Are there really open axioms?

Question

I still have doubts about the subject. I need help.

I do not think so. What exists, in fact, I think, are seemingly open axioms, with some implicit quantifier.

For example, when it is stated that x = x is an equality axiom, it seems to me that the following statement is implicit: for every x, x = x.

Am I right?

in math godel's incompleteness theorems might be something to look up about ( numberphile has a video on youtube). at least that's my first thought when thing open and axioms in the same place ( though I don't know the definitions myself). — , Sep 11 '17 at 22:53
@Paulo Argolo Where have you encountered the claim that there are open axioms? I am curious because I always think of a formula with a free variable as lacking a truth value. I would like to see a different way of thinking of it. — Jim H, Sep 11 '17 at 23:17
It is common to present the axioms for equational theories (e.g., the theory of groups) as a set of equations. These open formulas are intended to be read as universally quantified. — Rob Arthan, Sep 11 '17 at 23:22
@Jim H . As for the open axioms, see, for example, Principles of Mathematical Logic, by Hilbert & Ackermann, pages 68, 69, 107, 108, etc. A link: https://academiaanalitica.files.wordpress.com/2016/10/hilbert-david-the-principles-of-mathematical-logic.pdf — Paulo Argolo, Sep 11 '17 at 23:32
Yes; they are really open axioms. In the formula $(x=x)$ there is no quantifiers; thus, the variable $x$ is free. — Mauro ALLEGRANZA, Sep 12 '17 at 05:59
And YES: they are equivalent to the "closed" version, e.g. $\forall x \ (x=x)$. We can derive the first from the second through the valid formula: $\forall x \ (x=x) \to (x=x)$. And we can derive the second from the first through the suitable instance of the generalization theorem: "if $\Gamma \vdash \varphi$, then $\Gamma \vdash \forall x \varphi$ ($x$ not free in $\Gamma$). — Mauro ALLEGRANZA, Sep 12 '17 at 06:01
@Mauro ALLEGRANZA: I am curious where you encountered "the generalization theorem". In what kind of proof system? — beroal, Sep 12 '17 at 07:04
@beroal - Every one... in different forms. It is "built in" in Natural Deduction (see $\forall$-intro) and it is provable in Hilbert -style proof systems. — Mauro ALLEGRANZA, Sep 12 '17 at 07:08
IMHO, mathematicians use open axioms in a Hilbert system because, well, they are allowed. They use because they can. Nothing will change if you close axioms (add a universal quantifier for every free variable). You are asking what do open axioms "really" mean. It is a soft question until you specify what meaning you are talking about. Is it model theory? — beroal, Sep 12 '17 at 07:09
Some proof systems, like Mendelson's one has the Gen inference rule (originally due to Frege and Russell). — Mauro ALLEGRANZA, Sep 12 '17 at 07:11
Regarding the "meaning" of formula with free variables, see the post: meaning of free variables. — Mauro ALLEGRANZA, Sep 12 '17 at 07:13
@Mauro ALLEGRANZA: “It is "built in" in Natural Deduction” If it is built in, it is not a theorem. “it is provable in Hilbert -style proof systems” Sorry, the article that you linked does not claim that generalization is provable in Hilbert-style proof systems. The phrase "From (7) by Generalization" refers to the inference rule (also known as Gen).This is explained at the beginning of the article. If it is an inference rule, it is not a theorem. — beroal, Sep 12 '17 at 07:13
I cannot tell what is different from this post and a previous one, and so I have marked this one as a duplicate. Surely, if the answer to the previous post is "yes", then the answer to this post is trivial. — Carl Mummert, Sep 12 '17 at 11:51

score 1 · Answer 1 · answered Sep 12 '17 at 05:27

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Here’s where formal semantics really helps. I’ll assume we’re working in classical first-order logic with its standard model theory.

Assuming your axioms have a model, there is therefore an interpretation function in your model which among other things maps (or "assigns") each variable to a specific value. So let’s say the domain of your model is the natural numbers, and the interpretation function happens to map $x$ to the number $2$. So your open axiom really means $2 = 2$ in the semantics. So there’s no way to prove $3 = 3$ with that axiom. If you could, first-order logic would be unsound, which it isn’t, by the Soundness Theorem.

Here’s another way to go at it. Try using your axiom to prove $3 = 3$. You can’t apply your axiom because you can’t specialize $x$ to $3$ as you’d expect; only the universally quantified version of your axiom can do it: $\forall x(x = x)$.

So maybe you could write a theory with open axioms, but as you can see they aren't much use and don't capture what you're trying to capture.

answered Sep 12 '17 at 05:27

ShyPerson

1,720

But in every sound and complete proof system fo predicate logic with equality we can derive $\forall x (x=x)$ from $(x=x)$ (see e.g. Universal generalization). – Mauro ALLEGRANZA Sep 12 '17 at 07:27
@MauroALLEGRANZA Are you sure? The rule for $\forall$-intro is that the variable must not be present in an assumption, and axioms are about as assumptuous as it gets. Being able to derive $\forall x ~:~ x = x$ from $x=x$ would result in being unable to have a triple of axioms defining zero ($Z$) and one ($I$) like ${\forall x ~:~ x \cdot Z = Z,~ \forall x ~:~ x \cdot I = x,~Z \ne I}$ since the $Z$ variable (as well as the $I$ variable) would refer to a different variable in each axiom. – DanielV Sep 12 '17 at 12:24
@DanielV: this is one of the perpetually irritating issues. There are indeed systems in which $x = x \vdash (\forall x)[x = x]$, and others in which that logical implication does not hold. So different people learn different systems, and then they find the others surprising. Of course all the usual systems agree on when a set of sentences logically implies another sentence. But they differ for non-sentences. – Carl Mummert Sep 12 '17 at 13:33
Separately, in your set of axioms, the $Z$ is not a variable (as in $x = x$) but a constant (as in, part of the signature of the theory). In some systems, it is possible to generalize any variable, but not always possible to generalize a constant. – Carl Mummert Sep 12 '17 at 13:36
I think it is actually a bit more complicated than that even. For example, in a Hilbert System, it would be more correct to say $(\emptyset \vdash x = x) \vdash \forall x ~ (x = x)$, it is just happens that $(\emptyset \vdash x = x)$ doesn't hold since it isn't a FOL tautology. – DanielV Sep 12 '17 at 14:11
@DanielV: I am not familiar, in regular first order logic, with having a $\vdash$ on the left of the $\vdash$. When I wrote $x = x \vdash (\forall x)[x = x]$ I just meant that there is a derivation of $(\forall x)[x = x]$ from the axiom $x = x$. – Carl Mummert Sep 13 '17 at 02:50
@CarlMummert $(A \vdash B) \vdash C$ just means "if you have a proof $A \vdash B$ then you can infer $C$". Hilbert style logic tries to be a logic that only generates new theorems from existing theorems, but that breaks down when you have free variables b/c then the assumptions actually matter in addition to proven statements (unless you consider that a variable is unrestricted to be a "theorem"). – DanielV Sep 13 '17 at 02:55
To put it another way, the inference "from $A$, infer $\forall k~A$" only works if every proof of the logic is of the form $\emptyset \vdash P$. As soon as you introduce proofs of the form $Q \vdash P$, then $\forall$-intro has to be modified to take into account the presence free of variables in assumptions. So I actually somewhat doubt there is reliable logic where $x = x \vdash \forall x ~ x = x$ is present as directly stated. I think the correct answer to O.P. question is "$x = x$ is not a correct way of stating reflexivity" (except perhaps in something like PRA, not sure). – DanielV Sep 13 '17 at 03:02
@DanielV: The system presented in Mendelson's well known textbbook does include an inference rule that allows $(\forall x) A$ to be derived from $A$ without any restrictions. In the 4th edition the two inference rules he uses are on page 70: modus ponens and unrestricted generalization "Gen". He then modifies the deduction theorem (p. 74) to handle the resulting issues. But using Gen does not require the deduction theorem; his system is sound and complete and allows us to generalize any formula on any variable. – Carl Mummert Sep 13 '17 at 03:04
Then I suspect the logic described in Mendelson (I don't have a copy unfortunately) does not allow for non-tautological axioms, and $x = x$ is not a FOL tautology. – DanielV Sep 13 '17 at 03:10
@DanielV: instead, he changes the definition of what it means for a model to satisfy a formula with free variables; essentially $M \vDash \phi$ means that for every variable assignment $v$, $M$ satisfies $\phi$ under that assignment. So $M \vDash R(x)$ means that every element of $M$ satisfies $R$, hence in that case $M$ also satisfies $(\forall x)R(x)$, and so the Gen rule is sound. I don't work much in equational logic or universal algebra, but I believe open axioms are more common in those areas. – Carl Mummert Sep 13 '17 at 03:15
This is different, of course, from the convention where a model includes a variable assignment, in which case a model with a particular assignment could satisfy $R(x)$ without satisfying $(\forall x)R(x)$. In this other convention, of course, Gen would not be sound. – Carl Mummert Sep 13 '17 at 03:16

Are there really open axioms?

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