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I am a beginner in logic. I have a basic question about implication. If "$B$ follows from $A$", we can write $A\rightarrow B$; but what about "$B$ does not follow from $A$"? Someone might suggest $\neg(A\rightarrow B)$, but that's not what "$B$ does not follow from $A$" means. Because $\neg(A\rightarrow B)$ is logically equal to $A\wedge\neg B$. But saying that "$B$ does not follow from $A$", it might be the case that $B$ is undecidable given $A$. So, I wonder how this sentence can be analyzed using logical tools. Thanks in advance!

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    $\begingroup$ $¬(A → B)$ is not intuitionistically equivalent to $A ∧ ¬B$. Perhaps your problem is that classical logic is too degenerate to express what you want. $\endgroup$
    – Dan Doel
    Commented Dec 2, 2020 at 15:44
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    $\begingroup$ @DanDoel Even though we don't have $(A\rightarrow B)\leftrightarrow(\neg A\vee B)$ in intuitionistic logic, we still can derive $\neg\neg A\wedge\neg B$ from $\neg(A\rightarrow B)$. In the wikipedia page for intuitionistic logic, it lists the valid rule $(\neg A\vee B)\rightarrow(A\rightarrow B)$. Using contraposition, now derive $\neg(A\rightarrow B)\rightarrow\neg(\neg A\vee B)$ and the consequent is equivalently to $\neg\neg A\wedge \neg B$. This is a result as terrible as $A\wedge \neg B$ because it says that $A$ is not negatable and $\neg B$ is true. $\endgroup$
    – ferdinand
    Commented Dec 2, 2020 at 15:51
  • $\begingroup$ Typically the symbol used for this is $\implies$ (code is \implies) not $\to$. $\endgroup$
    – K.defaoite
    Commented Dec 2, 2020 at 15:54
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    $\begingroup$ I've used $A\not \implies B$ and I don't think anyone has ever called me on it. Technically it logistically doesn't mean anything because we can have any of the four truth values of $A$ and of $B$ but I use is as a warning "this line of reasoning is not valid". .. I don't know in formal logic what the answer to you question, but if we are concerned with only formal logic, then why would anyone ever need to make such a statement? $\endgroup$
    – fleablood
    Commented Dec 2, 2020 at 16:12
  • $\begingroup$ Intuitionistic logic isn't the only way to organize a system with more distinctions between formulas than classical logic. If the inuitionistic meaning of $¬(A → B)$ is still unacceptable, then one can try to craft something that is acceptable. That isn't really possible if one doesn't recognize that classical logic is merely a way to work, not the way to work, though. It seems like what is desired in this case is a sort of meta-reasoning, which could possibly be incorporated with modalities. $\endgroup$
    – Dan Doel
    Commented Dec 2, 2020 at 17:12

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For 'follows from', you should not use the material implication $\to$, but rather the logical implication, for which you can use $\Rightarrow$ or $\vDash$

The difference is that $\to$ is the truth-functional operator you are familiar with, but the $\Rightarrow$ is a meta-logical symbol that claims some logical relationship between two statements.

$P \to Q$ is true in some world as long as it is not true in that world that $P$ is true and $Q$ is false.

$P \Rightarrow Q$ is true is it impossible for $P$ to be true and $Q$ to be false, i.e. that there is no world in which $P$ is true and $Q$ is false.

It is the latter that we typically mean by 'follows from': no matter what the circumstances are (i.e. no matter what world we're dealing with), if $P$ is true, then $Q$ will be true.

So, to say that $Q$ does not follow from $P$, we can write $P \not \Rightarrow Q$ ... but we can not write $\neg (P \Rightarrow Q)$, because then we are mixing up logic with meta-logic. We should simply say that 'it is not the case that $P \Rightarrow Q$.

And indeed, just because it is not the case that $P \Rightarrow Q$, does not mean anything about the truth-value of $P$ or $Q$ when evaluated in some particular world. Notably, it does not mean that $P$ is true and $Q$ is false. Moreover, if in some specific world $P$ is true, $Q$ can still be true or false. So these are all things you were looking for.

In sum, the $\Rightarrow$ (or $\vDash$) is what you should use when thinking about logical implication, rather than the $\to$.

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    $\begingroup$ Can you provide an example of a case where $P \not \Rightarrow Q$ but it is not the case that $\neg (P \Rightarrow Q)$? I'm not sure I understand the distinction you're making. $\endgroup$ Commented Dec 2, 2020 at 17:45
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    $\begingroup$ @RavenclawPrefect As I tried to say at the end, you should avoid writing $\neg (P \Rightarrow Q)$: now you are mixing up a logic symbol with a meta-logic symbol in one expression: do not do that! You should really just stick to "It is not the case that $P \Rightarrow Q$", or write $P \not \Rightarrow Q$ $\endgroup$
    – Bram28
    Commented Dec 2, 2020 at 17:49
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$A \to B$ does not mean "$B$ follows from $A$". It means "either $A$ is false or $B$ is true in this particular situation". It does not make sense to say such a thing as "$A \to B$ is true" without specifying in which situation/model it is to have that truth value. To say that a $A \to B$ is false in a given situation is indeed logically equivalent to $A \land \neg B$ being true in that situation.

If what you want to say is "$B$ logically follows from $A$", then you're not talking about truth in a specific situation, but preservance of truth from the premises to the conclusion across any possible situation. In that case, what you really want is logical consequence, $\vDash$:

$A \vDash B$
= "In every situation in which $A$ is true, $B$ is true as well."

This permits for situations in which $B$ is false, as long as in those situations $A$ is false as well. It also permits for situations in which $B$ is true despite $A$ being false. What it rules out is the existence of situations in which $A$ is true but $B$ is false.

The negation of this is "$B$ does not follow from $A$":

$A \not \vDash B$
= "Not in every situation in which $A$ is true, $B$ is true as well."

This does not rule out the existence of situations in which $B$ is true or $A$ false or both of that, but it does assert the existence of at least one situation in which $A$ is true but $B$ is not. It also permits for, but does not entail, the possibility that $B$ is false not only in some but all situation in which $A$ is true.

If you want to say that $B$ is undecidable from $A$, i.e. that there exist situations in which $A$ is true and $B$ false but there also exist situations in which $A$ is true and $B$ true, you have to make two claims:

$A \not \vDash B$ and $A \not \vDash \neg B$

meaning that from $A$ you can neither definitely conclude $B$ nor $\neg B$.


We do have the following connection between the notions of material conditional and logical consequence (called the deduction theorem):

$$A \vDash B \text{ if and only if } \vDash A \to B$$

i.e. $B$ logically follows from $A$ iff the material conditional is $A \to B$ is tautological.
The crucial point here is that logical consequence commits the material conditional to being true in all situations; when the consequence is denied, $\not \vDash A \to B$, this merely negates its universalness, leading to the conditional being false in some situations -- which should be what you want. This is different from $\vDash \neg (A \to B)$, which would be a stronger claim meaning that $A \to B$ is false and thereby $A \land \neg B$ true in all situations, which, as you figured out, is not the intended state of affairs.

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Let $A$ and $B$ be propositions, the claim that $B$ doesn't follow from $A$ can be alternatively described as: either (i) $A$ implies $\neg B$ or (ii) $B$ is irrelevant to $A$ (i.e., given $A$, whether $B$ is true or not is not determined).

(i) can be easily expressed as (using your notation): $A\rightarrow\neg B$.

(ii) can hardly be expressed in the object language of formal logic. In order to express it, first define $p$ as a mapping from a proposition to a scale of probability ranging from 0 to 1, both ends included: $$p:\textsf{Prop}\rightarrow[0,1].$$ So, the probability of $A$ and $B$ shall be expressed as $p(A)$ and $p(B)$ respectively. Now, consider (ii), it amounts to say that the probability of $B$ given $A$ is equal to the probability of $B$ itself but is irrelevant to the probability of $A$: $$P(B|A)=P(B).$$ This might seem a little bit complicated, but I think it's the easiest way to capture (ii).

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  • $\begingroup$ "$B$ does not follow from $A$" does not mean that $\neg B$ follows from $A$. $\endgroup$ Commented Dec 2, 2020 at 17:50
  • $\begingroup$ @lemontree You are right. But i was using the terminology that Samual used in his post. Now I make it more explicit: $A$ implies $\neg B$. $\endgroup$
    – ferdinand
    Commented Dec 2, 2020 at 17:52
  • $\begingroup$ Yes, and precisely this is not a proper interpretation of the intended assertion. Your claim that "'$B$ does not follow from $A$' can alternatively be described as '$A$ implies $\neg B$' is misleading. $\endgroup$ Commented Dec 2, 2020 at 17:55
  • $\begingroup$ Your second claim that "$B$ is independent of $A$" can not be expressed in symbolic logic is also wrong -- it can be expressed via $\not \vDash$. $\endgroup$ Commented Dec 2, 2020 at 17:56
  • $\begingroup$ @lemontree Note that this is only the first clause in (i). I also give the second clause in (ii), which is exactly what Samuel means. $\endgroup$
    – ferdinand
    Commented Dec 2, 2020 at 17:56
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You might like to look at modal logic, which adds a new operator: $\square p$ means “$p$ is necessarily true”. The idea is that we can imagine alternate worlds which are plausible, but different from the world in which we find ourselves. For example, we find ourselves in a universe in which I am wearing a green shirt. Is the necessarily the case? No, of course not, we could easily have been in a universe in which I was wearing a blue shirt. But could we have found ourselves in a universe in which $2+2=5$? No, that is impossible; $2+2=4$ in every possible universe. “I am wearing a green shirt” is only locally true, but $2+2=4$ is globally true.

The exact behavior and semantics of $\square$ is a philosophical question, and there are different ways to define it, just like the way it is a philosophical question whether $p\to q$ is identical with $\lnot(p\land \lnot q)$. (Classical logic says yes, intuitionistic logic says they are different.) One typical axiom is $\square p\to p$, which says that if $p$ is true in every possible universe, it is true in the particular universe in which we live. Another is $$\square(p\to q) \to (\square p\to \square q).$$

Here $\square(p\to q)$ means that $q$ necessarily follows from $p$: in every possible universe where $p$ is true, $q$ is true in that universe also.

Suppose we agree that $\square p\to p$, so by contrapositive $\lnot p\to\lnot \square p$. This says that if $p$ is false (in our universe), then it is not necessarily true. But it might still be true in some other universe. We cannot conclude from $\lnot\square p$ ($p$ is not necessarily true) that $\square\lnot p$ ($p$ is necessarily false).

Now what you asked for is available: in modal logic, $\square(a\to b)$ means that $b$ necessarily follows from $a$, and $\lnot\square(a\to b)$ means that $b$ does not necessarily follow from $a$—but it might follow anyway, locally if not globally. Modal logic usually has a second operator $\diamond p$ which means that $p$ is possibly true; usually $\diamond p$ is taking to be an abbreviation for $\lnot\square\lnot p$, so that $\lnot\diamond p$ and $\square\lnot p$ are two ways of writing that $p$ is impossible. So consider:

$$\square(p\to q)\\\\ p\to q\\\\ \diamond(p\to q)\\\\ $$

The first one says that from $p$ we can deduce $q$ in every possible universe. The second only says that from $p$ we can deduce $q$ in the contingent universe we are actually in. The last one only says that there is some universe, not necessarily this one, in which we can deduce $q$ from $p$. Applying some simple transformations to the last one we can turn it into $$\lnot\square(p\land\lnot q)$$ which says that it is not necessarily the case that $p$ is true and $q$ is false.

Stanford Encyclopedia of Philosophy article on modal logic
Wikipedia

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$A\; \implies \; B $ says that if $ A $ is true, we are sure and certain that $ B $ is true and cannot be false.

$ A $ does not imply $ B $ says that even if $ A $ is true, we can say Nothing about $ B $.

For example, $(p\vee q)\wedge q $ does not imply $ p$.

$ p $ could true and false as well.

The best way to understand the "implication" begins by the notion of "Argument".

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    $\begingroup$ But your answer does not solve Samuel's question because he exactly knows what you have said and asks how the second line of your answer can be formally analyzed. I think that's what he wants. Not just an example, but a general way to formalize that. $\endgroup$
    – Emini Jask
    Commented Dec 2, 2020 at 15:56
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In a lot of cases of trying to say "it doesn't follow that...." the issue is that there is implicit quantification that needs to be negated. For example, let $S$ be "it is (S)aturday" and $W$ be "it is the (W)eekend". From $S$ follows $W$, so we can say $$S \to W$$ But just because it is the weekend doesn't mean it is Saturday. But as you identified, $\lnot(W \to S)$ doesn't say what we want** because it is propositionally equivalent to $W \land \lnot S$. What is missing is that your full statement is "for every day, if it is the weekend then it is Saturday", which is correct to negate as "there is a day which is a weekend but is not Saturday":

$$\lnot \forall d. W(d) \to S(d)$$ $$\exists d. W(d) \land \lnot S(d)$$


** Except sometimes it does, because you could have meant for example "if today is the weekend then today is Saturday" because you have ruled out that it isn't Sunday. In this case $\lnot (W \to S)$ is correct.

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