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In some books the word "if" is used in definitions and it is not clear if they actually mean "iff" (i.e "if and only if"). I'd like to know if in mathematical literature in general "if" in definitions means "iff".

For example I am reading "Essential topology" and the following definition is written:

In a topological space $T$, a collection $B$ of open subsets of $T$ is said to form a basis for the topology on $T$ if every open subset of $T$ can be written as a union of sets in $B$.

Should I assume the converse in such a case? Should I assume that given a basis $B$ for a topological space, every open set can be written as a union of sets in $B$?

This is just an example, I am not asking specifically about this sentence.

fiftyeight
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    Your suspect is well-founded: in mathematical literature always "if" in definitions means "iff" ! In fact "iff" theorems can be used as alternative definitions. – Tony Piccolo Nov 14 '13 at 09:17
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    The problem is that if "iff" is true, then the "if" only still holds valid. The "iff" always is the stronger expression. – Alfe Nov 14 '13 at 11:43
  • IFF suggests a two way implication, IF doesn't. So A IFF B does also mean B IFF A. (A <-> B) A IF B does only mean A -> B. – schlingel Nov 14 '13 at 14:39
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    I seem to recall reading that this, like many other conventions in mathematics, was established by Bourbaki. Does anyone have a reference that confirms or refutes this? – ronno Nov 14 '13 at 14:57
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    I see that schlignel and Alfe contradict to the Tony Piccolo's statement. They say that if < iff while Tony says that if = iff. Right? I wonder because math is claimed to be the ideally exact science and rigorous thereby and it is the science about proofs. How can it's logic be based on such terrible uncertainty? – Val Nov 14 '13 at 16:05
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    @Val: in fact, mathematicians have no difficulty with this at all, but students often do. The same could be said for many other things: playing violin, writing essays, etc. It takes practice to absorb the techniques of a field. – Carl Mummert Nov 15 '13 at 12:09
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    @Val It's also worth noting that Tony didn't say "if=iff". He said that this is true for definitions, but not necessarily always. – mdp Nov 15 '13 at 12:14
  • I think this post should be made CW, given the nature of the discussion it spawned. – tomasz Nov 15 '13 at 21:49
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    Bill Adams, my graduate algebra professor at UMCP, Always, when presenting a definition, reminded us that definitions are always if and only if. I'll never confuse them! Thanks, Bill! – DaveUM Nov 28 '13 at 20:54

15 Answers15

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As it is a definition, the validity of the property (here "being a basis for the topology") must be defined by it in all cases. So implicitly all cases not mentioned do not have the property.

This convention is even stronger than the "if" meaning "iff" in definitions. Take as example the definition "an integer $p>1$ is called a prime number if it cannot be written as a product $p=ab$ of integers $a,b>1$". This says that $6$ is not a prime number, since $6=2\times 3$; this is an instance of the "iff" meaning in definitions. But it also says implicitly that $1$ is not a prime number, nor $-5$ nor $\pi$ nor $\exp(\pi\mathbf i/3)$ nor $\mathbf{GL}(3,\Bbb R)$, as none of these can be described as integers $p>1$; even a statement with "iff" would in itself not seem to state non-primality of those objects. But since it is a definition, anything that does not match its description is implicitly excluded from the property.

Without this implicit exclusion of cases not mentioned, it would be very hard indeed to give a complete definition of any property. Imagine (assuming the "everything is a set" philosophy) the ugliness of "a set $x$ is called a prime number if $x\in\Bbb Z$ and $x>1$ and ...".

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    If I remember correctly, iff has been first used by Halmos who preferred it for definitions only. Only later the (horrible) word iff became used as a synonym of if and only if also in theorems. – egreg Nov 14 '13 at 10:03
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    A lot of answers are good, I chose this one because the writer mentioned the stronger implication that a definition has, stronger then an "iff" statement – fiftyeight Nov 18 '13 at 17:20
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    It's a very nice answer, but I am not sure if some of the points are universally true. Personally, I would take the given definition of a prime to mean precisely that being prime is a property that an integer $p > 1$ may have, and that all other objects simply fall outside the range of the definition (and hence $\pi$ is neither prime nor non-prime). In fact, $-5$ is prime in $\mathbb{Z}$, and $1+i$ is prime in $\mathbb{Z}[i]$. Of course, this is mostly a matter of convention and/or preference, but I feel that $6$ is non-prime in a fundamentally different way than $\pi$. – Jakub Konieczny Nov 19 '13 at 20:46
  • Of course! Where would one start "factoring" pi? – DaveUM Nov 28 '13 at 20:58
  • @Feanor: Whether the adjective "prime" is used in other contexts than that of "prime number" (which $1+i$ is not according to the definition I gave) is irrelevant to this question. If one defines "prime number" then the term must be completely and definitely defined. If one adopts the given definition (better ones certainly exist, but I wanted a formulation containing "if"), then one cannot later in a whim extend the notion to cover more things. You must be able to say for instance "for every prime number $p$ the ring $\Bbb Z/p\Bbb Z$ is a field" without risk of saying nonsense. – Marc van Leeuwen Nov 29 '13 at 04:38
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    @MarcvanLeeuwen: I definitely agree that the definition of $p$ being a prime number is fixed once and for all after the above definition is given, but this is not quite my point. My point is: the definition you gave strongly suggests that we should only apply it to integers greater than $1$, and not to anything else. This is surely a matter of convention, and possibly personal taste. But, in any case, I would be very surprised if I saw a definition like that, and then saw it argued that $\pi$ is not a prime number. At the same time, saying that $6$ is not a prime is natural. – Jakub Konieczny Nov 29 '13 at 13:07
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    @MarcvanLeeuwen: It seems to me that there are two ways of seeing things here; all I say is of course my personal impression. One can see definitions like the one above to implicitly exclude everything that does not fit (e.g. $p = 1+i$), and then safely say things like "if $p$ is a prime (number), then $\mathbb{Z}/p\mathbb{Z}$ is a field". Or, one can see such definitions to only apply to numbers; then one has to quantify "if $p \in \mathbb{N}$ is a prime (number), then $\mathbb{Z}/p\mathbb{Z}$ is a field", but also one can say "if $p$ is prime, $p|ab$ then $p|a$" or $p|b$ in more generality. – Jakub Konieczny Nov 29 '13 at 13:17
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    I am having trouble understanding how the definition is stronger than an iff statement. Suppose we said "an integer $p>1$ is called a prime number $\iff$ it cannot be written as a product $p=ab$ of integers $a,b>1$". Could someone please enlighten me on how this definition does not exclude $\pi$ from being a prime number? Sure, $\pi$ cannot be written as a product $ab$, but doesn't the quantifiction "an integer" apply to the whole sentence, both in the iff sentence and in the definition? – Ovi Dec 01 '17 at 15:26
  • @Ovi The number 1 cannot be written as a product of integers a,b > 1, so by the right to left direction of the "if and only if" implication it should be a prime number and also > 1, which is clearly not the case. This is the sense in which the if in a definition is stronger than the word iff. We could still make it work within a formal logic but it would be too tedious. – MuchToLearn Jun 28 '18 at 22:14
  • @Ovi The phrase "a quadratic polynomial in $\Bbb R[X]$ is irreducible if and only if its discriminant is negative" is a (true) statement that claims nothing about complex polynomials, or about real polynomials of degree${}\neq2$, or about other things to which the predicate "irreducible" might apply. In particular it does not say "a polynomial is irreducible iff it is real, quadratic, and has a negative discriminant", which would be quite false. – Marc van Leeuwen Jul 02 '18 at 11:28
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    @MuchToLearn: I don't agree with you; when not used as definition, my sentence "an integer${}>1$ is a prime number if..." does not claim anything about $1$ being prime or not (nor about non-positive or non-integer numbers), and it wouldn't even with "if" replaced by "iff". It is only the context of this being a definition that allows deducing that being a prime number presupposes being an integer${}>1$. – Marc van Leeuwen Jul 02 '18 at 11:35
  • Something can be called a prime number even if it is not an integer p > 1, can't it? For instance 1+i is a prime number in the Gaussian integers. So isn't it effectively still ambiguous in your statement whether for instance the integer p=1 is a prime number or not? – Klaas van Aarsen Aug 17 '19 at 12:12
  • @KlaasvanAarsen I was just giving an example definition and it meaning, not pretending this is the best definition of prime number possible. But for what it's worth, if you accept this definition, then you are reserving the term "prime number" for certain natural numbers, and you cannot later allow it to include certain Gaussian integers (which by the way are not integers in the usual definition of that term; terminology is sometimes illogical). When we say Euclid proved there were infinitely many prime numbers (though he didn't say that), we certainly don't want them to be Gaussian ones. – Marc van Leeuwen Aug 17 '19 at 12:24
  • That's my point @MarcvanLeeuwen. We have to include somewhere that we're specifically only talking about natural numbers. And we have to effectively specify that everything we did not define for a prime number is not a prime number within that domain. Note that -5 is actually a prime number within the integers, but not within the natural numbers. – Klaas van Aarsen Aug 17 '19 at 12:38
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    @KlaasvanAarsen No it is not. In a ring (of which the integers are an archetypical example) one can define when elements are prime or irreducible, and both of these apply to $-5$ as element of $\Bbb Z$, but that does not make $-5$ a prime number. As a matter of sanity, one cannot use for a new notion a term that already means something else, though one can use something similar. (But I don't claim there are no examples of such terminology abuse.) – Marc van Leeuwen Aug 17 '19 at 12:53
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    @MarcvanLeeuwen, the wiki article is inconsistent. It starts with an introductory text that says that a prime number is a natural number (which is wrong imho). But in the later definition it says that 'A natural number is called a prime number ...'. That is the correct definition since it makes the definition within the particular domain. And yes, if the word 'prime number' is used without context, we generally assume that we are talking about the natural numbers. Can't trust introductory texts in wiki - they are not the definitions. – Klaas van Aarsen Aug 17 '19 at 13:48
  • @KlaasvanAarsen I think we are straying pretty far of topic, but I guess you really need to have the last word. That the WP article is not entirely consistent in style is no surprise, as there are usually many contributors, but at least this one does not contradict itself. And there is good news: you can contribute too. So why not change the article so that it becomes perfect (iyho)? – Marc van Leeuwen Aug 17 '19 at 16:37
  • True. And I just realized that the wiki definition was actually wrong so I fixed it just now. TBH, I'm still trying to understand why a definition is conventionally written with "if". Sorry for straying. – Klaas van Aarsen Aug 17 '19 at 17:03
  • @MarcvanLeeuwen Does it make a difference if we use instead of " an integer is called prime" just "an integer is prime"? – user599310 May 14 '21 at 22:47
  • Just to reiterate @Jakub's point: based on the conventional definitions "a natural number is prime if..." and "a function $f$ is continuous at point $a$ in its domain if...", saying that $\sqrt2$ is not prime or that the reciprocal function is discontinuous at $0$ is perhaps as meaningful as saying that $\sqrt2$ is agitated. – ryang Mar 09 '22 at 12:55
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As other pointed out, the "only if" part wouldn't make any sense as the process of defining the object is not over yet at the time you read the "only if".

But if it bothers you, you can always reformulate

An object $A$ is said to be new term if $P(A)$.

by

A new term is an object $A$ such that $P(A)$.

In your example, the reformulation would be :

A basis $\mathcal B$ of a topological space $T$ is a collection of open sets of $T$ such that every open set of $T$ is a union of elements of $\mathcal B$.

Pece
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    Good point the alternative formulation! – Marc van Leeuwen Nov 14 '13 at 10:47
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    This really is the key point. A definition is not at all the same kind of statement as a usual "if and only if", because one of the two sides is undefined until the definition is made. So a definition is really a statement that the newly defined term should be identified ("is") with a previously defined concept. It is conventional to use "if" rather than "if and only if" for this identification, but this is not the same "if" as the material conditional used to reason about mathematical theorems. – Carl Mummert Nov 15 '13 at 12:03
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    @CarlMummert You should really write that as answer. You captured the essential points, succinctly and correctly. Your statement also happens to be consistent with established standards of notation, which is helpful. You are (correctly) distinguishing between an axiom and a theorem i.e. hypothetical/ postulate, I believe. – Ellie Kesselman Nov 15 '13 at 12:10
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    @CarlMummert Surprise, surprise (not really :o) You are a professor of mathematics and computational logic! No wonder you express yourself so fluently, or rather, so elegantly. You will probably deny that there is any causality, while everyone else will continue thinking I am an unenlightened PhD fan grrl, but I have grown accustomed to that by now! – Ellie Kesselman Nov 15 '13 at 12:19
  • @Pece Is there a difference if we say "A new term is called an object A iff P(A)" ? – user599310 Feb 03 '21 at 17:32
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The deeper point here is that regardless whether we write "if" or "if and only if" in a definition, this is not the same as the usual meaning of "if and only if" as a material equivalence. The material equivalence of two propositions is the assertion that each one independently has a truth value, and the truth values are the same.

In a definition such as

a natural number $n$ is even if and only if $n$ is a multiple of $2$

we cannot read the "if and only if" as a material equivalence, because the left hand side does not have any truth value until after the definition is made. To view this definition as expressing an equivalence of two statements, we would need to already know that each side of the equivalence statement already has a truth value, but "$n$ is even" is (trivially) undefined before the definition is made.

Instead of expressing an equivalence, a definition tells us that the defined term is to be viewed as a synonym for the definition: a definition expresses an "is" relationship, not an "is equivalent to" relationship. For example, if I prove that $8$ is even, I do not need to apply modus ponens and the definition of an even number to conclude that $8$ is a multiple of $2$. Instead, I can directly "apply the definition": proving that "$8$ is even" already proves that "$8$ is a multiple of $2$", because the quoted phrases are synonyms for each other. One could also say that "$8$ is even" is an abbreviation of "$8$ is a multiple of $2$". In turn, that latter phrase is an abbreviation for "there is a natural number $m$ with $8=2m$".

In principle, one can remove all later definitions from an axiomatic theory, so that in the end one is left with synonymous statements that involve only the "undefined terms" of the theory.

For example, in Hilbert's axiomatic Euclidean plane geometry, there are only three undefined types of objects: point, lines, and planes. Every other statement, such as "the three angles in an equilateral triangle are all congruent", is in fact an abbreviation for a much longer statement about lines and points. This is not to say that the statement is equivalent to the longer statement about lines and points. The statement about angles has no truth value at all in this axiomatic framework except by virtue of its definition.

Carl Mummert
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Yes. This is an unfortunate convention but is firmly established.

hunter
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    Please cite your sources or provide additional details. Firmly established according to whom, or what? I usually wouldn't be so pointedly interrogative, as it can be considered rude. However, your answer received 14 up votes. That isn't your fault, but the burden of proof is yours. – Ellie Kesselman Nov 15 '13 at 12:26
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    @Feral Oink: thank you for your kind words above. Let me say in hunter's defense that on this site (unlike, say, Wikipedia), it's fine to write from personal knowledge. If you think someone is wrong, explain why. Or just downvote - that's fine too. But there is no obligation for someone to provide references for everything they claim. In this way, this site is a better reflection of real mathematical practice than Wikipedia. In principle, the vote count measures whether others agree with the answer. For example, I upvoted this answer earlier because it's correct, although it is a little terse. – Carl Mummert Nov 15 '13 at 12:50
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    I would agree with it too, only I don't believe it to be unfortunate. "If and only if" makes a definition sound like a theorem about two things being equal. If someone writes a definition that says “A is B iff C” it makes me wonder if I'm supposed to know what A is to begin with. It also blurs the distinction between the term that is being defined and the one which is used to define (which can be very confusing sometimes, much like theorems of form $A=B=C$). And not all languages have a shorthand expression such as iff, and “if and only if” spelled out is quite tedious to write on blackboard. – tomasz Nov 15 '13 at 21:43
  • @CarlMummert In many respects, I feel that StackExchange content is superior to Wikipedia, with or without references. This answer is just so very terse, thus my request. Your mere comment, and those made by others on prior answers, were lengthier and more substantive. But that is okay, as I should just wander back to English and Money and WebApps SE now, which is where I belong anyway. – Ellie Kesselman Nov 15 '13 at 22:25
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    @FeralOink: I understand your displeasure with this post, but is a citation really what it lacks? What kind of citation would make you happy, other than simply linking to another answer to this question offsite? It is an established convention. The fact so many people upvoted this answer seems to certify that, at the very least (and I agree, though I did not upvote). Are you questioning that? I just opened a few random textbooks from different branches of mathematics I had on my computer and they all used this convention (one used "provided" instead of "if"). – tomasz Nov 16 '13 at 01:04
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    @tomasz It doesn't need a citation necessarily. If this answer is so obvious, then why is it necessary to ask the question at all? Also, why is it necessary for all the other answers to have substantive content, if the answer is so trivially obvious? – Ellie Kesselman Nov 16 '13 at 01:57
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    @FeralOink: Some answers are only obvious when they are spelled out. Besides, to someone new to studying mathematics, it might seem like it's a coincidence of sorts, or a misunderstanding on his part, so it's enough for someone to say "yeah, you're right, that's the way it is". And once you're accustomed the answer is obvious, and as Carl Mummert said, the community certifies its veracity (of course, sometimes community certifies wrong answers and ignores right ones, but that's a different problem). The answer may be a little too concise, but I don't believe a citation would fix that at all. – tomasz Nov 16 '13 at 11:59
  • @tomasz It doesn't need a citation. It needs something, else there is a lack of consistency with the other good answers. It isn't fair to them. I don't especially care, but am puzzled that you continue to assert that on Math SE, it is sufficient for one to answer with "yeah that's the way it is" and get 16 up votes. And also that you would continue to claim that the answer "may be a little too concise". May be, a little, and too concise are qualifiers that would be confusing to someone new studying mathematics, as that would imply that this sort of answer were adequate. Enough! You win. – Ellie Kesselman Nov 17 '13 at 12:36
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    @FeralOink: I'm not saying that the answer is adequate (I don't really think so), the 17 upvoters who apparently appreciate a very pithy answer do say so. And I don't think you can hold the poster responsible for getting too many upvotes (unless the answer is flat out wrong, in which case he should delete it asap so as not to really confuse non-specialists). I never argued that the answer couldn't use some elaboration, just that there's no need for citation. Pardon me for avoiding more harsh formulations. :) I just don't think the answer is as awful as you seem to see it as. – tomasz Nov 17 '13 at 22:11
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My usual interpretation of this is that the "only if" is implicit. As an example, take the sentence "We call a gadget $A$ a widget if property $P$ holds for $A$", where the sentence is defining the word "widget". Because you just "invented" the word widget, it can only possibly apply to gadgets satisfying property $P$; any gadget that doesn't satisfy this property (or anything that isn't a gadget) cannot be a widget, because you didn't say it was.

So while the statement only gave the "if" statement, the fact that this statement is a definition implies the "only if" statement. (I'm not really claiming to have given a proof of anything, but merely some justification of why the convention is not as crazy or unfortunate as it might appear).

mdp
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    It is also relevant that the now-ubiquitous "iff" has only started to get used a lot since the 1950s. It is more sensible to drop "and only if" from every definition as a convenience than the mere second "f" -- and definition formulation practices probably predate the use of "iff". – Lord_Farin Nov 14 '13 at 10:03
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    @Lord_Farin That is true. I'm also on Serre's side in thinking that "iff" should only be used in blackboard-maths or scribblings, and not in print, so the convenience argument still holds for me. – mdp Nov 14 '13 at 10:06
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    The fact that it is a definition doesn't imply the 'only if' part. It is just a convention to leave out that part. Saying that an A is a B if P holds, but that an A is a B also if Q doesn't give a contradiction when P is contrary to Q. – Angelorf Nov 14 '13 at 17:07
  • @Angelorf That is correct, but I was only speaking loosely anyway. I agree that ultimately it is a convention, but was trying to give a justification as to why it is a sensible convention. It would be unusual to make two separate definitions of the form you describe. – mdp Nov 14 '13 at 17:15
  • @Angelorf In fact, the more I think about this, the more I think such things have to be made illegal, because you need the "only if" direction in proofs. If you write "$A$ is a $B$ if $P$ holds", but don't assume that $A$ can't also be a $B$ for other reasons, then you can't prove any theorem of the form "Let $A$ be a $B$. Then...". I suppose one way round this is to write if and only if in definitions, but another is to ban defining the same thing twice in non-equivalent ways (at least locally). – mdp Nov 15 '13 at 09:59
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    @Matt Pressland: of course someone does have to know how an object is defined in order to work with the definition. But that is not an "only if", it is just an application of the definition. In other words, defined terms are not "equivalent" to their definitions, they are synonyms for their definitions. The use of "if" in a definition is a convention, but it is not the same "if" as in "if $n|4$ then $n|8$". But the use of definitions is closely tied to the regular "if/then", and so one can get very far without ever worrying about the difference between an equivalence and a synonym. – Carl Mummert Nov 15 '13 at 12:07
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    @CarlMummert I am aware that there are different uses of if here. I think I am causing confusing by varying between being rigourous and hand-waving. My point was that if you choose to interpret the "if" in a definition as a logical implication, it is necessary that the converse implication also be true in order to prove any theorems. You need to be able to conclude "$P$ holds" from "$A$ is a $B$", and not just vice versa. – mdp Nov 15 '13 at 12:12
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    Yes, if you try to treat the "if" in a definition as a material implication, you have to read it as "if and only if". But, because the term being defined is not yet defined before the definition is made, it makes no sense to claim that the term is materially equivalent to something else. In other words, upon reflection, it is not coherent to read a definition as a material equivalence. This is why, historically, definitions were not viewed that way, but are viewed as an entirely different kind of statement. – Carl Mummert Nov 15 '13 at 12:45
  • @CarlMummert Sure - I don't think you disagree with what I am trying to say (although it's highly possible that I've failed to express what I'm trying to say well enough to make this clear). I agree with another commenter that it might be useful for you to collect some of your comments into an answer. Edit: Oh, you just did! – mdp Nov 15 '13 at 13:17
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Usually, in definitions you will not read iff, since the inverse inclusion makes no sense, for a definition establishes no relationship between two mathematical objects, but instead provides a name to one of both.

Nico
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I would say that the use of 'if' in such a sentence should be considered a non-mathematical use of 'if' as opposed to a mathematical if. This is because statements of the form 'x is called y' are meta-mathematical rather than mathematical.

A strict reading of 'if' in its logical meaning often does not make sense.

A function is called 'continuous' if it satisfies ...

Many functions which satisfy the property in question are never and have never been called 'continuous' (because no-one has ever had cause to study, name or classify them). 'Only if' would actually be more appropriate here.

jwg
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See Tarski's contribution in his Introduction to Logic and to the Methodology of Deductive Sciences, pt 1 "Elements of Logic. Deductive Method.", §II. "On the Sentential Calculus":

10. Equivalence of sentences

We shall consider one more expression from the field of sentential calculus. It is one which is comparatively rarely met in everyday language, namely, the phrase “if, and only if”. If any two sentences are joined up by this phrase, the result is a compound sentence called an EQUIVALENCE. The two sentences connected in this way are referred to as the LEFT and RIGHT SIDE OF THE EQUIVALENCE. By asserting the equivalence of two sentences, it is intended to exclude the possibility that one is true and the other false; an equivalence, therefore, is true if its left and right sides are either both true or both false, and otherwise the equivalence is false.

The sense of an equivalence can also be characterized in still another way. If, in a conditional sentence, we interchange antecedent and consequent, we obtain a new sentence which, in its relation to the original sentence, is called the CONVERSE SENTENCE (or the CONVERSE OF THE GIVEN SENTENCE). Let us take, for instance, as the original sentence the implication:

(I) if $x$ is a positive number, then $2x$ is a positive number;

the converse of this sentence will then be:

(II) if $2x$ is a positive number, then $x$ is a positive number.

As is shown by this example, it occurs that the converse of a true sentence is true. In order to see, on the other hand, that this is not a general rule, it is sufficient to replace “$2x$” by “$x^2$” in (I) and (II); the sentence (I) will remain true, while the sentence (II) becomes false. If, now, it happens that two conditional sentences, of which one is the converse of the other, are both true, then the fact of their simultaneous truth can also be expressed by joining the antecedent and consequent of any one of the two sentences by the words “if, and only if”. Thus, the above two implications—the original sentence (I) and the converse sentence (II)—may be replaced by a single sentence:

$x$ is a positive number if, and only if, $2x$ is a positive number

(in which the two sides of the equivalence may yet be interchanged).

There are, incidentally, still a few more possible formulations which may serve to express the same idea, e.g.:

from: $x$ is a positive number, it follows: $2x$ is a positive number, and conversely;

the conditions that $x$ is a positive number and that $2x$ is a positive number are equivalent with each other;

the condition that $x$ is a positive number is both necessary and sufficient for $2x$ to be a positive number;

for $x$ to be a positive number it is necessary and sufficient that $2x$ be a positive number.

Instead of joining two sentences by the phrase “if, and only if”, it is therefore, in general, also possible to say that the RELATION OF CONSEQUENCE holds between these two sentences IN BOTH DIRECTIONS, or that the two sentences are EQUIVALENT, or, finally, that each of the two sentences represents a NECESSARY AND SUFFICIENT CONDITION for the other.

For me Tarski's authority is beyond discussion.

Tony Piccolo
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    For those without Google Books access, here is the quote from Tarski: "It is worth noticing that mathematicians, in laying down definitions, prefer the words "if" or "in case that" to the phrase "if, and only if". Thus, to the definition of the symbol $\leq$, they would, presumably, give the following form: we say that $x \leq y$, if it is not the case that $x > y$. It looks as if such a definition merely states that the definiendum follows from the definiens, without emphasising that the relation of consequence also holds in the opposite directions, – Carl Mummert Nov 15 '13 at 12:55
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    (continued) "and thus fails to express the equivalence of definiendum and definiens. But what we actually have here is a tacit convention to the effect that "if" or "in case that", if use to join definiendum and definiens, are to mean the same as the phrase "if, and only if" ordinarily does." – Carl Mummert Nov 15 '13 at 12:56
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    While Tarski's authority is considerable, it does not mean we should blindly follow everything he said (especially considering he was quite a mean person from what I've heard). That said, the quote is quite sensible. ;) – tomasz Nov 15 '13 at 21:51
  • @CarlMummert What you quote is the next section (§11) after what I quoted in my edit to Tony Piccolo's answer. – Geremia Feb 06 '18 at 16:30
  • @TonyPiccolo You're right that Tarski explained iff very well. I assume you were referring to §10, because fityeight's question here is about interchangeability of iff vs. if statements; althugh §11, where Tarski relates iff to definitions, is certainly apropos, too. I could add a quote of §11, too, if you think it's necessary. – Geremia Feb 06 '18 at 16:32
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I want to add some comments on this topic and in particular regarding Tarski's reference [see Tony Piccolo's answer and Carl Mummert's comments].

Ref to George Tourlakis, Lectures in Logic and Set Theory : Volume 1, Mathematical Logic (2003), I.7. Defined Symbols [page 112-on] :

There are three possible kinds of formal abbreviations, namely, abbreviations of formulas, abbreviations of variable terms (i.e., “objects” that depend on free variables), and abbreviations of constant terms (i.e., “objects” that do not depend on free variables). Correspondingly, we introduce a new nonlogical symbol for a predicate, a function, or a constant in order to accomplish such abbreviations.

I will consider only the predicate case, in order to cope with Tarski's example, i.e. :

$x \le y =_{def} \lnot (x > y)$.

We have that :

if $Q(\overrightarrow{x}_n)$ is some formula, we then can introduce a new predicate symbol “$P$” that stands for $Q$.

Note. In the present description, $Q$ is a syntactic (meta-)variable, while $P$ is a new formal predicate symbol.

This entails adding $P$ to [the language] $\mathcal L$ (i.e., to its alphabet $\mathcal V_k$) as a new $n$-ary predicate symbol [obtaining a new language $\mathcal L_1$], and adding

$P\overrightarrow{x}_n \leftrightarrow Q(\overrightarrow{x}_n) \quad \quad$ (i)

to [the theory] $\Gamma$ as the defining axiom for $P$ [obtaining a new theory $\Gamma_1$].

Suppose that $\mathcal F$ is a formula over $\mathcal L_1$, and that the predicate $P$ whose definition took us from $\mathcal L$ to $\mathcal L_1$, and hence is a symbol of $\mathcal L_1$ but not of $\mathcal L$) occurs in $\mathcal F$ zero or more times. Assume that $P$ has been defined by the axiom (i) above (included in $\Gamma_1$), where $Q$ is a formula over $\mathcal L$.

We eliminate $P$ from $\mathcal F$ by replacing all its occurrences by $Q$. [...] This results to a formula $\mathcal F^*$ over $\mathcal L$.

Now we have two basic (meta)-theorems :

I.7.1 Metatheorem (Elimination of Defined Symbols: I). Let $\Gamma$ be any theory over some formal language $\mathcal L$ [page 116] .

(a) Let the formula $Q$ be over $\mathcal L$, and $P$ be a new predicate symbol that extends $\mathcal L$ into $\mathcal L'$ and $\Gamma$ into $\Gamma'$ via the axiom $P\overrightarrow{x}_n \leftrightarrow Q(\overrightarrow{x}_n)$. Then, for any formula $\mathcal F$ over $\mathcal L$, the elimination [of $P$] as above yields a $\mathcal F^*$ over $\mathcal L$ such that :

$\Gamma \vdash \mathcal F \leftrightarrow \mathcal F^*$.

(b) [consider the case of a $n$-ary ($n \ge 0$) function symbol “$f$”, where the case $n=0$ applies to constants].

And :

I.7.3 Metatheorem (Elimination of Defined Symbols: II). Let $\Gamma$ be a theory over a language $\mathcal L$ [page 118] .

(a) If $\mathcal L'$ denotes the extension of $\mathcal L$ by the new predicate symbol $P$, and $\Gamma'$ denotes the extension of $\Gamma$ by the addition of the axiom $P\overrightarrow{x}_n \leftrightarrow Q(\overrightarrow{x}_n)$, where $Q$ is a formula over $\mathcal L$, then $\Gamma \vdash \mathcal F$ for any formula $\mathcal F$ over $\mathcal L$ such that $\Gamma ' \vdash \mathcal F$.

(b) [consider the case of a $n$-ary ($n \ge 0$) function symbol “$f$”, where the case $n=0$ applies to constants].

According to this formal treatment of definition, it is the "defining axiom" :

$x \le y \leftrightarrow \lnot (x > y)$

that licenses the use of $\le$ in the "extended" theory.

1

In a definition, when we say that X is called by the name Y, if it satisfies certain properties, we usually mean that entities in the same universe of discourse and of the same kind as X are not called by the name Y if they do not satisfy those properties.

The predicate is_called(X, "Y") is not a proposition in the domain of discourse; it is in the layer of symbols that are being used to discuss the domain, the "meta-domain", if you will.

Something in the actual domain cannot be put in a logical relationship with something in the meta-domain; that is a level violation.

It is never sensible to say:

If a number is divisible by four, it is called "even".

because a definition is expected to be complete, and some numbers that are not divisible by four are also even. If we drop the "called", it's something completely different:

If a number is divisible by four, it is even.

Now the symbol "even" is not being quoted for the purposes of being defined; it is evaluated and replaced by the properties that it denotes, and so we have a logical relationship being expressed purely in the domain of discourse.

So, in a way, the "if" in a definition is related to "if and only if": clearly, a number is not called even if it is not divisible by two, therefore a number is called even if, and only if it is divisible by two. However, it is redundant and verbose. Moreover, definitions which use "if" can always be rewritten into ones which do not use if at all:

Integers divisible by two are called "even".

We can, and should, use "if and only if", whenever we give a definition in such a way that we are not quoting the name. Suppose I have never defined what it means to be "even". It is appropriate to say:

  1. Integers can have certain property: they can be called even under certain conditions.

  2. An integer is even if and only if it is divisible by two.

Here, the iff is necessary, because the second statement isn't asserting itself as a definition of the term. I didn't use the word "called" or similar.

One or more such statements can constitute an implicit definition of the term, when they specify enough rules that every integer can be classified as even or not even.

The use of iff in the second statement above informs the reader, "I am the only rule you need to deduce what set of properties are denoted by the term 'even'".

Kaz
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In a definition, the meaning of a new phrase or word is described with the help of known terms. After the definition, "if" becomes "if and only if".

R K Sinha
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There is a logical difference between saying:

  1. The sidewalk is wet if it rained.
    • "it rained" ⇒ "the sidewalk is wet"

and

  1. The sidewalk is wet iff it rained.
    • "the sidewalk is wet" ⇒ "it rained"
           &
    • "it rained" ⇒ "the sidewalk is wet" (← converse)

("• ⇒ •" meaning "If •, then •.")

(1) means that "it rained" being true is a sufficient but not necessarily the sole condition for "The sidewalk is wet" to be true; other conditions might make it true.

(2) means that "it rained" is the only way "The sidewalk is wet" can be true (necessary & sufficient condition).

So, are "iff" and "if" interchangeable? Not necessarily:

  • Given a true iff statement, replacing "iff" with "if" always results in another true statement.
    ("iff" is stronger; it says more; it's a shorthand for two conditional propositions; cf. (1) vs. (2) above.)

  • Given a true if statement, replacing "if" with "iff" doesn't necessarily result in another true statement.


The OED's definition of "iff" corroborates what I wrote above:

to introduce a condition that is necessary as well as sufficient, or a statement that is implied by and implies the preceding one.

Geremia
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  1. Think about the definition: "All x are A if they are B".

This actually says that all B's are A.

  1. Now, think about this: "All x are A if and only if they are B".

This says that all B's are A and all A's are B.

  1. Now, if you say "All x are A if they are B and they are C".

What is told here is if something is both B and C, then it is A.

  1. Last one, "All x are A if and only if they are B and they are C".

This means, if something is both B and C, they it is A, besides, if something is A, then it is both B and C.

Now what does it actually mean?

  1. $\forall x, x\in B \implies x\in A$

This is equivalent to: $B \subset A$

  1. $\forall x, x\in B \iff x\in A$

This is equivalet to: $B = A$

  1. $\forall x, (x\in B \land x\in C) \implies x\in A$

This is equivalent to: $x\in (B\cap C) \implies x\in A$

And thus, $(B\cap C) \subset A$

  1. $\forall x(a\in B \land x\in C) \iff x\in A$

And finally what this says is : $(B\cap C) =A$

Zafer Sernikli
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My answer to your question, in short, is: Strictly speaking, no, ‘if’ and ‘iff’ are not interchangeable, but nevertheless there is a very general tacit convention for using ‘if’ in definitions where ‘iff’ is meant.

The Tarski quote referred to by Tony Piccolo establishes (by authority) that using ‘if’ where ‘if, and only if’ is (or should be) meant, is a tacit convention. The answer by hunter mentions that it is an ‘unfortunate convention’, but gives no justification for judging it unfortunate. I concur with hunter and in this answer give arguments that I feel justify his claim.

  1. A first argument is the fact that you asked this question (and others did as well: there are several duplicates). When the convention would be to use ‘iff’, you would not have asked the question.
  2. A second argument is the fact that, in case you stumble upon a text in which the convention is not followed, misinterpretation of that text is possible.
  3. A third argument is that it limits expressiveness, or at least forces one to find alternate formulations to one that would seem the first to present itself.

    This argument requires a bit more explanation: In other answers, it was stressed that the ‘if’ (or ‘iff’) linking definiendum and definiens should not be confused with material implication $\Rightarrow$ (or equivalence $\Leftrightarrow$), but give meaning on a higher level. We also have this distinction with other mathematical terms, ‘equality’ $=$, for example. When equality is used to define an object, many authors use some distinctive variant of $=$ to make this clear, e.g., $:=$ or $\triangleq$; let me use the former. Once $a:=3$, i.e., I have defined $a$ to be $3$, I can state $a=3$, i.e., $a$ is equal to $3$.

    Now, of course it is possible to introduce similar notation for $\Rightarrow$ and $\Leftrightarrow$ (and I have come across texts where this was done, but don't ask me for a reference):

    • Stating $x\in X \mathbin{:\Rightarrow} x/2\in\mathbb{Z}$ means that you've defined (specified) that $X$ only contains even numbers (but not necessarily all of them).
    • Stating $x\in X \mathbin{:\Leftrightarrow} x/2\in\mathbb{Z}$ defines $X$ to be the set of even numbers.
    • Stating $x\in X \mathbin{:\Leftarrow} x/2\in\mathbb{Z}$ means that you've defined (specified) that $X$ contains at least all even numbers (and possibly other objects).

    As in general, using much mathematical notation reduces the readability, so one would like to be able to express the above definition variants using natural language. Two natural pairs of correspondences that present themselves are ($:\Leftrightarrow$, ‘iff’) and ($:\Leftarrow$, ‘if’). As the convention conflates ‘if’ with ‘iff’, the latter pair cannot be used, and alternative, less natural formulations for it have to be found. This is what I meant when stating that the convention limits expressiveness.

    One can argue that natural language involving ‘if’ in day-to-day usage is not strict enough for it to be safely used in a more formal mathematical context. This is a valid criticism, but it would also proscribe the usage of ‘if’ in definitions where ‘iff’ is meant (as per Tarski's quote referred to above). The usage of ‘iff’ (or ‘if, and only if’), however, is not affected by this criticism: its verbosity makes it robust against misinterpretation.

equaeghe
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"if" and "iff (if and only if)" have the difference. This difference is about their imperativeness. The "iff" is more imperative than just "if".

The analogous can be found in "needed and sufficient condition". "if" is analogue to "sufficient condition", whereas "iff" is analogue to "needed and sufficient condition".

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    It escapes me how iff is imperative (and I suppose ifff is outright dictatorial), there is nothing to command. "I will slap you if you don't obey me" sounds rather imperative; the possibility that I might even slap you if you do obey does not seem to make it less so. – Marc van Leeuwen Nov 14 '13 at 14:13
  • @MarcvanLeeuwen Ok, as it escapes you, I'll catch it for you. Saying "A takes place if B does" does not mean that A cannot take palce when B does not but, say, C does. Saying 'A takes place if and only if B does' means that A cannot take place when B does not. It is nearly the same relationship as within needed and sufficient conditions. – Sergey Kanaev Sep 22 '15 at 11:53
  • +1 for mentioning sufficient vs. necessary+sufficient conditions – Geremia Feb 06 '18 at 03:17
  • @MarcvanLeeuwen Given a true iff statement, replacing "iff" with "if" always results in another true statement; thus, the difference is emphasis (what Sergey is calling "imperativeness"). – Geremia Feb 06 '18 at 03:21