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I suspect the following holds: $$\vdash \phi(x,y,\ldots,z)\iff \vdash\forall x\forall y\cdots\forall z\phi(x,y,\ldots,z)$$ where $\phi$ is a formula with free variables $x,y,\ldots, z$. Here is my argument using Natural Deduction :

$(\Rightarrow):$ $\vdash\phi(x,y,\ldots,z)$ means $\vdash\phi(x',y',\ldots,z')$ for new variables $x',y',\ldots,z'$. Now we may use $\forall$-introduction:

$$\frac{\dfrac{\phi(x',y',\ldots,z')}{\forall x\phi(x,y',\ldots,z')}}{\dfrac{\vdots}{\forall x\forall y\cdots\forall z \phi(x,y,\ldots,z)}}$$

$(\Leftarrow):$ using $\forall$-elimination repeatedly:

$$\frac{\dfrac{\forall x\forall y\cdots\forall z \phi(x,y,\ldots,z)}{\forall y\cdots\forall z \phi(x,y,\ldots,z)}}{\dfrac{\vdots}{\phi(x,y,\ldots,z)}}$$

Is the proof correct?

Sam
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  • The first sequent isn't well-scoped. If you mean $x \vdash \phi(x) \iff \vdash \forall x. \phi(x)$ then that is just the defining adjunction for $\forall$. – Naïm Camille Favier Nov 21 '24 at 12:12
  • @NaïmFavier I do not understand what the $x$ being at the left side of the $\vdash$ means. – Sam Nov 21 '24 at 12:28
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    Yes it works... What is the difference having $n$ variables instead of one? We have only to reiterate the application of the Universal Quantifier rules: $(\forall \text E)$ has no restriction (we have only to prevent "capturing") and $(\forall \text I)$ has the restriction that we cannot generalize a variable taht is free in some assumption: but $\vdash \phi(x)$ has no assumptions. – Mauro ALLEGRANZA Nov 21 '24 at 12:29
  • In predicate logic we have the GENERALIZATION THEOREM If $\Gamma \vdash \phi$ and x do not occur free in any formula in $\Gamma$, then $\Gamma \vdash \forall x \phi$. An the other direction is simply INSTANTIATION: $∀ x α \vdash α[t/x]$, where t is substitutable for x in α. – Mauro ALLEGRANZA Nov 21 '24 at 12:33
  • @MauroALLEGRANZA you are right: I could have proved it for a single variable, from which the result (as I've written it) follows from repetition. – Sam Nov 21 '24 at 13:42
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    @NaïmFavier There's nothing wrong here. It's just a formula with free variables and we may be able to find a derivation of it - namely just in case its universal closure is also provable. That's the point of what's being shown here. – Natalie Clarius Nov 21 '24 at 15:17
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    Your suggested $x \vdash ...$ doesn't even make sense in the system the OP is working with. – Natalie Clarius Nov 21 '24 at 15:18
  • @NatalieClarius I am suggesting that OP leave these issues in the last century and start using a system that makes sense. – Naïm Camille Favier Nov 21 '24 at 15:25
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    I'm suggesting that you leave your commentary on your personal opinions aside and try to answer the actual question being asked, or if that's not what you came here to do, then just say nothing and move on. The OP is trying to understand this particular deduction system and within this context their question is perfectly valid. You've made it clear elsewhere that you don't like this system, but that's beside the point and not helpful here. It's likely to only cause confusion at this stage. – Natalie Clarius Nov 21 '24 at 15:39
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    @Sam, your proof is correct. – Natalie Clarius Nov 21 '24 at 15:39

2 Answers2

1

Your proof is correct. But there is some sentences have to be formalized.

($\Rightarrow$) Suppose that $\vdash \phi(x, y, ..., z)$. We want to show that $\vdash \forall x \forall y ... \forall z \phi(x, y, ..., z).$ Let $x', y', ..., z'$ be new variables. By assumption, we have $\vdash \phi(x', y', ..., z')$. We can then use $\forall$-introduction repeatedly to obtain $\vdash \forall x \forall y ... \forall z \phi(x, y, ..., z)$.

($\Leftarrow$) Suppose that $\vdash \forall x \forall y ... \forall z \phi(x, y, ..., z)$. We want to show that $\vdash \phi(x, y, ..., z)$. We can use $\forall$-elimination repeatedly to obtain $\vdash \phi(x, y, ..., z)$.

QED

  • In the ($\Rightarrow$) direction, we use the fact that $\vdash \phi(x', y', ..., z')$ holds for any new variables $x', y', ..., z'$. This is because $\vdash \phi(x, y, ..., z)$ means that $\phi(x, y, ..., z)$ is provable without any assumptions about the variables $x, y, ..., z$. Therefore, we can substitute any new variables for $x, y, ..., z$ and the proof will still be valid.
  • In the ($\Leftarrow$) direction, we use the fact that $\forall x \forall y ... \forall z \phi(x, y, ..., z)$ implies $\phi(x, y, ..., z)$ for any variables $x, y, ..., z$. This is because $\forall x \forall y ... \forall z \phi(x, y, ..., z)$ means that $\phi(x, y, ..., z)$ holds for all possible values of $x, y, ..., z$. Therefore, it must hold for the particular values of $x, y, ..., z$ that we are interested in.
Crimson-bear
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0

Yes it works.

There is no "conceptual difference" for the $n$ variables case wrt the one quantifier case. We have only to reiterate the application of the Universal Quantifier rules: $(∀ \text E)$. The rule has no restriction (we have only to prevent "capturing") and $(∀ \text I)$ has the restriction that we cannot generalize a variable that is free in some assumption: but $\vdash \varphi (x)$ has no assumptions.

In predicate logic we have the GENERALIZATION THEOREM: "If $\gamma \vdash \varphi$ and $x$ does not occur free in any formula in $\Gamma$, then $\Gamma \vdash ∀x \varphi$.

And we have INSTANTIATION: $∀xα \vdash α[t/x]$, where $t$ is substitutable for $x$ in $\alpha$.

Mauro ALLEGRANZA
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