Though searching for previous questions returns thousands of results for the query "for any" "for all", none specifically address the following query:

I'm reading a textbook in which one definition requires that some condition holds for any $x,\ x' \in X,$ and right afterwards another definition requires that some other condition holds for all $x,\ x' \in X.$

Is there a difference between 'for any' and 'for all'?

  • 3
    $\begingroup$ I think it is, most often, a matter of choice. The symbol $\forall$ sometimes is read "for every", sometimes "for each", sometimes "for any". $\endgroup$
    – Siminore
    Jun 27, 2013 at 9:02
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    $\begingroup$ @user79202 I understand your distinction, but wouldn't one, in the latter case (where you write "If for any $x\in X\ P(x)$, then $Q$") rather say "If $\exists x\in X$...")? So that would mean that in those cases where $\exists$ is not applicable, "for any" and "for all" are equivalent? $\endgroup$
    – Bernd
    Jun 27, 2013 at 9:05
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    $\begingroup$ @Siminore So "a matter of choice" means that they are actually equivalent? Sorry... $\endgroup$
    – Bernd
    Jun 27, 2013 at 9:06
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    $\begingroup$ In the context of your question they are equivalent. user79202's example is fine. In the first example "for any" and "for all" mean the same. In the second example "for any" and "there exists" mean the same, you could just as well use $\exists$ there. $\endgroup$
    – m_l
    Jun 27, 2013 at 9:13
  • $\begingroup$ Well, since language has all sorts of quirks, I am not comfortable to state anything general. It might just be that it is the author's job to make sure, we get his intentions right... $\endgroup$
    – user79202
    Jun 27, 2013 at 9:13

4 Answers 4


This seems to depend on the context: "For all $x \in X \ P(x)$" is the same as "For any $x∈X \ P(x)$" On the other hand "If for any $x∈X \ P(x)$, then $Q$" means that the existence of at least one $x\in X$ with $P(x)$ implies $Q$, so $P(x)$ doesn't need to hold for all $x \in X$ to imply $Q$.

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    $\begingroup$ what about " if $x \not \in E_a$ for any $a$ hence $x \in E^c_a $ for any $a$"? $\endgroup$ Jan 25, 2018 at 16:13
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    $\begingroup$ I think $P(x)$ does still need to hold for al $x\in X$ though. $\endgroup$
    – asn32
    Feb 13, 2018 at 16:56
  • $\begingroup$ @CharlieParker Your sentence could possibly mean only $$(∀a \,\:x\not\in E_a) \implies ∀p\,\: x \in E^c_p $$ or $$∀a \;(x\not\in E_a \implies x \in E^c_a).$$ The former is a counterexample to this answer, whereas the latter is likely the intended reading. $\endgroup$
    – ryang
    Jun 30, 2023 at 14:49

"Any" is ambiguous and it depends on the context. It can refer to "there exists", "for all", or to a third case which I will talk about in the end.


Oxford Dictionary:

  1. [usually with negative or in questions] Used to refer to one or some of a thing or number of things, no matter how much or how many.
    [as determiner] ‘I don't have any choice’
    which means there does not exist a choice
  2. Whichever of a specified class might be chosen.
    [as determiner] ‘these constellations are visible at any hour of the night’
    which means for all hours

Also in Oracle database I remember a query to return the employees where employee.salary > any(10,20,30,40) which means the salary of the returned employee must be bigger than 10 or 20 or 30 or 40 which means "there exist" one salary in that tuple such that the employee.salary is bigger than it.

Same ambiguity is in math, since math did not come from nothing, rather it is a notation system for the language.


Some times "any" is used for the meaning of "any" and not "exist" or "all". For example, in the definition of the little o asymptotic notation we have:

$o(g(n))$ = { $f(n)$: for any positive constant c>0, there exists a constant $n_0>0$ such that $0 \leq f(n) < c g(n)$ for all $n \geq n_0$ }

Here "any" means "any" which is two things "there exists" and "for all" how??

  1. If you take any as "for all c" then the meaning is wrong because $n_0$>0 is attached to some choice of c and for each c there may be a different $n_0$>0. And you can not find a fixed $n_0$ for all c that satisfies the remaining because c can go very close to zero like c=0.000....00001

  2. If you take any as "there exist c" then the meaning is wrong also because for some c the remaining may apply but for another c the remaining may not apply.
    Example: let $f(n)=n$ and $g(n)=2n$:
    If $c=1$ then $n < 1 \times (2n)$ for $n \geq n0>0$
    But if $c=0.1$ then $n > 0.1 \times (2n)$ for all $n\geq n0>0$

So here "any" means "any" which is for all but one at a time , so in the little o asymptotic notation "any" means for all c>0 pick one at a time and the remaining should be satisfied.

Conclusion: Either do not use "any" in Math or explain to the reader what it means in your context.

  • 1
    $\begingroup$ Another common example for why to avoid "any" in mathematics is "pick any woman you want to be your wife". This doesn't mean you can choose every woman. $\endgroup$
    – DanielV
    May 7, 2019 at 10:59
  • $\begingroup$ @DanielV correct "any" is confusing $\endgroup$ May 7, 2019 at 11:07
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    $\begingroup$ I think your example for the third meaning is wrong. There, "for any" simply does indeed mean "for all". You can state that statement with existential quantifiers, and you would use $\forall c$, and with no confusion involved. I believe a third meaning does not exist. $\endgroup$ Dec 24, 2019 at 10:34
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    $\begingroup$ indeed the third meaning is nonsense. "for all c there is an n_0" means, because of the order of these statements, that the n_0 may depend on c. $\endgroup$
    – peter
    Feb 7, 2021 at 13:07
  1. Here's a comparison of the various translations of $$∀x{∈}F\; P(x).$$

    • For all elements $x$ in $F,\,P(x)$ holds” sometimes sounds like the property $P$ might belong to $F$ as a whole rather than to its individual members: “for all members of the family, they have a house” (1 house in total? or 5?). Contrast with “for each member of the family, they have a house” (definitely 5 houses in total). (“For all $x$” is Mathlish.)

    • For every element $x$ in $F,\,P(x)$ holds”, despite ‘every’ too having a collective sense, says that the property $P$ is common to the members of $F.$

    • For each element $x$ in $F,\,P(x)$ holds” directly attributes the property $P$ to individual members of $F.$

    • For any element $x$ in $F,\,P(x)$ holds” doesn't strongly communicate that the property $P$ belongs to each and every member of $F;$ nevertheless, “for any $x$” typically means “given an arbitrary $x$”, which, logically, is synonymous withfor every $x$” and “for each $x$”.

  2. The phrase ‘not any’ actually means ‘not some’ instead of ‘not every’.

    For example, observe that the sentences within each group below are equivalent to one another:

    • She does $\color\red{\textbf{not}}$ have $\color\red{\textbf{any}}$ disease.

      She does $\color\red{\textbf{not}}$ have $\color\red{\textbf{some}}$ disease.

      $\color\red{\boldsymbol{\lnot \exists}} x\, Dx$

      $\forall x\, \lnot Dx$

      For every disease, she does not have it.

    • She does not have every disease.

      $\lnot \forall x\, Dx$

      $\exists x\, \lnot Dx$

      For some disease, she does not have it.

  3. Consider this logical relationship: \begin{align} &&\forall x\;\big(Px\to Q\big)\tag{$\color{brown}{1a}$}\\ &\equiv{}&\big(\exists x\,Px\big)\to Q\tag{$\color{brown}{1e}$}\\&\large\not\equiv{}&\big(\forall x\,Px\big)\to Q\tag{$\color{blue}{2a}$}\\&\equiv{}&\exists x\;\big(Px\to Q\big).\tag{$\color{blue}{2e}$}\end{align}

    For each/every/any $x,$ if $x$ has property $P,$ then $Q$ is true. $\color{brown}{(1a)}$
    If some $x$ has property $P,$ then $Q$ is true.
    $\boxed{\textbf{Any }x}$ has property $P$ implies that $Q$ is true. $\quad\quad(\Large❔)$
    $\boxed{\textbf{If for any }x},\,x$ has property $P,$ then $Q$ is true.$\quad\quad(\Large❔)$
    $\boxed{\textbf{If any }x}$ has property $P,$ then $Q$ is true. $\quad\quad\quad\quad(\Large❔)$
    If each/every $x$ has property $P,$ then $Q$ is true. $\color{blue}{(2a)}$
    For some $x,$ if $x$ has property $P,$ then $Q$ is true. $\color{blue}{(2e)}$

    When the word ‘any’ is embedded in a conditional's antecedent, the idiomatic meaning is typically—though not always—sentence $\color{brown}{(1e)}$ instead of sentence $\color{blue}{(2a)};$ that is, ‘if any’ frequently means ‘if some’ instead of ‘if every’; sometimes, its intended meaning is ambiguous.

    For example, Wikipedia's definition of set disjointedness “A collection of two or more sets is disjoint $\boxed{\text{if any}} \,$two distinct sets of the collection are disjoint” requires external clarification whether three sets are disjoint

    • if some pair of distinct sets is disjoint


    • if every pair of distinct sets is disjoint;

    it turns out from reading between the lines two sections down that Wikipedia intends the ‘if every’ reading, even as the ‘if some’ reading is probably more immediate. No wonder authors disagree between these two contrasting definitions!

    Furthermore, compare

    • $\boxed{\text{If}}\,$the dog understands $\boxed{\text{any}}\,$command, then it is a genius.


    • $\boxed{\text{If}}\,$the dog understands $\boxed{\text{any}}\,$command, then it has been trained.

Summary: ‘for each’ and ‘for every’ are excellent translations of the universal quantifier; the word ‘any’ sometimes corresponds to ‘∃’ instead of ‘∀’, and in technical writing should be used judiciously.

  • $\begingroup$ your second equivalence is clearly wrong. $\endgroup$
    – peter
    Mar 3, 2023 at 20:28
  • $\begingroup$ yes. it seems to me that implication to the left is true, to the right not. to see this you can consider $Q:\Leftrightarrow\forall x:P(x)$. $\endgroup$
    – peter
    Mar 4, 2023 at 12:38
  • 1
    $\begingroup$ @peter $∃x(Px{\implies}∀xPx)$ is a first-order tautology (it's equivalent to $∃x(Px{\implies} ∀yPy)$), and $\Big((\forall x\,Px){\implies} Q\Big)\equiv\Big(\exists x\;(Px{\implies} Q)\Big)$ is indeed a logical equivalence. (Click on the links.) $\quad$ EDIT: I wrote a proof here. $\endgroup$
    – ryang
    Mar 4, 2023 at 13:12
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    $\begingroup$ interesting. i see no flaw in your proof, but its still hard to wrap my head around it. vacuous truth in the rhs is what my intuition overlooked. $\endgroup$
    – peter
    Mar 4, 2023 at 18:46
  • $\begingroup$ P.S. I rearranged the answer, so Peter's "second equivalence" now refers to the first equivalence. $\endgroup$
    – ryang
    May 25, 2023 at 8:58

Despite its score, ryang's post is totally wrong. [I'm a native English speaker and logician. And the downvotes on this post demonstrate the failure of the SE voting system.]

He incorrectly claims that "the phrase ‘not any’ actually means ‘not some’ instead of ‘not every’.", and says that the following are equivalent and examples of this claim:

(1) She does not have any disease.
(2) She does not have some disease.

This is bogus, as is obvious to any native English speaker. Consider another pair:

(3) He does not understand any complicated sentence.
(4) He does not understand some complicated sentence.

This pair obviously mean completely different things, contrary to ryang's claim.

He also incorrectly attempts to connect his incorrect claim with the FOL equivalence between "∀x ( P(x) ⇒ Q )" and "∃x ( P(x) ) ⇒ Q". This is quite obvious once you consider the following:

(5) If any student fails the course, he/she must retake the course.

Obviously, the pronoun "he/she" refers to "any student" that "fails the course". So one is NOT permitted to express this in anything close to the form of "∃x ( P(x) ) ⇒ Q"!

~ ~ ~

The true answer is that ryang is NOT describing actual English, and hence is not at all an answer to Bernd's question (and one wonders how many upvoters actually know proper English).

In proper English, "any" and related determiners "anyone" and "anything" function by inducing a ∀-quantifier at the global level, but not passing any indirect-statement boundary (including modal boundary) unless forced to. The determiners "some" and "every" and "each", on the other hand, stay as local as possible. And "some" means "any" when forced to extend beyond a local region.

The correct analysis of the above key examples are as follows:

(3) He does not understand any complicated sentence.
(3') ∀X∈ComplicatedSentences ( he does not understand X )

(4) He does not understand some complicated sentence.
(4') ∃X∈ComplicatedSentences ( he does not understand X )

(5) If any student fails the course, he/she must retake the course.
(5') ∀X∈Students ( X fails the course ⇒ X must retake the course )

Here are more examples:

(6) If some student fails the course, he/she must retake the course.
(6') ∀X∈Students ( X fails the course ⇒ X must retake the course )

(7) If every student of Paul fails some course, he will be fired.
(7') ( ∀X∈StudentsOf(Paul) ∃C∈Courses ( X fails C ) ) ⇒ ( Paul will be fired )

(8) If any student of Paul fails some course, he will be fired.
(8') ∀X∈StudentsOf(Paul) ( ∃C∈Courses ( X fails C ) ⇒ ( Paul will be fired ) )

(9) If everyone agrees on any additional rule, we will adopt it.
(9') ∀R∈AdditionalRules ( ∀X∈People ( X agrees on R ) ⇒ We will adopt R )

(10) If this job can be done by anyone, we would not need to hire any expert.
(10') ∀E∈Experts ( ( ∀X∈People ( this job can be done by X ) ) ⇒ we would not need to hire E ) )

In particular take a good look at (9) and (10), which demonstrates that my explanation is correct. In (9), "any additional rule" is quantified at the global level. In contrast, in (10), "done by anyone" is quantified only within the "can ..." boundary, but "any expert" is quantified at the global level.

Finally, the use of "if" instead of "iff" (if and only if) in a definition in conventional mathematics is an abuse of terminology and once you fix that then it becomes obvious that the quantifier represented by "any" does not pass the "iff". This fact about "iff" being a hard boundary for quantification is a convention but also partially logically motivated, but it will take too long to explain why...

  • $\begingroup$ A quick note regarding your critique: $\quad$ 1. My example "(2) She does not have some disease" is clearly an awkward, almost stilted, somewhat ambiguous sentence, and contriving to include it was not to reflect idiomatic English but merely to contrast 'not any' (¬∃x) with 'not every' (¬∀x). $\quad$ 2. Your beef over my answer's third point makes no sense, since its examples genuinely alternate between (∃x Px)⇒Q and (∀x Px)⇒Q, whereas your example "(5) If any(/some/a) student fails the course, they must retake the course" is clearly of the form ∀x(Px⇒Qx). $\endgroup$
    – ryang
    Jun 15, 2023 at 21:09
  • $\begingroup$ 3. I imagine you're referring to this; no its downvotes weren't mine. $\quad$ 4. None of your remaining examples (side note: methinks those hanging quantifiers complicate and distract from your core messages) contradict my answer, to which only example (8) is relevant; your example (8) is in fact logically equivalent to the form (∃x (∃c Fxc)) ⇒ Q, precisely illustrating my answer's third point that in the sentence (for any x, Px) ⇒ Q, 'any x' can mean 'some x'. $\endgroup$
    – ryang
    Jun 16, 2023 at 6:00

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