0

In the book of first order logic I'm reading they say that if $(\varphi_1,...,\varphi_n)$ is a proof from a set of formulae $\Sigma$, then so is $(\varphi_1,...,\varphi_n,\forall x \varphi_j)$ for all $1\leq j \leq n$. They assume that as an axiom but it doesn't make sense to me. Why if $\varphi$ is true then so is $\forall x \varphi$? For example if $\varphi$ is $x=1$ that wouldn't mean that $\forall x : x=1$ , or would it?

Maybe what I don't understand is: why $(\Sigma \vdash \varphi \implies \Sigma \vdash \forall x \varphi)$ is true but $(\varphi \implies \forall x \varphi)$ is not true?

The book is this: https://www.springer.com/cda/content/document/cda_downloaddocument/9781447121756-c2.pdf?SGWID=0-0-45-1192238-p174141200

Ruben a
  • 35
  • 1
  • 4

2 Answers2

1

A proof of $\varphi$, with $x$ as a free variable, treats $x$ as an arbitrary element of its domain of discourse. This is exactly how we obtain a proof of $\forall x ~ \varphi$—introduce a variable $x$ and, knowing only those properties of $x$ that hold of all elements in its domain of discourse, prove $\varphi$. This is essentially the content of the $\forall$-introduction rule (a.k.a. universal generalisation).

If you could prove $x=1$ knowing nothing about $x$ than that it is an element of its domain of discourse, then it would indeed be true that $\forall x (x = 1)$ is true. But of course this is only going to be true if the domain of discourse of $x$ is $\{ 1 \}$. (Or $\varnothing$, I suppose.)

Now if you're doing first-order logic, it's likely that your domain of discourse is some ambient set theoretic universe, in which case $x=1$ will almost certainly not be provable, but formulae like $\varnothing \subseteq x$ and $x \in \{ x \}$ are provable.

David C. Ullrich
  • 89,985
Clive Newstead
  • 64,925
  • The book I'm using doesn't seem to mention anything about a domain of discourse though.. And isn't set theory an extension of first order logic? – Ruben a Dec 30 '18 at 16:11
  • @Rubena: Regarding your second question, a model of a first-order theory is a set together with certain structure (constants, functions, etc.). This means that a model $V$ of (say) ZFC set theory is externally a set, even though internally it is a proper class. But then if you're doing first-order logic, you must be working within some meta-theory, which in all likelihood is itself a model of the theory of sets. – Clive Newstead Dec 30 '18 at 16:36
  • Regarding your first question, first-order logic is just a formal system for manipulating symbols, so the variable $x$ doesn't necessarily refer to an element of any set—it's just a symbol. But semantically the variable $x$ must refer to an element of a set, and it is this set that is the domain of discourse of $x$. – Clive Newstead Dec 30 '18 at 16:38
  • But then what's the difference between $x=1$ and $\forall x (x=1)$ ? If $x=1$ then all elements in the domain of discourse of $x$ are equal to 1? Then how does it differ from $\forall x(x=1)$ ? – Ruben a Dec 30 '18 at 16:48
  • The formula $x=1$ has a free variable, but the formula $\forall x ~ x=1$ does not. – Clive Newstead Dec 30 '18 at 16:49
  • 1
    @Rubena No! It's not true that if $x=1$ then all elements in the domain must equal $1$. Nobody has said that. You keep on leaving out that crucial hypothesis. What's true is that if it's possible to prove $x=1$ without assuming anything about $x$ then all elements in the domain must equal $1$. – David C. Ullrich Dec 30 '18 at 16:53
  • You seem to be taking for granted that we all realize that $x$ must not occur free in $\Sigma$; in fact it seems clear that the problem is precisely that the OP is leaving this out. – David C. Ullrich Dec 30 '18 at 17:15
0

Edit: It turns out that the system you're dealing with is very different from the systems I'm familiar with. Hence more or less everything below is wrong, with regard to that system. I'd simply delete the whole thing but I can't since it's "accepted"...

The reason your version makes no sense is it's false. You're leaving out a crucial hypothesis; what's true, and what it had better say in the book, is this:

If $\Sigma\vdash\phi$, where $x$ does not occur free in $\Sigma$, then $\Sigma\vdash\forall x\phi$.

That makes much more sense. Informally, saying $x$ does not occur free in $\Sigma$ means that the formulas in $\Sigma$ do not say anything about $x$. So saying $\Sigma\vdash \phi$ where $x$ does not occur free in $\Sigma$ means "you can prove $\phi$ without assuming anything about $x$".

Now it makes sense. If you can prove $\phi$ from a set of assumptions that don't mention $x$, then $\phi$ follows from your assumptions regardless of what $x$ is, so $\forall x\phi$ "should" also follow from the same assumptions.

Taking $\phi$ to be $x=1$ is a bad example, because it's hard to imagine a system where you can prove $x=1$ without assuming anything about $x$. Instead imagine you have some axioms describing how the real numbers work, and you want to prove $\forall x(x>1\to x^2>1)$. If you can prove $x>1\to x^2>1$ without assuming anything special about $x$ then $\forall x(x>1\to x^2>1)$ follows. (That is how people prove $\forall x\phi$ in actual math: prove $\phi$ without assuming anything about $x$.)

Hmm. An example of a $\Sigma$ where $\Sigma\vdash x=1$ but $x$ is not free in $\Sigma$ would be $\Sigma=\{\forall y (y=1)\}$. Yes, $\forall y (y=1)$ implies $x=1$, and it also implies $\forall x (x=1)$.

Another example, illustrating how if $x$ does occur free in $\Sigma$ then all bets are off: It's true that $x=1\vdash x=1$. But $x=1\not\vdash \forall x(x=1)$. If you leave out that "where $x$ does not occur free in $\Sigma$" then you're going to conclude that in fact $x=1\vdash\forall x(x=1)$, which is nonsense; luckily $x=1\vdash\forall x(x=1)$ does not follow from the correct version of the result we're talking about.

David C. Ullrich
  • 89,985
  • 1
    But the exposition of FOL used by the OP is "Mendelson-like" : no restriction on Gen rule. The restriction is on the Deduction Th. So $(x=1) \vdash \forall x (x=1)$ is formally correct in the system; what is not is to apply DT to it to get : $\vdash (x=1) \to \forall x (x=1)$. – Mauro ALLEGRANZA Dec 30 '18 at 18:13
  • Oh. All I know about logic is what I read in Enderton years ago; the notion of a system were $x=1\vdash\forall x(x=1)$ makes no sense to me. As sort of a check on whether it's "$x=1$" or $"\vdash$" that's totally different: what about $x=1\models\forall x(x=1)$? – David C. Ullrich Dec 30 '18 at 18:30
  • Only browsed the text... it seems to me that is not defined. But $\mathfrak A \vDash \varphi(x)$ iff $\mathfrak A \vDash \varphi(x) [h]$ for every evaluation $h$ and thus $\mathfrak A \vDash \varphi(x)$ iff $\mathfrak A \vDash \forall x \varphi(x)$. So, no problem... – Mauro ALLEGRANZA Dec 30 '18 at 18:55
  • @MauroALLEGRANZA but then in "Mendelson-like logic" what is the difference between $(x=1)$ and $\forall x (x=1)$ ? What do those formulas mean? – Ruben a Dec 30 '18 at 19:18
  • @Rubena - consider $\mathbb N$ as domain of the interpretation. $\mathbb N \nvDash \forall x (x=1)$, obviously, while for $(x=1)$ we need a variable assignment $h$. If $h(x)=1$ then $\mathbb N \vDash (x=1)[h]$, while for $h'$ such that e.g. $h'(x)=0$ we have that $\mathbb N \nvDash (x=1)[h']$. So the semantics is Ok: we are licensed to asserts $\mathfrak A \vDash \varphi(x)$ only if every variable assignment $h$ gives the same truth value. – Mauro ALLEGRANZA Dec 30 '18 at 19:37