Does "if" turn a free variable into a bound variable?

Question

A closed formula expresses a proposition (which is a truth-apt concept, so a concept that is either be "true" or "false"). A closed formula is a Boolean-valued formula with no free variables.

A free variable is a variable that "specifies places in an expression where substitution may take place and is not a parameter of this or any container expression", it has not been bound by a variable-binding operator like

$$ \sum _{x\in S}\quad \quad \prod _{x\in S}\quad \quad \int _{0}^{\infty }\cdots \,dx\quad \quad \lim _{x\to 0}\quad \quad \forall x\quad \quad \exists x $$

So for example (in the following all numbers, variables, and function values are real numbers)

\begin{align} f(x) = 5, f(x) = 2*x \qquad\qquad\qquad\qquad\qquad (1) \end{align}

is not true or false, as long as we don't know what $x$ is. But if we say,

$$ \text{if } x = 2, f(x) = 2*x \text{, then } f(x) = 5 \qquad\;\;\qquad (2) $$

this becomes truth-apt (in this case this is false). So, I would think that "if" is a variable-binding operator. Is this correct?

Another way of writing "if" is by using $\Rightarrow$, e.g.,

$$ x = 2, f(x) = 2*x \Rightarrow f(x) = 5 \qquad\qquad\qquad (3) $$

which I would think is a closed formula which is false.

$f(x) = 5, f(x) = 2*x$ looks implausible if $x$ is restricted to integers as there is an implicit $\exists x \cdots$. Similarly $x^2<0$ looks implausible if $x$ is restricted to reals and difficult to interpret if $x$ can take any complex value. — Henry, May 11 '22 at 14:38
@Henry Given that we're talking about free vs bound variables, we shouldn't add implicit quantifiers. And $x$ is still free in $x \ne x$ even if it's false no matter how you fill in $x$. — eyeballfrog, May 11 '22 at 14:48
@eyeballfrog if you say $x \not=x$ has $x$ free then perhaps you could say $x=2 \implies x=3$ also has $x$ free, so adding if would not have made $x$ a bound variable — Henry, May 11 '22 at 15:01
@Henry That's true. In fact, I'd say that the if isn't binding the variable, but rather the implicit $\forall$ that is coming with it. — eyeballfrog, May 11 '22 at 15:03
@Henry: When I wrote the questions, I had in mind, that all numbers are real numbers. In (1) I did not want to write $\exists$ as this would have bound $x$. I wanted to express that (1) could be true if $x=2.5$, but (1) does not say what $x$ is. — Make42, May 11 '22 at 15:05
It's worth noting that $$\forall x,;(x=2\implies x=3)$$ is false, but $$\exists x,;(x=2\implies x=3)$$ is true. — mr_e_man, May 11 '22 at 15:07
@eyeballfrog: Do you mean that $if x=2$ is an abbrevation of $\forall x, \text{ where} x=2$? — Make42, May 11 '22 at 15:10
@mr_e_man: Does your first statement mean "for all x, where x equals 2, x must equal 3", which is equivalent to "if x equals 2, then x equals 3" (which is false)? — Make42, May 11 '22 at 15:12
@Make42 though $\exists x,;(x=2\implies x=3)$ is true for example when $x=4$ — Henry, May 11 '22 at 15:13
@Henry: How so? How do I have to translate this statement into English? "there exists an x, such that if x equals 2, then it (also) equals 3" seems false to me and I do not see how to bring in the $x=4$. "there exists an x, namely x=4, such that if x equals 2, then it (also) equals 3" sounds weird to me. — Make42, May 11 '22 at 15:17
@Make42 Sometimes the "if" word is thought of as an informal replacement for the universal quantifier. For instance, when I teach undergraduate Foundations of Mathematics, I instruct students to search for "hidden quantifiers", and one place I tell them such quantifiers can be found is in "if" statements. But in formal logic, "if P then Q" should be treated as synonymous with P$\implies$Q, and free variables in P and Q remain free in P$\implies$Q. — Lee Mosher, May 11 '22 at 15:17
Yes, the first statement means "For all $x$, if $x$ equals $2$, then $x$ equals $3$." The second statement is indeed weird. $(4=2\implies4=3)$ is true. — mr_e_man, May 11 '22 at 15:19
@mr_e_man: But $\Rightarrow$ also means "from ... follows ...", so how does $4=3$ follow from $4=2$ or does $\Rightarrow$ have a different meaning here than in other areas of math? (This is the first time I am doing mathemathical logic.) Or do I have to say that from "false" (left side) follows "false" (right side)? (This also sounds weird.) — Make42, May 11 '22 at 15:25
@mr_e_man: I am reading up on the material you linked to. I think I am lacking some general understanding. I always thought the $\implies$ just mean that "if the left hand side (the condition) is true, then the right hand side (the conclusion) is true" and that if the condition is not true, nothing is said about the right hand side. Also I always assumed the whole statement (the implication) to be true, so basically I always only thought about the first case in the truth table. I used $\implies$ in mathematical proofs, but not in the field of logic. Vacuous truths still confuses me, I guess. — Make42, May 11 '22 at 16:55
@Make42 Your previous comment is exactly correct, except "I always assumed the whole statement (the implication) to be true, so basically I always only thought about the first case in the truth table." Don't you mean that a true conditional (i.e., an implication) corresponds to three rows, not just the first one, of its truth table (this doesn't preclude the fact that a false antecedent confers no truth information about its consequent). I don't think you've been fundamentally misunderstanding the issues raised on this page, just getting used to the nitty-gritty. — ryang, May 18 '22 at 04:24
@ryang: I meant, I never thaught about an "x\Rightarrow y" to be true or false. I always used it in such a way that it was assumed that x is true and then I showed that y follows from x and then I continued using the true y. For example I had some assumption(s) x and then I showed that some theorem is true in such a case where x is true. — Make42, May 19 '22 at 09:45
@Make42 Yes, you are reading ⇒ in informal writing as → corresponding to those three rows (that is, as an assertion that the logical conditional is true), which is exactly what it means. Such is the nitty-gritty I meant. — ryang, May 19 '22 at 09:56
@ryang: Ok, but why would that be the first three rows of https://en.wikipedia.org/wiki/Truth_table#Logical_implication? I assume that $x$ is true (so the third and fourth rows are disregarded) and I consider $x\Rightarrow y$ to be true (so the second row is disregarded). That is why I wrote, that I only considered (in the past) the first row. What did you mean? — Make42, May 19 '22 at 10:01
@Make42 While that wiki page is treating both symbols as equivalent, I'm pointing out that in mathematics, the metalogical symbol ⇒ is actually understood to mean that the logical operator → is true. You know that nothing useful can be said about $Q$ when $(P$ is false and $(P\implies Q))$ *because the two rows of → corresponding to $P=$ false* are telling you that in this case $Q$ could swing either way. — ryang, May 19 '22 at 10:35
The only row that is irrelevant to ⇒ is the one where → is false. My previous two comments should now be clearer. So, this formal logic business does not actually conflict with your informal reading/understanding/practice of logic. A related answer that I wrote. — ryang, May 19 '22 at 10:51
@ryang: Yes, that was very helpful! I wasn't aware of the distinction between ⇒ and →. I also read up on the stuff that you linked to (esp. the articles by Asaf). Bram28 writes that ⇒ and ⊨ are the same (https://math.stackexchange.com/a/3931759/340174), while you make a distinction (https://math.stackexchange.com/a/3884866/340174). What is the distinction? — Make42, May 19 '22 at 11:28

ryang · Accepted Answer · 2022-05-11T16:55:22.483

2

\begin{align} f(x) = 5\text{ and } f(x) = 2x\tag1 \end{align} is not true or false, as long as we don't know what $x$ is.

$$\text{if } x = 2\text{ and } f(x) = 2x \text{, then } f(x) = 5\tag2$$ this becomes truth-apt (in this case this is false). So, I would think that "if" is a variable-binding operator. Is this correct?

Has $f$ been pre-defined such that $f(x)=2x\,?$ If so, then the open formula $(2)$ is actually true for all values of $x$ except $2.5.$ What is false though is the closed formula $$\text{for each $x,\Big($if } x = 2\text{ and } f(x) = 2x \text{, then } f(x) = 5\Big)\tag{2a}.$$ Informally, open conditional formulae like $(2)$ are typically treated as closed formulae like $(2a);$ it is this implicit universal quantification—not the conditional operator per se—that binds the free variable $x.$

On the other hand, if $f(x)$ actually equals, say, $-\dfrac1x,$ then both formulae $(2)$ and $(2a)$ are definitely true in the standard interpretation.

edited May 11 '22 at 16:55

answered May 11 '22 at 16:46

ryang

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Thanks for the answer! I tried to read as much material (that was related to the comments under the question) as possible and as seemed sensible. Now, I come back to your answer: Do you mean with "Has $f$ been pre-defined such that $f(x)=2x$?" for example "$x=4$, if $x=2$ and $f(x)=2x$, then $f(x)=5$" or "if $4=2$ and $f(x)=4$, then $4=5$" - that this is true? But in that case "if $2.5=2$ and $f(x)=5$, then $5=5$" would also be "if false, then true", which would also be true, but you write that it would be false for $x=2.5$. What do I get wrong? – Make42 May 14 '22 at 22:13
Also, what do you mean by your last paragraph? Do you want to substitute my $f(x)=2x$ with $f(x)=-\frac1x$? Or do you want to write "for each x, ( if $x=2$ and $-\frac1x=2x$, then $f(x)=5$)", which would be true because "$x=2$ and $-\frac1x=2x$" is "$x=2$ and $x=\sqrt{-\frac12}$" which is false? – Make42 May 14 '22 at 22:18
@Make42 1. All I'm saying that is your formulae $(1)$ and $(2)$ had not been sufficiently contextualised, since you hadn't talked about what $f$ is. $f$ could also vary, just like $x$ does. So, to make the discussion cleaner, we let $f(x)=2x$ in the first part of my Answer. $\quad$ 2. The latter: in this second part of my Answer, we let $f(x)=-\frac1x.$ $\quad$ 3. The point is that while formulae $(2)$ and $(2a)$ are semantically different, formula $(2)$ is typically understood as a shorthand for formula $(2a),$ and it is this apparent inconsistency that is giving you pause. – ryang May 14 '22 at 23:27
Ok, interesting. What do you mean by saying that $f$ has not been contextualized? It is a real-valued function with the reals as domain. Was that the missing contextualization or do you mean something else? I am not sure why I need to per-definied $f$, since I clearly defined it in each (1), (2) and (3), in the same way as in (2a), namely the part "$=2$ and $()=2$" has always been the same.

Make42

May 16 '22 at 12:40

@Make42 As pointed out in my previous comment, the first part of my Answer is based on $f$ being defined as $f(x)=2x,$ while the second part of my Answer (its final paragraph) is based on $f$ being defined as $f(x)=-\frac1x.$ $\quad(1),(2),(2a)$ are merely assertions/statements that may or may not be true; on the other hand, in my first sentence of this comment, I have given two separate definitions for $f;$ in other words, I have provided two separate contexts with which to interpret $(1),(2),(2a).$ – ryang May 16 '22 at 13:09

Ok, so do I understand correctly, that the $f(x)=2x$ in (1),(2),(2a) is not a definition what $f(x)$ is, but a statement that can be true or false? Thus, one needs a definition of $f$ beforehand to bind $f$ and to make (1),(2),(2a) truth-apt? – Make42 May 16 '22 at 13:54

Yes, but that alone doesn't automatically make $(1),(2),(2a)$ truth-apt, as I pointed out in my first sentence of the Answer. In any case, don't get bogged down by all this sidetracking: the main point of this Q&A page is the boldfaced portion. – ryang May 16 '22 at 14:00

1

Ok, I think I got the main points. Actually, the "sidetracking" and "reading up" got me to learn a lot, so thanks for that, too. Also, I think that the question of when a variable becomes bound is also important for $f$ and I did not realize, that there was an issue in the first place. Regarding that I will ask a separate question then. – Make42 May 16 '22 at 14:04

score 0 · Answer 2 · answered May 17 '22 at 19:54

Actually, we do not need variable-bounding in this instance; quantifier-free fragment of first-order logic with equality (see Ian Chiswell and Wilfrid Hodges' Mathematical Logic, chapter 5) or equational logic will do.

We can take the signature as

$\sigma_{eq} = \langle \mathcal{A}, a, b, c,\ldots f, g, h,\ldots =, R, S,\ldots\rangle$

where $\mathcal{A}$ is a set as the domain of discourse; $a, b, c,\ldots$ are individual constants; $f, g, h,\ldots$ are (term-forming) functions; $=, R, S,\ldots$ are relations (representing predicates in the domain). '$=$' is the equality relation, it can be put into the logical vocabulary into the signature.

The functions can be represented as

$\phi$ as $=\!(x, 2)$, $\psi$ as $=\!(f(x), 2)$, $\upsilon$ as $=\!(f(x), 5)$.

Natural deduction system with the '=' introduction rule

$\dfrac{}{t=t}\;\;\; (=I)$

and '=' elimination rule

$\dfrac{s = t\;\;\;\;\phi(s/x)}{\phi(t/x)}\;\;\; (=E)$

together with the equality axioms of reflexivity, symmetry, and transitivity, will suffice.

A significant point in this perspective is that free variables are not merely syntactic elements. They are also semantic constituents that serve roughly as names or parameters in a fashion comparable to logical names in computer science for which we have not specified definite physical names.

Does "if" turn a free variable into a bound variable?

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