Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question
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 9:17
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
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
@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. –  Matthew Pressland Nov 15 '13 at 12:14
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

14 Answers 14

up vote 33 down vote accepted

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 ...".

share|cite|improve this answer
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
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
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 $\pi$. –  Feanor Nov 19 '13 at 20:46
@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. –  Feanor Nov 29 '13 at 13:07
@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. –  Feanor Nov 29 '13 at 13:17

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 $P(A)$.


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 $\mathcal B$.

share|cite|improve this answer
Good point the alternative formulation! –  Marc van Leeuwen Nov 14 '13 at 10:47
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
@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
@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

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

share|cite|improve this answer
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
@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
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
@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 1:04
@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 1:57

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).

share|cite|improve this answer
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
@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. –  Matthew Pressland Nov 14 '13 at 10:06
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. –  Matthew Pressland Nov 14 '13 at 17:15
@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

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.

share|cite|improve this answer
Also an excellent answer. If I could have chosen two answers I would choose this one as well –  fiftyeight Nov 18 '13 at 17:22
That's an excellent point, and also one which is easy to overlook. –  Feanor Nov 29 '13 at 13:22
It's really amazing. –  user 170039 Jul 9 at 12:11
I think was far and away the nicest answer on this thread. Very nice. +1 –  Daniel W. Farlow Aug 6 at 5:16

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.

share|cite|improve this answer

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.

share|cite|improve this answer

See Tarski's contribution in his Introduction to Logic and to the Methodology of Deductive Sciences p. 36

For me Tarski's authority is beyond discussion.

share|cite|improve this answer
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
(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
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

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'".

share|cite|improve this answer

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.

share|cite|improve this answer
It's really great. –  user 170039 Jul 9 at 7:44

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".

share|cite|improve this answer
  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 : $(B\cap C) =A$

share|cite|improve this answer

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.

share|cite|improve this answer

"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".

share|cite|improve this answer
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. –  Sergy Kanaev Sep 22 at 11:53

protected by Michael Greinecker Nov 15 '13 at 6:18

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.