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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider the statement

$(1)$     If I'm in Richmond, then I'm in Virginia.

This statement is an implication $p \implies q$, where $p$ is "I'm in Richmond" and $q$ is "I'm in Virginia." Incidentally, this statement is true.

Now consider the statement

$(2)$     If I'm in Springfield, then I'm in Virginia.

This implication is false, because there are Springfields in other states, for example, Illinois. So if we let $r$ be the statement "I'm in Springfield", what is the correct way of showing that statement $(2)$ is false?

I suppose it should be $(r \not\implies q)$ or rather $\lnot(r\implies q)$, but this is equivalent to $\lnot(\lnot r \lor q)$, which is equivalent to $(r \land \lnot q)$, which means "I'm in Springfield and I'm not in Virginia," which is clearly not the case either because there is definitely a Springfield in Virginia.

I think $a\not\implies b$ doesn't mean "$a$ doesn't necessarily imply $b$," I think it means "$a$ never implies $b$." Is this correct? If so, how should I represent the statement, "It is not the case that if I'm in Springfield then I'm in Virginia," in other words, "Being in Springfield doesn't necessarily imply that I'm in Virginia."?

share|cite|improve this question
But there is more than one Richmond, for example, there is a Richmond in North Yorkshire, England. – martini Oct 1 '12 at 11:51
Having lived in Richmond, Indiana, I must point out that the first statement is also false. – Brian M. Scott Oct 1 '12 at 11:58
No, it is not correct that $a\not\to b$ means ‘$a$ never implies $b$’; it means that $a$ does not necessarily imply $b$, i.e., that it is possible (but not necessary) for $a$ to be true and $b$ to be false. – Brian M. Scott Oct 1 '12 at 12:01
up vote 5 down vote accepted

I think your problem could depend on the following: Consider your statement $r \land \neg q$, which you seem to read as: "I'm in Springfield and I'm not in Virginia" (whereever you are exactly, but it can be true, as you can be in some Springfield which is not in Virginia. So you use a generalisation implicit, let 's make it explicit.

There seems to be some dependence on the place where you actually are. Let's make this dependence more clearly and write the implicit generalisation which you seem to use explicit

$(2)$ Whichever place I am, if I'm in Springfield, then I'm in Virginia.

As you say, this is not true, proven for example by Springfield, Illinois. The negation of $(2)$ is then

$\neg (2)$ There is some place, where I'm in Springfield, but not in Virginia.

Again by the town above, this is true. To find the negation, note that $(2)$ has now the form $\forall x\; \bigl(r(x) \to q(x)\bigr)$ with $r$ and $q$ as above, $x$ being a variable which ranges over the places in the world. The negation is $\exists x\; \neg\bigl(r(x) \to q(x)\bigr)$, hence $\exists x\; \bigl(r(x) \land \neg q(x)\bigr)$.

So when using implicit generalisation, take care of it, when you do negation.

share|cite|improve this answer

When we say that $a$ implies $b$ we say that $a$ cannot occur without $b$ occurring as well.

When we say that $a$ does not necessarily imply $b$ we say that $a$ may occur with $b$ or without $b$ occurring as well.

To say that $a$ never implies $b$ is to say that $a$ in fact implies the negation of $b$.

The reason for this distinction is that we may assume more information in the background. For example if you add the assumption "Whenever you talk about a city which may lie in Virgina, you will always refer to the one in Virginia." is an additional assumption which make your example an implication, because there is a Springfield in Virginia (at least for sake of argument).

share|cite|improve this answer
Interesting, a downvote! – Asaf Karagila Oct 1 '12 at 22:15

The negation of $r \Rightarrow q$ is $\neg (r \Rightarrow q)$, as you said. Imagine a person in Springfield, Florida saying "I'm in Springfield and I'm not in Virgina". This would be true. The problem is that you actually have to different statements:

$(2a)$ If I, TestSubject…, am in Springfield, I, TestSubject…, am in Virginia.

And on the other hand:

$(2b)$ If someone is in in Sprinfield, they are in Virginia.

Where $(2b)$ would be translated to $\forall x \big(S(x) \Rightarrow V(x) \big)$, whereas $S(x)$ means "$x$ is in Springfield", and $V(x)$ means "$x$" is in Virginia". $(2b)$ is wrong, since there is a person in Springfield, Florida. On the other hand $(2a)$ is true because the material implication "$\Rightarrow$" actually just reads "Assuming the left side is true, the right side also happens to be true". So applied to your sentences, since the left side is true, and the right side happens to be true, also, $(2a)$ is correct.

share|cite|improve this answer
Ok then, change all the "I am"s to "one is"s. Is your answer still valid? – chharvey Oct 1 '12 at 22:14
@TestSubject528491 If you do that, I would argue that the logical structure of the statements would be that of $(2b)$, so $\forall x \big(S(x) \Rightarrow V(x)\big)$. You don't have atomic propositions there, but rather "parameterized" ones. The statements $r$ and $q$ use the same yet unspecified person to be verified, so the translation "$r \Rightarrow q$" isn't correct. You must use first order logic to capture the structure and formulate its negation: $\exists x : \big(S(x) \land \lnot V(x)\big)$. – k.stm Oct 2 '12 at 7:59
@TestSubject528491 The other possibility, of course, is that your statements are parameterized by the location you're (one is) in, which is what martini suggested in his thorough answer. And it also makes a difference, for example if all other Springfields were ghost towns and noone was living there. Then the person-ranges-version of the statement would actually be true, while the location-ranges-version would be false. – k.stm Oct 2 '12 at 8:15

Your last statement is not correct for propositional calculus, all the rest is.

$a\not\to b$ is shorthand for $\lnot(a\to b)$ and it is the negation of '$a$ necessarily implies $b$', that is, '$a$ doesn't necessarily implies $b$', in other words: 'there can be a situation when $a$ but not $b$'.

The 'being in Springfield of Illinois' is a situation (evaluation) such that $r\land \lnot q$ holds there. On the other hand, it is also true that $r\land\lnot q$ is false in Springfield of Illinois, but evaluated there, $r\to q$ is false as well.

How to say.. you implicitly introduced a variable and the truth of $p$ and $q$ depends on that variable ('my location', in this case). So, in this sense, the evaluation of $p\to q$ may also depend on that variable. So, this is somehow making the propositional calculus rather a first order logic, or some kind of modal logic: there are implications such as $\Box (p\to q)$ or $p\to\Box q$ which can better express.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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