Frames characterised by the formula $ p \to \square_2 \diamond_1 p$

Question

I am trying to solve exercises from the book Modal Logic by Patrick Blackburn, Maarten de Rijke and Yde Venema. I am having a problem to solve one of the exercises in the section 3.1, the exercise is 3.1.1 in page 131. Please help me.

Exercise : Consider a language with two diamonds $\diamond_1$ and $\diamond_2$. Show that the formula $\alpha := p \to \square_2 \diamond_1 p$ is valid on precisely those frames where the two relations $R_1 \text{ and } R_2$ satisfy $\forall xy (R_2xy \to R_1yx)$

Now I am able to show one part of the problem which is whenever in a frame $R_1 \text{ and } R_2$ satisfy $\forall xy (R_2xy \to R_1yx)$, $\alpha$ is valid there.

But I have stuck in the opposite direction. Suppose $\alpha$ is true in a frame and let $R_2xy$. I can not find a suitable approach to show $R_1yx$. I have though about breaking into cases when $M \text{ (corresponding model) },x \vDash p$ and $M,x \vDash \lnot p$. But it is not helping me. I am completely stuck here.

Perhaps I am missing something.Please help me to solve this problem. I will be very grateful. Thnx in advance.

Answer 1

Hint

We will prove the converse ...

Assume that there are two "worlds" $x,y$ such that :

$R_2xy$ and $\lnot R_1yx$.

Assume that $x \vDash p$ and $y \nvDash p$; in this way, $y \nvDash \diamond_1 p$, because there are no worlds "accesible$_1$" to $y$ where $p$ is true.

But $p$ is false in a world "accesible$_2$" to $x$ (i.e. in $y$) and thus : $x \nvDash \square_2\diamond_1 p$.

In conclusion :

$x \vDash p$ and $x \nvDash \square_2\diamond_1 p$

i.e.

$x \nvDash p \to \square_2\diamond_1 p$.

Answer 2

I think I have figured out a solution from the idea of Mauro ALLEGRANZA's answer. I hope it is correct. Please comment and rectify if there is any mistake. Thanks.

To show the converse part (i.e. whenever a frame $F$ satisfies $\alpha$, it satisfies the condition $\forall xy (R_2xy \to R_1yx)$) we go by contradiction.

Assume if possible there is a frame $F$ which satisfies $\alpha$ but $\exists xy$ such that $R_2xy$ and $\lnot R_1yx$

Then consider a valuation $V$ such that $V(p)=\{x\}$ and since $F$ satisfies $\alpha$ hence $\alpha$ is true at $x$ under $V$ i.e.

$F,V,x \vDash p \to \square_2 \diamond_1 p$

$\Rightarrow F,V,y \vDash \diamond_1 p$ as $R_2xy$

$\Rightarrow \exists z : R_1yz \text{ and } F,V,z \vDash p$

But $\lnot R_1yx$ means $z \ne x$ and $F,V,z \vDash p$ means $z\in V(p)$ which is a contradiction as $V(p)=\{x\}$

Hence the proof.