I was watching this video on statements. There is an example:

$x + \frac12 = 2$

It's an open statement as the truth value could be T or F depending on the value of $x$.

$∃x: x + \frac12 = 2$ and $x ∈ ℤ$

Now the statement is closed statement (proposition) as the truth value is F.

But later the professor said:

$x + \frac12 = 2$ and $x ∈ ℤ$

is an open statement. But it's not clear to me how it's an open statement as for any value of $x$ the statement is F?

Update from the author of the video:

$x$ is a free (unquantified) variable here, so by definition this is an open statement. But you've identified something that's a source of confusion. Many online sites say that a statement is open if its truth value depends on what values the variable(s) take on. That usually coincides with the definition I indicated above. But this example points out that the two definitions are not the same. While it's easy to slip into the "is the truth value known" definition (as I seemed briefly to do at around 19:40), the definition of an open statement that works best is that it's a statement with one or more free variables, as I point out in several places in this video.

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    $\begingroup$ You are right, I don't think it's very clear from that video segment you link to. It seems the lecturer are mixing free and open. I don't know what open means, but I think you should just ignore it for now, and if it pops up later, try to find out what he meant. $\endgroup$
    – Pål GD
    Commented Oct 11, 2021 at 20:53
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    $\begingroup$ The property of being open is syntactic, not semantic. It just means that there are free variables, in this case $x$. $\endgroup$ Commented Oct 11, 2021 at 22:12
  • $\begingroup$ @user21820, thanks for your editing. I restored the title to plaintext, since it is discouraged to use MathJax in the title. One reason is that questions with MathJax in their titles will not be shown in the "Hot Network Questions" list. $\endgroup$
    – John L.
    Commented Oct 15, 2021 at 2:15

4 Answers 4


$$x=1$$ $$∃x: x=1$$ The first is an open statement, since no value for $x$ is given. $x$ is called a free variable here.
The second is a closed statement, because it talks about all possible values of $x$. $x$ is not a free variable here.

An open statement can be true or false depending on what the values of its free variables are.
A closed statement is either always true or always false. (But we might not know which) $$x=x+1$$ This is an open statement since there is no $∃x$ there. It just so happens that it is always false, but that is not relevant to openness.

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    $\begingroup$ true in "integers modulo 1" though! $\endgroup$ Commented Oct 12, 2021 at 8:21
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    $\begingroup$ And true for some values of floats, though that gets dangerously close to applied CS ;) $\endgroup$
    – jpa
    Commented Oct 12, 2021 at 13:26
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    $\begingroup$ @jpa if you're going to take “some values of floats” into account, then $\forall x : x = x$ is false. At that point you might as well give up trying to prove anything. $\endgroup$ Commented Oct 12, 2021 at 19:39
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    $\begingroup$ @leftaroundabout: This is getting rather pedantic, but -- even if we define $+$ as floating-point addition, we would still probably use the regular $=$ from first-order logic, rather than redefining $=$ as IEEE 754 floating-point equality. (It's extremely common to define $+$ differently in different contexts, but messing with $=$ takes some chutzpah!) $\endgroup$
    – ruakh
    Commented Oct 13, 2021 at 7:40

Whether it's an open statement or not depends on the structure of the statement, no on whether you can prove the truth value.

Look at the second statement. There are three things you could substitute: You could substitute any number for the "1/2", you could substitute any number for the "2", and you could substitute any set of numbers for the "Z".

All these changes don't change the structure of the statement, so it remains an open statement with any of those changes. Of course, depending on the values you substitute, there might be a solution for x, or there might be none, or multiple solutions, but that doesn't change the structure of the statement which remains open.


Generally in first order logic open statement simply means a formula with free variable(s) and thus is not a (closed) sentence. How can you be sure that its closed statement is false if it turns out that the actual statement reads $\lnot \exists x: x + 1/2 = 2$ AND $x ∈ Z$?

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    $\begingroup$ The second half of your post ("How can you be sure that its closed statement is false if it turns out that ...") does not make any sense. $\endgroup$
    – user21820
    Commented Oct 12, 2021 at 6:25
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    $\begingroup$ @user21820 This answer appears to be directly copied from a comment. (by the same user). mohottnad: The first half of your answer is fine as an answer, but perhaps you should rephrase the second part to make it more clear it is a rhetorical question, rather than an actual question. $\endgroup$
    – Discrete lizard
    Commented Oct 12, 2021 at 7:34
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    $\begingroup$ @user21820 thx for your time and offer critique! When I saw OP's question the first thing I thought is free formula doesn't have truth value, a basics of FOL which actually is what the video author intends to criticize (see OP added the author's own response by now) which IMO is unnecessary, as my second half example is used to remind oneself as long as a formula is open, its truth value may always be left to be judged until it's closed as a complete sentence. (I understand this particular one seems a negation of a valid theorem, but still...) Welcome to constructive suggestions... $\endgroup$
    – cinch
    Commented Oct 12, 2021 at 15:43
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    $\begingroup$ That is meaningless. The truth-value of an open formula is either defined as that of its universal closure or not at all. This has absolutely nothing to do with your so-called example. $\endgroup$
    – user21820
    Commented Oct 12, 2021 at 16:41
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    $\begingroup$ @mohottnad: Would you also argue that we can't be sure that $\exists x: x + 1/2 = 2 \land x ∈ Z$ is false, because the full statement might turn out to be $\lnot \exists x: x + 1/2 = 2 \land x ∈ Z$? $\endgroup$
    – ruakh
    Commented Oct 13, 2021 at 7:46

The derivative of $3x+5$ is a function, even though it's always equal to $3$. (The statement "the derivative of $3x+5$ is $3$" is really shorthand for "The derivative of $f:x\rightarrow 3x+5$ is $f':x\rightarrow3$".) The derivative of $3x+5$ at $x=2$ is not a function, it's a real number.

An open statement can be viewed as being a function of the free variables. Whether that function has different values for different inputs is irrelevant to whether it's a function/open statement. If an expression has no free variables, then it's called a closed statement (technically, it can be considered to be a nullary function, so one could argue that the analogy between functions and open statements should be between non-nullary functions and open statements).


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