Not Necessarily Imply

Question

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 $(r \kern.6em\not\kern -.6em \implies q)$ 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 $a\kern.6em\not\kern -.6em \implies b$ 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."?

Answer 1

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.

Answer 2

Let $A,B,A(x),$ and $B(x)$ represent logic formulae like $\big(P\land Q(x)\big)$ and $\forall x\,P(x).$

1. The statement $$A{\kern.6em\not\kern-.6em\implies} B\tag{NI}$$ says that $A$ does not imply $B;$ by definition, it means that $\big(A\to B\big)$ is false, that is, that $$A \text{ is true and $B$ is false}.$$

2. In mathematics, typically, the statement $$A(x) \Longrightarrow B(x) $$ implicitly says that $A$ universally implies $B,$ that is, $$\text{for each }x,\big(A(x) → B(x) \big).$$ So, the statement $$A(x) {\kern.6em\not\kern-.6em\implies} B(x)\tag{NUI}$$ implicitly says that $A$ does not universally imply $B,$ that is, $$\text{for some }x,\text{ it is false that }\big(A(x) → B(x)\big);$$ in particular, $A$ universally does not imply $B$ means that $$\text{for each }x,\text{ it is false that }\big(A(x) → B(x)\big).\tag{UNI}$$

The above statements are in the context of a particular interpretation (that is, with reference to the assigned meanings of $A,B,A(x)$ and $B(x),$ for example, $A:=$ ‘Amy is in Springfield’ and $B(x):=$ ‘$x$ is in Virginia’); in other words, we are discussing the situational (synthetic or analytic) truth rather than the logical truth of $\big(A\to B\big)$ and $\big(A(x)\to B(x)\big).$

3. If we want to analyse $\big(A\to B\big)$'s truth value as the interpretation varies, there are three cases:

   • $\big(A\to B\big)$ is valid, i.e., is true in every interpretation; we say that $A$ logically implies $B$ and write $$A\models B;$$
   • $\big(A\to B\big)$ is satisfiable but invalid, i.e., is true in some interpretations but false in others;
   • $\big(A\to B\big)$ is unsatisfiable; i.e., in every interpretation, $A$ is true and $B$ false; we say that $A$ logically does not imply $B$ and write $$\models \lnot\big(A\to B\big).\tag{LNI}$$

   In the second and third cases, $\big(A\to B\big)$ is invalid; we say that $A$ does not logically imply $B$ and write $$A\;\;{\Large\not\models}\: B.\tag{NLI}$$

4. Analogously:

   • $A(x)$ logically universally implies $B(x)$ $$\models\forall x\,\big( A(x)\to B(x)\big);$$
   • $A(x)$ does not logically universally imply $B(x)$ $${\Large\not\models}\;\forall x\,\big( A(x)\to B(x)\big);\tag{NLUI}$$
   • $A(x)$ logically does not universally imply $B(x)$ $$\models\lnot\forall x\,\big( A(x)\to B(x)\big);\tag{LNUI}$$
   • $A(x)$ logically universally does not imply $B(x)$ $$\models\forall x\,\lnot\big( A(x)\to B(x)\big).\tag{LUNI}$$

   $(\text{LUNI})$ is a special case of $(\text{LNUI}),$ which is a special case of $(\text{NLUI}).$

In first-order logic, “necessarily imply” might mean universally imply or logically imply or logically universally imply (Types of ‘imply’). Regarding the given statements:

$(1)$ Being in Richmond universally implies being in Virginia.

$(2)$ Being in Springfield does not universally imply being in Virginia.

Answer 3

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).

Answer 4

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.

Answer 5

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.