The statement is from the book Linear Integral Equations by Rainer Kress:

A compact operator cannot have a bounded inverse unless its range has finite dimension.

Here are my questions:

  • How should I understand the word "unless" here? If one interprets the statement in the form of $A\rightarrow B$, what should be $A$ and $B$ respectively?
  • Is there an operator such that its range has finite dimension, but it does not have a bounded inverse? There can be two cases for "not have a bounded inverse":
    • It does not have inverse.
    • It has an inverse but the inverse is unbounded.

For the first question, I think we can translate a statement of the form "$P$ unless $Q$" as "if not $Q$, then $P$", or: $\neg Q → P$, according to this note for formal logic. But I'm not sure if the "unless" above is the same as the one here, which is the very reason why I ask the second question.

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    $\begingroup$ @QiaochuYuan: I don't buy that interpreation. The sentence "$A$ is false unless $B$" certainly asserts $A\to B$, or equivalently $\neg B\to\neg A$, but you seem to want it (also?) to assert $B\to A$? I don't think that's how the word is used in English; one can assert "it's not a real party unless someone pukes" but still concede that sometimes people puke without there being a real party on (say, if they have the flu). $\endgroup$ Mar 17, 2012 at 22:25
  • 2
    $\begingroup$ @Henning: ah, you're right. I confused myself. $\endgroup$ Mar 18, 2012 at 7:18

3 Answers 3


As a general warning (mostly addressed to other attempts to answer this), attempts to translate natural language structures into formal propositional logic are prone to failure because (a) the logical structure is not always clear or unambiguous from the surface structure alone (roughly, the words used) and (b) propositional operators, like negation ("not"), implication ("implies"), and disjunction ("or"), are truth-functional, which means that when you use an operator to build a complex proposition, the truth-value (e.g., true, false, indeterminate) of the resulting proposition is completely determined by the truth-values of the arguments to the operator.

To see why (b) is a problem, consider that natural language can and does use things other than the truth-values of the arguments when determining the truth-value of a complex proposition. For example, speakers can use context (previous discourse, world knowledge), presuppositions generated by the arguments, and pragmatic implicatures. For example, the sentences

1) She jumped on her horse and rode into the sunset.

2) She rode into the sunset and jumped on her horse.

are equivalent if you only consider the truth of the propositions

3) She jumped on her horse.

4) She rode into the sunset.

which is all that a translation into propositional logic will permit. However, (1) and (2), as English sentences, are not equivalent because they also convey that the events happened in the order in which they are stated. So (3) and (4) can both be true but (2) can be true or false depending on the order in which (3) and (4) happened, i.e., on the truth-value of the additional sentence

(5) She jumped on her horse before she rode into the sunset.

For an example more related to point (a), consider that this dependence on order is weaker or absent in

(6) John kissed Mary and Bob kissed Roger.

(7) John looks like Bob and sounds like Roger.

because the same people are not involved in (6) and the predicates are stative rather than active in (7), nether of which has to do with the truth of the constituent propositions.

For another example, the word "or" in English sometimes has an inclusive meaning (true if either or both of the arguments are true) and sometimes has an exclusive meaning (true only if exactly one of the arguments is true). Consider

8) Do you want your coffee with cream or sugar?

9) Do you want your coffee hot or cold?

You can answer "both" to (8) (which is inclusive) but not to (9) (which is exclusive), at least not without causing problems. The form is the same, and the meaning of the whole is determined by the lexical meanings of words in the arguments and facts about the world.

The scope of operators in natural language is not always clear from surface form either. Consider

10) John thinks that Mike's wife is tall, but John must be confused. Mike's wife is not tall. Mike's wife is short.

11) John thinks that Mike's wife is tall, but John must be confused. Mike's wife is not tall. Mike isn't even married.

The scope of "not" in "Mike's wife is not tall" is determined (in this case, disambiguated) by the context -- it's exactly the same sentence on the surface but with two different logical meanings. The negation can alternatively be interpreted as different operators altogether.

Classical propositional operators also fail to capture relevance, counterfactuals, and other things that natural language includes. There's just all kinds of failures everywhere. So attempts to say "this word always translates to this operator" or "any sentence of the form [blah] has this meaning" are just begging for counterexamples.

To address the question more directly, for future reference or a more general approach, try thinking of a sentence where the structural and logical relationships seem the same but where the truth dependencies are clearer to you. For example,

12) A compact operator cannot have a bounded inverse unless its range has finite dimension.

seems the same as

13) You cannot have a blue cup unless you have a cup.

Analyze it by setting:

p = You have a blue cup.

~p = You don't have a blue cup.

q = You have a cup.

~q = You don't have a cup.

~p unless q = You don't have a blue cup unless you have a cup.

To me, this is equivalent to p implies q or q is necessary for p. Construct a truth table to help:

$$\begin{array}{|c|c|c|c|} \hline p &q &\neg p &\neg p \text{ unless } q \\ \hline t &t &f &t \\ t &f &f &f \\ f &t &t &t? \\ f &f &t &t? \\ \hline \end{array}$$

To me, this is indeterminate when p is false, but if that's not an option, then I would call it true in that case since it more strongly just rules out the case where p is true but q is false.

If you want to express this as $A \rightarrow B$, then A is "x has a bounded inverse" and B is "the range of x has finite dimension", where x is a compact operator and universally quantified. I would translate it as something like

$$\forall x\ [(Cx \wedge Bx) \rightarrow Dx]$$

Thankfully, this agrees with the answer given by Harald Hanche-Olsenbased, presumably based on the known facts. So this meaning of "unless" in this case, where the form is, say, (not C) unless D, is $(C \rightarrow D) \Leftrightarrow (\neg D \rightarrow \neg C) \Leftrightarrow (\neg C \vee D)$. But other meanings are still possible. See the edits below.

Cheers, Rachel

Edit: As a further example of the perils, here's an alternative use for "unless". See how time (as tense) can make a difference:

(14) The box is empty unless John filled it. [John filled it --> ~(The box is empty)]

(15) The box is empty unless John locked it. [John locked it --> ~(The box is empty)]

(16) The box will be empty unless John fills it. [~(John fills it) --> The box will be empty]

(17) The box will be empty unless John locks it. [~(John locks it) --> The box will be empty]

Additionally, there is a sense in which (16-17) imply (14-15). If you say on Monday that the box will be empty unless John locks it (today), then on Tuesday, it's true that the box is empty unless John locked it (yesterday).

If you think of the form of (14-17) as P unless Q, the idea of (14-15) is that Q being true causes P to be false. But the idea of (16-17) is that Q being false causes P to be true. So both types of sentence only rule out one case. Sentence (15) rules out the case that John locked the box and it's empty (i.e., (15) means ~(P & Q) <=> (~P v ~Q)). Sentence (17) rules out the case that John doesn't lock the box and it's not empty (i.e., (17) means ~(~P & ~Q) <=> (P v Q)). Similarly for (14) and (16).

There are probably other complications that I haven't thought of. Language is complicated. Hence formal logic.

Edit: To answer the followup question: This is going to be complicated, so I hope you're really interested. :^)

Let's first get some relevant facts out of the way about how these operators work in formal logic. The following formulas/propositions are equivalent to each other, which I'm symbolizing with $\Leftrightarrow$, meaning they are indistinguishable with respect to truth-values; when one is true, the other will be true, and when one is false, the other will be false too, i.e., they have the same meaning:

$$\begin{align} (\phi \rightarrow \psi) &\Leftrightarrow (\neg \psi \rightarrow \neg \phi) \Leftrightarrow (\neg \phi \vee \psi) \Leftrightarrow \neg (\phi \wedge \neg\psi) \\ \neg\neg \phi &\Leftrightarrow \phi \end{align}$$

So this means: when you have two negations, you can drop them, just like negation on numbers; you can switch back and forth between implication, disjunction, and conjunction; when you have an implication, you can negate both of the arguments, swap their positions and end up with an equivalent formula. However, other combinations of negating and swapping do not yield an equivalent formula, so $\phi \rightarrow \psi$ and $\neg \phi \rightarrow \neg \psi$ are not equivalent, and neither are $\neg \phi \rightarrow \psi$ and $\phi \rightarrow \neg \psi$.

What I think about "unless" in English is that it has at least two different meanings. This was the point of examples (14-17). Think of how you would use it in real conversation:

(18) You're gonna be late unless you leave now.

I think (19) is a good paraphrase of (18):

(19) If you don't leave now, you're gonna be late.

And if you let A = you're gonna be late and B = you leave now, then (19) is more clearly $\neg B \rightarrow A$. And here, you are not asserting that only one thing will make the person late. Right? You're not ruling out the possibility that they leave now and don't still end up being late for another reason. You're only ruling out the possibility that they don't leave now but still manage to arrive on time. In particular, (18) does not imply

(20) If you leave now, you won't be late.

Right? This would translate to $B \rightarrow \neg A$. Additionally, (19) and (20) together mean that $\neg A$ and $B$ are equivalent (in our formal interpretation, at least). Why? Because (19) translates to $\neg B \rightarrow A$, and as mentioned above, this is equivalent (by negating both, swapping, and dropping the double-negation) to $\neg A \rightarrow B$. Since (20) translates to $B \rightarrow \neg A$, (19) and (20) together mean that $\neg A$ and $B$ imply each other, which is the definition of equivalence. Now, the translation of implication (if, then) in English to implication ($\rightarrow$) in formal logic is not perfect. It fails to capture many relationships. So even the translation of (19) to $\neg B \rightarrow A$ doesn't do justice to the exact meanings in English. But I think it captures the most important aspect, which is ruling out the possibility of both $\neg A$ and $\neg B$ being true.

To complicate matters, there is another use in English. I thought at first that the difference had to do with whether you were making a prediction or asserting that some fact is already the case. But you can still get both meanings even with predictions:

(21) You'll be on time unless you take the long way.

Let A' = you'll be on time and B' = you take the long way (there are primes there, in case you missed them). Consider two paraphrase analogs of (19) and (20):

(22) If you don't take the long way, you'll be on time. [$\neg B' \rightarrow A'$]

(23) If you take the long way, you won't be on time. [$B' \rightarrow \neg A'$]

Both (22) and (23) are implied by (21). Sentence (21) is equating $\neg A'$ and $B'$, in contrast to what (18) does. Why might this happen? Sentence (21) differs from (18) in whether or not the first part (A or A') is assumed to already be true or to be the "default" case (what determines default behavior is another question; it might be world knowledge, convention, prior discourse, or signaled in another way). In (18), the person is not assumed to already be late, and lateness is not default case. So A is assumed to be false by default. The first part of (21), A' (the negation of A), is assumed to be true by default. Yes? I think this default business might be what signals the different interpretations of "unless".

The original question was more complex because the first part was negated. Because negation interacts with lots of things in natural language, you really need to see them in context in order to make a better judgement. Add negations to (18) and (21), giving two versions of each since we have to consider default behavior.

(24) Don't worry about being late. You're not gonna be late unless you leave now.

(25) You need to leave now. You're not gonna be on time unless you leave now.

(26) You need to take the long way there. You won't be on time unless you take the long way.

(27) Don't worry about being late. You won't be late unless you take the long way.

I think (24) has a meaning analogous to (21), but (25) is a paraphrase of (18).The same goes for the others: (27) is a paraphrase as (21), but (26) has a meaning analogous to (18). This further suggests that being on time is assumed to be the default, even regardless of how it's expressed (as being on time or not being late). And note that this effect is strong enough to give the somewhat counterintuitive meanings of (24) and (26).

This doesn't necessarily answer the question of how "unless" behaves when you're not making predictions or in math, where you might not have preferred defaults. But the following show that you can still get different interpretations, and you have to consider the meaning in context. See how the first sentence can be clarified in different ways, so these are all different meanings:

(28) x equals y unless w equals z, and when w equals z is the only time x doesn't equal y. [ $x \neq y \leftrightarrow w=z$ ]

(29) x equals y unless w equals z, and x might not equal y under other conditions as well. [ $w=z \rightarrow x \neq y$ ]

(30) x equals y unless w equals z, in which case, x may or may not equal y, and this is the only time when x may not equal y. [ $x \neq y \rightarrow w=z$ ]

(31) x equals y unless w equals z, in which case, x may or may not equal y, and x may not equal y under other conditions as well. [useless]

The "cannot" in the original question was a clue because negation usually signals a contrast, e.g., with expectations or defaults, and "cannot" is emphatic in this context. So people will tend to say

(32) x cannot equal y unless w equals z.

when they mean that w must equal z in order for x to equal y, i.e., $x=y \rightarrow w=z$. I think this is analogous to (33):

(33) You won't be on time unless you leave now.

It doesn't seem that there is a default regarding equality in math. (There might not be many defaults at all in math.) This is a little difficult to check with English because we don't have a word for "not equal", and the presence of "not" in the phrase has an effect on the interpretation. But substitute equality with either order "less than" or "greater than", which I don't think have defaults either. The presence of "not" in (32) is what signals that the thing being negated is the default, and so it triggers the same interpretation as in (18), where the first part is not the default.

The lesson again is to beware of ever saying "this word or construction always translates to [blah]".

  • $\begingroup$ That post is nicely written. Thanks! $\endgroup$
    – shuhalo
    Mar 18, 2012 at 10:27
  • $\begingroup$ @Rachel: As I understand, do you mean "$A$ unless $B$" is "$A\rightarrow \neg B$" instead of "$\neg B\rightarrow A$"? $\endgroup$
    – user9464
    Mar 19, 2012 at 18:31
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    $\begingroup$ @Jack, I think it can mean either of those. I edited in further explanation to my answer. The reason that my answer is so horribly long is to try to convince you that there isn't a single logical translation of "A unless B". You cannot rely on a superficial analysis of the structure. You must also consider the meaning, and you must consider the meaning in context, which includes prior discourse, common knowledge, etc. And even then, you might still be left with ambiguity. That's natural language. $\endgroup$
    – Rachel
    Mar 20, 2012 at 6:58
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    $\begingroup$ @Rachel: That is a nice survey on logic, however, you're missing something: What you're talking about is logic and its (consistent or not) structure in general. Here we're confronted with a concrete statement, i.e. you cannot put the full truth table - wrong! The statement (p=t,q=f) simply can never happen. By the way, in general logic (p=f,q=t) gives (!p+q=t) and (p=f,q=f) gives (!p+q=t). ...all in all this is far overload and missleading $\endgroup$ Feb 2, 2014 at 10:54
  • $\begingroup$ Wow, thank you for this awesome writeup. This cleared up so many of my confusions about formal logic. I was always trying to force the meaning of sentences in english onto some formal logic equivalent and going crazy in the process. $\endgroup$ Jun 22, 2016 at 12:27

"$P$ unless $Q$" means that, in general, $P$ holds but with the only possible exception when $Q$ holds. So, if $Q$ holds, then $P$ does not have to hold (so, we do not know whether $P$ holds or not in this case), but if $Q$ does NOT hold, then $P$ must hold for sure.

This is being equivalent to have said: $$\neg P\Rightarrow Q$$ or, equivalently, $$\neg Q\Rightarrow P$$ or $$P\vee Q.$$

So, basically, it is equivalent to saying that "at least one of $P$ or $Q$ must hold".

  • $\begingroup$ I can't think of an example for this interpretation that doesn't involve predictions about the future. Can you give one? $\endgroup$
    – Rachel
    Mar 18, 2012 at 15:57
  • $\begingroup$ Yes, the original question (I mean, the statement in it). $\endgroup$
    – Vadim
    Feb 5, 2014 at 3:23

It means that if the compact operator has a bounded inverse, then the range has finite dimension.

A bounded projection operator onto a finite dimensional subspace is one example of a compact map with no inverse. On the other hand, if the range is finite dimensional and an inverse exists, then the inverse is bounded. I could quote the open mapping theorem for that, but that's overkill: Any linear operator defined on a finite dimensional normed space is bounded. (In particular, all norms on a finite dimensional space are equivalent.)


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