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Let $(T,>)$ be a frame of minimal temporal logic, i.e. a frame as defined in Kripke semantics where the relation is a partial order relation $>$ defined on the set $T$ of worlds, called instants.

I read that the implication$$FA\land FB\rightarrow F(Fa\land B)\lor F(A\land B)\lor F(A\land FB)$$ is valid if and only if $>$ is a total order towards the future (D. Palladino, C. Palladino, Logiche matematiche). I think that it means that $FA\land FB\rightarrow F(Fa\land B)\lor F(A\land B)\lor F(A\land FB)$ is valid in instant $t'$ under any interpretation if and only if $\{t\in T:t>t'\}$ is totally ordered.

Analogously, $$PA\land PB\rightarrow P(Pa\land B)\lor P(A\land B)\lor P(A\land PB)$$ is valid if and only if $>$ is a total order towards the past.

I have no problem in showing to myself why the validity of the two formulae implies each of the two kinds of total order, but I cannot find an interpretation of the propositional constant useful to prove the converse implications. Could anybody explain how to prove them, which I think could be done by a proper interpretation "trick"? I thank you all very much!

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To show that the first formula is valid if $(T,>)$ is a total order towards the future (i.e., has no branching towards the future), assume that for a valuation $\sigma$ and a point $x \in T$ you have $(T,>),\sigma,x \Vdash FA \land FB$. Then there are $y,z \in T$ with $x<y$ and $x<z$ such that $(T,>),\sigma,y\Vdash A$ and $(T,>),\sigma,z \Vdash B$. But since $(T,>)$ is a total order towards the future you have ($z<y$ or $y=z$ or $y<z$). In the first case $(T,>),\sigma, z \Vdash (FA \land B)$, in the second case $(T,>),\sigma, y \Vdash (A \land B)$, and in the last case $(T,>),\sigma,y \Vdash (A \land FB)$. Thus in any case $(T,>),\sigma, x \Vdash F(FA \land B) \lor F(A\land B) \lor F(A\land FB)$. Hence the whole implication is valid.

For the other direction, if $(T,>)$ is not a total order towards the future, then there are $x,y,z \in T$ with $x<y$ and $x<z$ but such that $y$ and $z$ are incomparable, i.e., neither $y<z$ nor $z<y$ nor $y=z$ hold. Now construct the valuation $\sigma$ such that $\sigma(A) = \{y\}$ and $\sigma(B) = \{z\}$. Then $FA \land FB$ holds in $x$, but $F(FA \land B) \lor F(A\land B) \lor F(A \land FB)$ does not. Thus the implication $FA \land FB \to F(FA \land B) \lor F(A \land B) \lor F(A \land FB)$ is not valid.

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  • $\begingroup$ Very, very clear! Thank you so much! $\endgroup$ – Self-teaching worker Mar 3 '15 at 12:19

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