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Take the statement $\forall x \in \mathbb{R} (y=5x)$ where $y$ is free and $y \in \mathbb{R}$, I think that this statement means "For all real $x$, $5x$ equals some real number $y$." Now I know that there is no such number $y$ such that every number when multiplied by $5$ equals $y$. Therefore the statement is false, but can open statements have a fixed truth value?

Nav Bhatthal
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    A simpler example is just $y=5$. – TonyK Aug 08 '23 at 18:27
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    How does this relate, can you elaborate please? – Nav Bhatthal Aug 08 '23 at 18:27
  • Yes, the presence of the free variable alone is what makes it open. – NDB Aug 08 '23 at 18:29
  • It's a formula whose truth depends on the value of a variable. See this link for a discussion. – TonyK Aug 08 '23 at 18:29
  • So an open statement (one with atleast one free variable) does not have a definite truth value? How can we choose $y$ such that $\forall x (y=5x)$ is true? – Nav Bhatthal Aug 08 '23 at 18:30
  • It could have a definite truth value, as in your example. Truth value is irrelevant to the quality of being open. Being an open statement is an entirely syntactical property. Imagine a machine that parses the formula detecting variables that are not bound. It doesn't require any semantic information. – NDB Aug 08 '23 at 18:30
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    So the truth value depends on the choice of free variable, but if for all choices, the truth value is the same, then the open statement has a fixed truth value? Just in most cases the truth value can change from true to false? @NDB – Nav Bhatthal Aug 08 '23 at 18:32
  • the open sentence does not have a truth value – RyRy the Fly Guy Aug 08 '23 at 20:35
  • @TonyK: the link you cite shows the opposite of what you claim. An open sentence is one whose truth value might depend on the value assigned to one of its free variables. But if $P(x)$ is false for some $x$ for any choice of values for any other free variables in $P(x)$, $\forall x(P(x))$ is false. – Rob Arthan Aug 08 '23 at 21:28
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    @Mathguy It seems fairly arbitrary whether we consider an open sentence like this to "have a truth value" or merely to represent a constant function from variable assignments to truth values. Is there something you're trying to do that depends on this distinction? – Karl Aug 08 '23 at 23:21

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(Edit: I agree with Rob Arthan's comments and answer indicating that we allow proofs of open sentences and can think of such proofs as establishing truth values for open sentences. Alternatively, though, we can insist that open sentences and their proofs represent "templates" for closed ones, or functions on the domain. This answer takes that perspective.)

I'd say that an open sentence doesn't have a truth value until you specify values for the free variable(s) or add a quantifier binding them (i.e. $\forall y$ or $\exists y$). In your example, the resulting closed sentence will always be false, but that doesn't mean the open sentence itself is false. (This is analogous to the fact that a constant function is not the same thing as its unique value; it's a function, not a value.)

Sometimes for conciseness, though, we interpret an open sentence as having an implicit "forall" quantifier binding each free variable. The intended meaning should be clear from context.

Karl
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  • This doesn't make sense, over $\Bbb{R}$, $\forall x \in (y = 5x)$ is provably false. This is not analogous to the fact that a constant function is not the same thing as its unique value. The provability of $\phi \Leftrightarrow \psi$ does not require $\phi$ and $\psi$ to have the same free variables. – Rob Arthan Aug 08 '23 at 21:23
  • @RobArthan Isn't a proof involving free variables usually understood technically as a meta-proof? Or e.g. as the sub-proof of a surrounding forall-introduction step that binds the free variable? Maybe I'm wrong that "most formal systems" make this distinction. Really this whole question is a bit pedantic because there's no actual confusion about how to use these kinds of sentences. – Karl Aug 08 '23 at 23:06
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    in a word, no: I don't read the definitions of the notion of proof in any of the standard texts as implying that there should be no free variables. – Rob Arthan Aug 08 '23 at 23:24
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Yes it is perfectly possible for a formula with free variables to be identically true or identically false. In your case, using properties of $\Bbb{R}$ and the algebraic operations on it, we can show that $\forall x \in \Bbb{R}(y = 5x)$ implies a contradiction:

\begin{align*} &\mbox{A. Assume $\forall x \in \Bbb{R}(y = 5x)$.}\\ &\mbox{B. Specialising A to $1$, we have $y = 5$.}\\ &\mbox{C. Specialising A to $-1$, we have $y = -5$.}\\ &\mbox{D. From B, D and algebra, we have $y \neq y$. Contradiction!}\\ \end{align*}

Most open formulas represent truth values that do depend on the values assigned to the free variables, but that is not always the case.

Rob Arthan
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  • I have seen online that to check the truth of an open formula we must choose values for the free variables, in your approach you do not do this (but choose to look at the bounded variables instead), are both approaches logically sound? – Nav Bhatthal Aug 09 '23 at 16:07
  • Also, what does it mean to assume $\forall x (y=5x)$? We do not have a restriction on $y$ (as its free), are you assuming $\exists y \forall x (y=5x)$? And then showing that no such $y$ exists, therefore the open formula is always false? (you cannot find $y$ to make it true). – Nav Bhatthal Aug 09 '23 at 17:28
  • I am assuming that some formula with a free variable holds. $y$ is (effectively) universally quantified at the meta-level in my proof, but you don't need to look at it like that: I am just following the rules of first-order logic using axioms that hold in the field $\Bbb{R}$. I am certainly not assuming $\exists y\forall(x(y = 5x)$ and in fact what I am doing amounts to a disproof of that sentence. – Rob Arthan Aug 09 '23 at 20:55
  • You want to show that $\forall x (y=5x)$ is never true for any possible choice of $y$. I get that. That sounds a lot like you want to show $\forall y \forall x (y=5x)$ is false, I also get that, but my question is, how can you use the "$y$ doesn't equal $y$" argument, when $y$ is clearly free and can take any value? Is it not better to use $\forall y \forall x (y=5x)$, set $x$ to be $1$ then say $\forall y (y=5)$ is never true? – Nav Bhatthal Aug 10 '23 at 06:04
  • i.e On line B you have already reached a contradiction, because $y=5$ cannot be true for all choices of $y$. – Nav Bhatthal Aug 10 '23 at 06:11
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    No, $y$ might be $5$ on line B. You should read carefully through a proof of the soundness of your favourite deductive system for first-order logic. You will see that the formulas that appear in intermediate steps in a deduction aren't either universally or existentially closed: they can involve free variables that refer to unspecified values, i.e., they are universally quantified at the meta-level outside the deduction, but not quantified in the individual steps in the deduction. – Rob Arthan Aug 10 '23 at 21:25
  • Can I ask what you mean my meta level? I've not seen anyone else mention it before, and I can't find it on the internet. Is my proof in the first comment of mine after your first comment not valid then? – Nav Bhatthal Aug 11 '23 at 06:07
  • See https://en.wikipedia.org/wiki/Semantic_theory_of_truth. – Rob Arthan Aug 11 '23 at 20:53