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There must be a flaw in the following argument, but I don't see it at the moment. Who can point it out?

In first-order logic, suppose that a structure $\mathfrak{U}$ is a model of the formula ($x = 3$):

$$ \mathfrak{U} \vDash x = 3 $$

Then, by the definition of a model, for any assignment function s it is true that

$$ \mathfrak{U} \vDash (x = 3) [s]$$

So the universe of $\mathfrak{U}$ has only one element, which equals the constant 3. And so the structure is also a model of $\forall x (x = 3)$:

$$ \mathfrak{U} \vDash \forall x (x = 3) $$

We have just seen that any model of $x = 3$ is also a model of $\forall x (x = 3)$. So

$$ x = 3 \vDash \forall x (x = 3) $$

Choose any complete deduction system for first-order logic. Then by completeness we have

$$ x = 3 \vdash \forall x (x = 3) $$

There is a natural deduction rule

$$ \frac{\Gamma, A \vdash B}{\Gamma \vdash A \to B} $$

So we can apply it to obtain

$$ \varnothing \vdash x = 3 \to \forall x (x = 3) $$

or

$$ \vdash x = 3 \to \forall x (x = 3) $$

But that statement seems to be invalid. For example, it's not true in a structure whose universe has more than one element under an assignment that maps x to 3.

What am I missing here?

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  • $\begingroup$ Since it otherwise contains an unbound variable, isn't $x = 3$ implicitly $\forall x(x =3)$? $\endgroup$ Jun 12, 2022 at 13:42
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    $\begingroup$ The flaw is, unsurprisingly, when you say "presumably it is true...". $\endgroup$ Jun 12, 2022 at 14:26
  • $\begingroup$ Should be $\exists x (x=3)$ $\endgroup$ Jun 12, 2022 at 15:19
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    $\begingroup$ Depending on the particular formulation of first logic, the exact error you have may vary, however the one I know best is the version in which the Deduction Theorem ("natural deduction rule") requires that in the proof to which you apply the theorem, you don't use Generalization on on a variable free in the hypothesis. Semantically, your last formula is not logically valid, but the second formula IS correct since: if the formula $x=3$ is TRUE (not just satisfied) in some structure, then it's generalization is also true in that structure ("true" means satisfied by all assignments). $\endgroup$
    – Ned
    Jun 12, 2022 at 16:29
  • $\begingroup$ @AlexKruckman: I edited my question to make my argument more precise, and removed the dodgy-sounding text "Presumably it is true". $\endgroup$ Jun 12, 2022 at 16:36

1 Answer 1

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So, $ x = 3 \models (\forall x \mathop. x = 3)$ is not true, and it is not true for the following reason.

$ \Phi \models \Delta $, where $\Phi$ and $\Delta$ are sets of formulas, means the following:

$$\Phi \models \Delta \;\; \textit{if and only if} \;\; \text{for all $(M, v)$, if $M, v \models \Phi$, then $M, v \models \Delta$} $$

$M, v \models \varphi$ means that $\varphi$ holds in the structure $M$ with the variables interpreted by $v$, a mapping from variable names to their denotations. Note that we are quantifying over $M$ and $v$ at the same time.

The relation that you appear to be using in place of $\models$ is well-defined, and we might write it $\Phi \models^g \Delta$. I have seen it called the global consequence relation of first-order logic, but it is not commonly used and certainly isn't the default. I'm only mentioning the global consequence relation at all to stress the importance of running back to the definitions in logic. I first heard about the idea of a global consequence relation from sequitur's answer to a question I asked about modal logic.

$$ \Phi \models^g \Delta \;\;\textit{if and only if}\;\; \text{for all $M$ such that $M \models \Phi$, it holds that $M \models \Delta$} $$

In the above definition $M \models \Phi$ holds if and only if $M, v \models \Phi$ for all $v$.

In the family of proof systems that use the $\vdash$ notation, $\vdash$ is the syntactic analogue of $\models$ not $\models^g$.

Also, as you have noticed, it is indeed not true that $x = 3 \models^g (\forall x \mathop. x = 3)$ implies $\models^g (x = 3) \to (\forall x \mathop. x = 3)$.

The equivalent of implication introduction in some hypothetical proof system in $\models^g$-land would give us $\models^g (\forall x \mathop. x = 3) \to (\forall x \mathop. x = 3)$. Basically, because the free variables are quantified over early, we have to protect statements that contain them.

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  • $\begingroup$ Thanks for this helpful reply! I got the definition of ⊨ that I was using from the textbook "A Friendly Introduction to Mathematical Logic, 2nd Edition" by Christopher Leary (tinyurl.com/friendlylogic). In section 1.9 "Logical Implication", it says "Definition 1.9.1. Suppose that ∆ and Γ are sets of L-formulas. We will say that ∆ logically implies Γ and write ∆ |= Γ if for every L-structure U, if U |= ∆, then U |= Γ." I had no idea that this definition was not standard! I just checked Enderton's "A Mathematical Introduction to Logic" and saw that his definition agrees with yours. $\endgroup$ Jun 12, 2022 at 17:07
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    $\begingroup$ @AdamDingle I briefly checked the book you linked and they are indeed not constraining $\Gamma$ and $\Delta$ to be sets of sentences. Their choice of definition of $\models$ appears to be consistent with the proof calculus they're using, which looks like a Hilbert system (in section 2.4). This sort of makes sense, we want $\vdash$ and $\models$ to "line up" because of the completeness theorem. My calling $\models^g$ nonstandard may have been too strong. I don't actually know how widespread it is. I just know that I haven't come across a book yet which explicitly covers both relations. $\endgroup$ Jun 12, 2022 at 17:20
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    $\begingroup$ The global consequence relation was used in many mathematical logic textbooks (without the little "g" notation) as the primary notion of logical consequence [Mendelson's text for example]. However, I've been away from this stuff for a long while so I don't know what's found in newer treatments. $\endgroup$
    – Ned
    Jun 12, 2022 at 17:54
  • $\begingroup$ As a followup (over a year later), I just noticed that in Rautenberg's textbook "A Concise Introduction to Mathematical Logic, 3rd Edition" he discusses the relationship between ⊨ and ⊨g on pp. 80-81 in section 2.5 "Logical Consequence and Theories". I found his explanation to be helpful. $\endgroup$ Apr 30, 2023 at 8:43

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