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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.

Make42
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  • $f(x) = 5, f(x) = 2*x$ 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
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    @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
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    @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
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    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
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    @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
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    See this and this. – mr_e_man May 11 '22 at 15:28
  • @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
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    @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
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    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
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    @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
  • Bram28 is an explanation wizard! 2. This is why I distinguish between those two metalogical symbols. To elaborate: the ⇒ (implication) that we use in mathematics is actually weaker than logical implication (⊨); when ⇒ is being encountered, the context informs us whether ⊨ is also meant.
    • – ryang May 19 '22 at 11:50