Blackburn, $1.3.3(c)$: Consider the basic temporal language and the frames $(\mathbb Z, <), (\mathbb Q, <), (\mathbb R, <)$ (the integer, rational, and real numbers, respectively, all ordered by the usual less-than relation $<$). Check the validity of $$(Ep \land E\lnot p \land A(p\to Hp) \land A(\lnot p\to G\lnot p)) \to E(Hp\land G\lnot p)$$ in each of these frames, where $$E\phi:= P\phi \lor \phi\lor F\phi$$ $$A\phi:= H\phi\land \phi\land G\phi $$
My work:
Consider a frame $\mathfrak F$ and the formula $\phi:= (Ep \land E\lnot p \land A(p\to Hp) \land A(\lnot p\to G\lnot p)) \to E(Hp\land G\lnot p)$ where $\phi_1 := (Ep \land E\lnot p \land A(p\to Hp) \land A(\lnot p\to G\lnot p))$ and $\phi_2:= E(Hp\land G\lnot p)$. Consider an arbitrary world $w$, s.t. $\mathfrak F,w\Vdash \phi_1$. This basically means that for some $w_1$, $p$ holds, and $p$ holds for all $w<w_1$ and for some $w_2$, $\lnot p$ holds, and $\lnot p$ holds for all $w>w_2$. Showing $\mathfrak F,w\Vdash \phi_2$ amounts to asking:
$$\text{Can we find a world $w_0\in\mathfrak F$, such that $p$ holds $\forall w<w_0$ and $\lnot p$ holds $\forall w<w_0$?}$$
I think we can always do this if $\mathfrak F = (\mathbb Z,<)$, because we will always be able to find an integer $z\in\mathbb Z$ s.t. $p$ holds at $z$, and $\lnot p$ holds at $z+1$. Then just pick $w_0 = z$.
For $(\mathbb Q,<)$ and $(\mathbb R,<)$ I'm not able to prove the claim rigorously, but I think $\phi$ is valid in both these frames too (is that correct?). This is because once we find $w_1,w_2$ as defined earlier, we can iteratively (infinite times, so I'm suspicious) check what subset of the number line is assigned to worlds between $w_1,w_2$ by the valuation $V$ in question. For $w_1<w_3<w_2$, if $w_3\in V(p)$ then we can replace $w_1$ by $w_3$. If $w_1\in V(\lnot p)$, then we can replace $w_2$ by $w_3$.
How does this sound? Is it good/rigorous enough? How can I refine this argument?
Thanks a lot!
Notations for reference:
$H$ denotes "has always been the case", $G$ denotes "always going to be the case". Let me know if you've any trouble with the notation, I'd be happy to clear it up.
$(Ep\wedge E\neg p\wedge A(p\to Hp)\wedge A(\neg p\to G\neg p))\to E(Hp\wedge G\neg p)$ is valid in $(\mathbb{Z},<)$ and $(\mathbb{R},<)$, and it is not valid in $(\mathbb{Q},<)$.
I could send you an email if you need the proof.