Expressing "does not imply''

Question

In ordinary discourse, when we say that "$A$ implies $B$", we shall formalize it by writing the following: $$A\rightarrow B$$ But when we say that "$A$ does not imply $B$", we cannot formalize it as the following: $$\neg(A\rightarrow B)$$ Because by material implication, we have the following equivalences for the second formula: $$\neg(A\rightarrow B)\Longleftrightarrow\neg(\neg A\vee B)\Longleftrightarrow A\wedge\neg B$$ But the ordinary meaning of the sentence "$A$ does not imply $B$" is that $B$ does not follow from $A$: if $A$ is true, $B$ is either false or undecidable. I wonder how "$A$ does not imply $B$" is formally expressed in the object language (not at the meta-level). Thanks!

The point here is that the "ordinary discourse" version of "A implies B" is not the mathematical $A \to B$, and really has no equivalent in mathematical logic. — Robert Israel, Nov 13 '20 at 17:07
But $(A \to B)$ is False exactly when $A$ is True and $B$ False. Thus, $\lnot (A \to B)$ is True exactly when $A$ is True and $B$ False. — Mauro ALLEGRANZA, Nov 13 '20 at 17:56
If you agree with this, we have that $¬(A→B)$ is True exactly when $A$ is True and $\lnot B$ is True. — Mauro ALLEGRANZA, Nov 13 '20 at 17:57

Taroccoesbrocco · Accepted Answer · 2022-04-16T07:06:12.963

7

The formalization of a sentence in ordinary discourse claiming that "$A$ implies $B$" or "$A$ does not imply $B$" is outside the languages of propositional or first-order or higher-order logic. It is in the language of logical consequences, which roughly consists of pairs of (sets of)$-$propositional of first-order or higher-order$-$formulas.

When we use the expression "$A$ does not imply $B$" in ordinary discourse, we can mean two distinct things:

$\models \lnot (A \to B)$, which means that $\lnot (A \to B)$ is a valid (or provable) formula. This amounts to say $\models A \land \lnot B$, i.e. every structure satisfies $A$ but not $B$.
$A \not \models B$, which means that $B$ is not a logical consequence of (or is not provable from) $A$, so there exists a structure where $A$ is true and $B$ is false. This amounts to say $\not \models A \to B$.

The difference is essentially whether we mean "not" as part of the language of propositional logic, or as a negation of the relation of logical consequence. Usually, the intended meaning is 2.

Note that meaning 1 is stronger than meaning 2, in the sense that, given the formulas $A$ and $B$, if Point 1 holds then in particular Point 2 holds.

edited Apr 16 '22 at 07:06

answered Nov 13 '20 at 17:14

Taroccoesbrocco

16,964

Thanks! I was thinking about that, perhaps also $\not\rightarrow$? But these are expressions in the meta-language, right? Is there any way to express it in the object language? – Fred Nov 13 '20 at 17:17
@Fred - In ordinary discourse, usually if you affirm something like "$A$ implies $B$" you mean that $A \to B$ is provable, which can be expressed as $\models A \to B$ or equivalently $A \models B$. I mean, our claims are not in the language of logic but in the language of logical consequences, which roughly consists of pairs of (sets of) logical formulas. You can choose as an object language the former or the latter, with the latter you are more faithful to what we mean in ordinary discourse. – Taroccoesbrocco Nov 13 '20 at 17:45
Could 1) be expressed as ∀x(A(x) ∧ ¬B), and 2) as ∃x(A(x) ∧ ¬B) ? Here x is some structure and A(x) means it satisfies A. – Inertial Ignorance Nov 14 '20 at 02:07
@InertialIgnorance: Why not say that "A implies B" means "∀w ( A@w ⇒ B@w )" where "A@w" means "A holds in world w". Then "A does not imply B" means "∃w ( A@w ∧ ¬(B@w) )", which is not the same as "A but not B" (which means "∀w ( A@w ∧ (¬B)@w )". This can be expressed more succinctly by a kind of modal logic (either S4 or S5), but it is different from provability logic. Specifically, in this world-based modal logic we have $⬜{⊥}⇒⊥$ and hence $⬜(¬⬜{⊥})$, so we cannot also be able to prove $⬜(¬⬜{⊥})⇒⬜{⊥}$ otherwise we get $⬜{⊥}$ and thereby $⊥$... – user21820 Nov 14 '20 at 02:36
@InertialIgnorance - Your proposal essentially rewrites my answer with a different notation. Note that the language you used is not a language for first-order logic because you are interpreting the symbols in a specific way (indeed, you assume that the domain of quantifiers is the set of structures), while a mere language for first-order logic should be non-interpreted. – Taroccoesbrocco Nov 14 '20 at 03:05
@Taroccoesbrocco Yeah I just noticed you explicitly said "every structure" and "there exists a structure", so I was rewriting the same idea (just in predicate notation). Although aren't you also interpreting the symbols as well, since you're using predicate keywords "every" and "there exists" as you reference structures. What other domain could you have here? – Inertial Ignorance Nov 14 '20 at 03:24
@InertialIgnorance - "aren't you also interpreting the symbols as well?" Yes, and indeed I'm claiming that the correct formalization of a sentence "A implies B" is outside propositional or first-order language, exactly for the reason that you said. I just wanted to stress that the fact that your notations is "first-order-language-like" does not imply that you are able to formalize a sentence "A implies B" in a first-order language. – Taroccoesbrocco Nov 14 '20 at 03:35
"we can mean two distinct* things: 1..... 2.*" Isn’t meaning #1 strictly stronger than, instead of distinct from, meaning #2 ? – ryang Apr 15 '22 at 20:11
@ryang - Of course, meaning 1 is strictly stronger than meaning 2. In particular, the two meanings are different (just as $X \subsetneq Y$ implies that $X \neq Y$). – Taroccoesbrocco Apr 16 '22 at 04:09
@Taroccoesbrocco I do agree with your answer (in fact I have a similar one at Not necessarily imply); my point is that introducing the two meanings as "distinct" (when they actually overlap) may detract the reader away from the fact that meaning #1 implies #2. – ryang Apr 16 '22 at 04:31
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@ryang - I see your point. I was considering that the implication is obvious, but I guess you're right, it would be better to say it explicitly. I added a sentence in my answer. Feel free to edit my answer, if you want to say something more. Thanks! – Taroccoesbrocco Apr 16 '22 at 07:08

score 1 · Answer 2 · answered Nov 14 '20 at 02:23

There is a way to express it in the object language, using modal operators. "$A$ implies $B$" in English often means "$⬜(A⇒B)$", which can be interpreted as "we can justify that $A⇒B$". So "$A$ does not imply $B$" in natural language often means "$¬⬜(A⇒B)$". Note that $¬⬜(A⇒B)$ does not imply $⬜¬(A⇒B)$, because it may be that both $¬⬜(A⇒B)$ and $¬⬜¬(A⇒B)$ hold.

For example, $x,y∈ℝ$ and $x≤y$ do not imply $x<y$, equivalently $¬⬜( x,y∈ℝ∧x≤y ⇒ x<y )$. And this is not equivalent to $⬜¬( x,y∈ℝ∧x≤y ⇒ x<y )$, which can be interpreted as "we can justify that $x,y∈ℝ$ and $x≤y$ and $x≥y$" (which is clearly nonsense).

There is in fact a connection between this formalization of English "implies" and meta-logic: This "$⬜$" can be taken as the provability operator from provability logic (where "$⬜$" is given a higher precedence than other logical operators), and so "$⬜(A⇒B)$" does in fact express "$⊢A⇒B$" internally, and is stronger than "$⬜A⇒⬜B$" (which expresses "$(⊢A)⇒(⊢B)$" internally. A distinguishing example is that PA (Peano Arithmetic) proves $⬜(¬⬜{⊥})⇒⬜{⊥}$ but (we strongly believe) does not prove $⬜(¬⬜{⊥}⇒⊥)$ (where "$⬜Q$" can be expressed in PA using Godel coding).

Expressing "does not imply''

2 Answers2