say that there is a model of $\Gamma$ and $\lnot \forall x \lnot A(x)$, i.e. there is a model of $\Gamma$ and $\exists x A(x)$.

Taroccoesbrocco
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  • Thanks a lot! I see! – ferdinand Oct 21 '20 at 21:13
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    @ferdinand - You're welcome! If everything is clear, please mark it as accepted, in the interests of site hygiene. Otherwise, don't hesitate to ask for more explanation! – Taroccoesbrocco Oct 22 '20 at 06:14
  • Please also take a look at Kelly's question, which appears to be similar to mine. But I am not sure how to answer it. Here is the link: https://math.stackexchange.com/questions/3879093/first-order-de-morgans-laws-for-nvdash – ferdinand Oct 24 '20 at 21:46
  • @ferdinand - It seems to me that there is already a nice answer there. – Taroccoesbrocco Oct 24 '20 at 22:47