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With this deduction rule, in the premise of the rule: the term to be substituted for a variable must be arbitrary (refer to an arbitrary d $\in$ D).

What constitutes arbitrary and not arbitrary?

  1. $ P(a) \quad\quad Premise$
  2. $ \forall x P(x) \quad (1), \forall-I: a/x$

Would term john not be considered arbitrary and thus line 2 be incorrect?

  1. $ P(john) \quad Premise$
  2. $ \forall x P(x) \quad (1), \forall-I: john/x$
0implies0
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  • It might be worth noting that in some formalizations, the fundamental encoding of propositions is using de Bruijn indices or something similar, and the version with variable names is considered to be merely syntactic sugar for human readability. In such a formalization, the ${\forall}I$ rule might look something like: if $\Gamma {\uparrow} \vdash \phi$ then $\Gamma \vdash \forall \cdot, \phi$. Here $p{\uparrow}$ is notation for $p[v_0 := v_1, v_1 := v_2, \ldots]$ and then $\Gamma{\uparrow} := { p{\uparrow} \mid p \in \Gamma }$. – Daniel Schepler Nov 22 '20 at 00:15
  • i.e. $\Gamma\uparrow$ denotes renumbering all the free variables in assumptions in the context $\Gamma$ to "make room for" a new variable. – Daniel Schepler Nov 22 '20 at 00:16
  • @DanielSchepler: While de Bruijn indices are certainly an interesting way of dealing with the issue of variable scoping, I don't think it's at the appropriate level here? =) – user21820 Nov 22 '20 at 06:54

2 Answers2

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First of all, I hope you understand the intuition behind this:

Just because some specific object has some property obviously does not mean that all objects from the domain have that property.

However, if an arbitrary object from the domain has some property, then all objects do.

And just to be clear: by 'arbitrary' object we mean: we know and have assumed nothing about this object other than that it is some object from the domain.

Now, how exactly this is being formalized in a specific formal system depends on a lot of formal details. In some systems, variables are used to denote arbitrary objects, but in other systems, 'temporary constants' are being used, typically in combination with certain kinds of subproofs.

So, if you ask me if whether you can apply $\forall \ I$ to infer $\forall x \ P(x)$ from $P(John)$, I really cannot answer that; it all depends on the specifics of the system you are using.

Bram28
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  • Thanks for this, I understand why it needs to arbitrary. The intent of my question was how do I know if a term is arbitrary or not. From your definition of arbitrary: I think I see now that the term john is not just an object from the domain, it is a specific object from the domain. – 0implies0 Nov 20 '20 at 00:53
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    Yes, i figured that that was what you were asking, but just wanted to be sure. In the second part I addressed what I thought was your real question: how do you know, just by looking at some syntactical expression, whether or not it is used to denote some arbitrary object? And again, the answer is an emphatic "you don't .... unless you figure out how that expression is being used in the formal system as a whole". And sure, a string like $John$ certainly suggests that it denotes a specific individual ... but I can tell you that there are systems where it can be used for an arbitrary object. – Bram28 Nov 20 '20 at 01:08
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    Yes! You phrased my question better than I did. I see, in the system I am using lowercase character sequences like john, ann etc. will be constant and only refer to a single object in the domain. – 0implies0 Nov 20 '20 at 01:15
  • The natural deduction system itself would be the same as the one I was using for propositional logic with the addition of a pair of rules for {$\forall, \exists, = $} (https://math.stackexchange.com/questions/3885408/prove-a-vdash-neg-neg-a-natural-deduction). The notation is not widely used, I don't think. – 0implies0 Nov 20 '20 at 01:19
1

The $(\forall \text I)$ rule is:

if $\Gamma \vdash \varphi[x/a]$, then $\Gamma \vdash \forall x \varphi$, provided that parameter $a$a is “fresh” in the sense of having no other occurrences in $\Gamma , \varphi$

The proviso is consistent with the intuitive meaning of the rule: if $\varphi$ holds of an object $a$ whatever, then it holds of every object.

The proviso is needed in order to avoid the fallacy: John is a Philosopher, therefore everything is a Philospher.

In your wrong proof above, you have committed exactly these fallacy: the parameter $a$ [in your case: John] must not occur in $\Gamma$. In your case $\Gamma = \{ P(\text {John}) \}$.

In conclusion, the issue is: how can you prove $\vdash P(\text {John})$?

Example: consider the first-order language of arithmetic with individual constants $0$ and $1$ and let $\mathsf {PA}$ the collection of first-order Peano axiom.

We have: $\mathsf {PA} \vdash (0 \ne 1)$,

Now, applying $(\forall \text I)$ to it, using $0$ as $\text {John}$, we conclude with: $\mathsf {PA} \vdash \forall x (x \ne 1)$.

Where is the mistake ?

Mauro ALLEGRANZA
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