Strong Kleene interpretation

Question

Consider: $\\$ $\Box(\phi \wedge \psi) \rightarrow \Box(\phi) \wedge \Box(\psi)$

I guess this yields by the reflexivity axiom for intensional predicate logic? But I was wondering whether it is also aquired under a "strong Kleene"(SK) interpretation of the truth definition for the same logic?

As I understand it, the truth condition for the above implication, on SK, is such that if it is true, then: $V(\Box(\phi \wedge \psi))=0$ (false) (in all accessible worlds) OR (but not AND?) $V(\Box(\phi) \wedge \Box(\psi))=1$ (true) (in all accessible wolds) on SK.

The left-hand side conjunction is false if either of its conjucts $\phi$ and $\psi$ is and it is true if $V(\phi)=1$ (in all accessible worlds) and $V(\psi)=1$ (in all accessible worlds). But if that conjunction is made false, by atleast one of its conjuncts being false, the implication is thus made true. But then, the right-hand side conjunction($ \Box(\phi) \wedge \Box(\psi)$ cannot hold on this account (as is admitted on weak Kleene interpretation atleast) for necessity $(\Box)$ requires that both $\phi$ and $\psi$ be true in all accessible worlds, right?

To sums things up, if you find my representation of SK sound and just, does SK in this way (by not allowing for the left-hand side of the implication to be $0$ and the right-hand side expression to be $1$ if true, or vice versa) make $\Box(\phi \wedge \psi) \rightarrow \Box(\phi) \wedge \Box(\psi)$ valid? It seems to be a consequence of the strong implication (OR, but not AND (see above)?) that we must get $1$ and $1$ or $0$ and $0$ as truth values for the implication if true.

Does this seem right or should validation of $\Box(\phi \wedge \psi) \rightarrow \Box(\phi) \wedge \Box(\psi)$ by SK be done differently?

I feel slightly confused. Would much appreciate some help to render this more clear.

$\\$ Thanks!

Answer 1

I'll refer to :

• L.T.F. Gamut, Logic, Language, and Meaning, Volume 2 : Intensional Logic and Logical Grammar (1991), page 54 :

Definition 2 Let $M$ a model and $w \in W$; then the truth conditions are :

[...]

(iii) $V_{M,w} (\phi \rightarrow \psi) = 1$ iff $V_{M,w} (\phi)=0$ or $V_{M,w} (\psi)=1$

(iv) $V_{M,w} (\square \phi)=1$ iff for every $w' \in W$ such that $wRw'$ : $V_{M,w'} (\phi)=1$.

Clause (iv) is too strict because $\square \phi$ can be true in a world only if all the constants occuring in $\phi$ refer to entities which are present in all worlds accessible from that world.

Thus a "relaxed" condition is proposed [page 55] :

(iv'') $V_{M,w} (\square \phi)=1$ iff $V_{M,w} (\phi)=1$ and $V_{M,w'} (\phi)=1$ for every $w' \in W$ such that $wRw'$, for which $V_{M,w'} (\phi)$ is defined.

But in this case we have still problems; in particular, the valid principle of modal propositional logic :

$\square (\phi \land \psi) \rightarrow \square \psi$

is no longer valid. For $\square (\phi \land \psi)$ to be true it is only required that $\phi \land \psi$ be true in all worlds which have in their domains the references of the constants occurring in either $\phi$ or $\psi$.

It follows [see page 56] a description of a counterexample, based on a model $M$ with three worlds $\{ w_1, w_2, w_3 \}$ such that $w_1Rw_2$ and $w_1Rw_3$ and $D_{w_1}=D_{w_2}= \{ a,b \}$ and $D_{w_3}= \{ a \}$.

Let $c_1, c_2$ two constants such that $I_M(c_1)=a$ and $I_M(c_2)=b$ and let $P$ a predicate such that $I_{w_1}(P) = I_{w_2}(P) =\{ a,b \}$ and $I_{w_3}(P) = \emptyset$.

Finally, consider $Pc_1$ as $\phi$ and $Pc_2$ as $\psi$.

Now in $M$ we have $V_{w_1} (\square(Pc_1 \land Pc_2))=1$ according to (iv''), since $V_{w_1} (Pc_1 \land Pc_2) = V_{w_2} (Pc_1 \land Pc_2)=1$, while $V_{w_3} (Pc_1 \land Pc_2)$ is undefined. But on the other hand, $V_{w_1} (\square Pc_1)=0$, since $V_{w_3} (Pc_1)=0$.

Now the main point.

Problems like these would seem to suggest that a "strong" interpretation of the connectives would be preferable to the "weak" interpretation given them in definition 2. Under the weak interpretation, the truth value of a conjunction is undefined if that of either of its conjuncts is undefined, even if the other conjunct is false. Under the strong interpretation, on the other hand, a conjunct is false if either of its conjuncts is, even if the other conjunct is neither true nor false.

[...]

Returning to the example and recalculating the truth values under the strong interpretation of the connectives, we see that because $V_{w_3} (Pc_1)=0$ $V_{w_3} (Pc_1 \land Pc_2) = 0$ instead of being undefined. As a result, we have $V_{w_1} (\square(Pc_1 \land Pc_2))=0$ instead of $V_{w_1} (\square(Pc_1 \land Pc_2))=1$. So the model is no longer a counterexample to the validity of $\square (\phi \land \psi) \rightarrow \square \psi$.

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Note : the reference to "strong Kleene" interpretation is to the treatment of the conectives (in particular : $\land$) in Kleene three-valued logic.