# Logical consequence in all structures in Kripke semantics

I read* the following definition of logical consequence in all structures within Kripke semantics:$$X\models A\iff\text{ for every } (W,R),\text{ if }(W,R)\models X,\text{ then }(W,R)\models A$$ $$\iff\text{ for every } (W,R),\quad(W,R),X\models A$$where $X$ and $A$ are sets of formulae and $(W,R)$ is a structure with a set of worlds $W$ and an accessibility relation $R$.

I do not understand why the definition of the first line is the same as the second one.

I know that $(W,R)\models X$ means that for any interpretation $I$, $(W,R,I)\models X$, i.e., for any interpretation $I$ and for any world $u\in W$, $(W,R,I),u\models X$. I also know that $(W,R),X\models A$ means that, for any interpretation $I$, $(W,R,I),X\models A$, i.e., for any interpretation $I$ and for an world $u\in W$, $(W,R,I),u\models X\Rightarrow(W,R,I),u\models A$.

Using quantifiers I would say that the first line means that, for every $(W,R)$,$$\forall I\forall u\in W\quad(W,R,I),u\models X\quad\Rightarrow\quad\forall I\forall u\in W\quad(W,R,I),u\models A$$which I am not sure to be licitly identified with the second line, which I would understand to mean that, for every $(W,R)$,$$\forall I\forall u\in W\quad[(W,R,I)\models X\Rightarrow (W,R,I)\models A].$$I suspect I am misunderstanding something, but I am not sure what. Could anybody explain what the two definitions of $X\models A$ mean and why they are the same? Thank you very much for any answer!

EDIT: No proof of the written equivalence has been received as of Mar 1'15. Nevertheless the two "long comments" seems to point to some typographical problem in the statement of the equivalence, and therefore, in the absence of a further proof, or in the presence of the argument that the equivalence would not hold in general without additional hypothesis, I have decided to chose the longer "long comment" as an answer, heartily thanking both Carl and Mauro for their invaluable help.

Anyhow, since $(W,R),X\models A$ means that $A$ is a logical consequence of $X$ in structure $(W,R)$, it is an obviously acceptable definition that $A$ is a logical consequence of $X$ in all structures if and only if, for every $(W,R)$, $(W,R),X\models A$.

• @the downvoter: thank you very much if you can suggest how to amend this question! Any problem with the formatting, with my English grammar, is non-classical logic off topic? – Self-teaching worker Feb 25 '15 at 19:04
• You have to verify the definitions of local vs global log cons; see Dale Jacquette (editor), A Companion to Philosophical Logic (2008), page 425. – Mauro ALLEGRANZA Feb 25 '15 at 20:41
• In the above ref (page 425) there is a condition (in term of the modal "necessity" operator) under which the two are equivalent: I suppose that you have to verify if the condition (tacitly) apply in your system. – Mauro ALLEGRANZA Feb 26 '15 at 8:21
• The "interpretation" $I$ here is for free variables - is that correct? Perhaps the statement is only intended to cover sentences, in which case the $I$ should not be needed. – Carl Mummert Feb 26 '15 at 12:27
• The definitions you are using seem to be quite unusual - are they the ones from the book? It seems strange that your Kripke models do not include the propositional variable interpretation already, and that instead you quantify over all possible interpretations of the propositional variables. In your definitions, can you give an example of a Kripke model that does satisfy any formula (e.g. a model that satisfies the propositional formula $X \land Y$)? I do not have access to the book you mentioned. – Carl Mummert Feb 26 '15 at 12:33

This is only a long comment ...

I'm not familiar with this kind of issue, and I'm a little bit perplexed (like you).

Forget for a moment about Kripke's model.

In classical logic, we have two "similar" relations :

we say that a formula $\psi$ is logical consequence of a formula $\varphi$, in symbols : $\varphi \vDash \psi$ when :

for all models $\mathcal M$, if $\mathcal M \vDash \varphi$, then $\mathcal M \vDash \psi$ --- (1).

This relation is different from :

if $\vDash \varphi$, then $\vDash \psi$,

which is :

if $\mathcal M \vDash \varphi$, for all $\mathcal M$, then $\mathcal M \vDash \psi$, for all $\mathcal M$ --- (2).

The two are linked by the relation :

(1) implies (2);

we can prove it easily by the property of the universal quantifier with respect to $\to$ :

$∀x(α → β) → (∀xα → ∀xβ)$.

The converse does not hold in general.

In modal logic we have two definitions [see Dale Jacquette (editor), A Companion to Philosophical Logic (2008), page 425] :

What does logical consequence mean for modal languages? Just like we have local and global notions of truth and validity, we have two consequence relations for modal formulas. A piece of terminology: if $\mathcal S$ is a class of models, then a model from $\mathcal S$ is simply a model $\mathcal M$ in $\mathcal S$; if $\mathcal S$ is a class of frames, then a model from $\mathcal S$ is a model based on a frame in $\mathcal S$.

DEFINITION 3 Let $\mathcal S$ be a class of models or a class of frames. Let $\Sigma$ and $A$ be a set of modal formula and a single formula. We say that $A$ is a (local) semantic consequence of $\Sigma$ over $\mathcal S$ (notation: $\mathcal S \vDash_{\mathcal S} A$) if for all models $\mathcal M$ from $\mathcal S$, and all states $w \in \mathcal M$, if $\mathcal M,w \vDash \Sigma$, then $\mathcal M,w \vDash A$.

As an example, suppose that we are working with $\mathsf{Tran}$, the class of frames $(W,R)$ in which $R$ is a transitive relation. Then $\{ \diamond \diamond p \} \vDash_\mathsf{Tran} \diamond p$, but $\diamond p$ is not a local consequence of $\{ \diamond \diamond p \}$ over the class of all frames.

DEFINITION 4 Let $A, \Sigma$ and $\mathcal S$ be as in Definition 3. Then $A$ is a global semantic consequence of $\Sigma$ over $\mathcal S$ (notation: $\Sigma \vDash^g_{\mathcal S} A$) if for all structures (i.e. models or frames) $S \in \mathcal S$, if $S \vDash \Sigma$, then $S \vDash A$.

The local and global notions are different, yet there is a systematic connection between them. One can show that, for $\Sigma$ a set of formulas and $F$ a class of frames, $\Sigma \vDash^g_F A$ is equivalent to $\{ \square^{n} B \mid B \in \Sigma, n \in \omega \} \vDash_F A$.

• The question doesn't seem to have anything parallel to "if $\vDash \phi$ then $\vDash \psi$". The two forms in the question are more parallel to: (1) for all $M$, if $M \vDash \phi$ then $M \vDash \psi$ and (2) for all $M$, $M \vDash \phi \to \psi$ – Carl Mummert Feb 26 '15 at 12:30
• Thank you for the remarks! – Self-teaching worker Feb 26 '15 at 13:17
• @Self-teachingDavide - you are welcome ! It seems to me (see the example in the quotation above) that the authors suppose tacitly some "constraint" on the class of frames... – Mauro ALLEGRANZA Feb 26 '15 at 13:20

This is too long for a comment, although it does not completely solve the problem.

It is more typical to view a Kripke frame $F$ as coming with a variable interpretation. So let us say that a K-frame is a triple $(W,R,I)$ such that $W$ is a set of worlds, $R$ is an accessibility relation, and $I$ gives a valuation of propositional variables on each world of $W$. We say that $F \vDash \phi$ if the formula $\phi$ holds at every world of $W$, using the normal definition of truth-in-a-world in Kripke frames.

Attempting to follow the definitions in the question, let us say that $F,\phi \vDash \psi$ if every world of $F$ that satisfies $\phi$ also satisfies $\psi$. (If we are talking about intuitionistic Kripke frames, this is not the same as $F \vDash \phi \to \psi$.)

With that definition of a K-frame, consider these statements quantifying over K-frames $F$ and formulas $\phi,\psi$:

1. If $F \vDash \phi$ then $F \vDash \psi$

2. $F,\phi \vDash \psi$.

Then we have that (2) implies (1), by a relatively simple argument.

But (1) does not imply (2), because (1) is true in any K-frame $F$ with $F \not \vDash \phi$, and we can make a K-frame with $F \not \vDash \phi$ such that $F,\phi \not \vDash \psi$. For example, we can make $\phi$ false in one world, and in another world make $\phi$ true and $\psi$ false.

So I am struggling to make sense of the particular definitions in the book that was mentioned in the question. I do not see how to easily arrive at the claim mentioned. Unfortunately, I don't have the book, and probably could not read it if I did. But something unusual is indeed going on.

• Thank you for the remarks! $F\models\phi$ expresses what my book would call the validity of $\phi$ in model $F$ and $F,\phi\models\psi$ represents the fact that $\psi$ is a logical consequence of $\phi$ in model $F$. – Self-teaching worker Feb 26 '15 at 13:26
• Carl, I suspect that the issue is related to the (Nec) rule; if I'm right, the rule is "validated" by the fact that : if $\vDash \varphi$, then $\vDash \square \varphi$. But (Nec) is similar to (Gen), i.e. $\varphi \to \square \varphi$ is not valid. Thus (I suspect) how to define : $\varphi \vDash \square \varphi$ ? – Mauro ALLEGRANZA Feb 27 '15 at 9:29