For every intuitionistic kripke frame $F = (W, R)$ and every formula $\phi$ and points $x,y \in W$ if $x \models \phi$ and $x R y$ then $y\models\phi$

Question

I need to show that:

For every intuitionistic kripke frame $F = (W, R)$ and every formula $\phi$ and every points $x,y \in W$ if $x \models \phi$ and $x R y$ then $y \models \phi$

I know it's done by induction:

$\bullet \phi = p$. Then $ p \in V(x)$, but V is upward closed so $p \in V(y)$, so $y \models \phi$

$\bullet \phi = \phi_1 \lor \phi _2$. -trivial using the hypothesis (also for $\land$)

$\bullet \phi = \phi_1 \rightarrow \phi _2$. Here is my problem. I can't seem to continue the formal proof here. Any help?

Answer 1

$\phi = \phi_1 \rightarrow \phi_2$

We will show that $y \models \phi_1 \rightarrow \phi_2$

Let $z \in W$ and $y R z$ and let $z \models \phi_1$

But $x R y$ and $y R z$ and R is transitive. So $xRz$

But $x \models \phi_1 \rightarrow \phi_2$ and $xRz$ and $z \models \phi_1$

So $z \models \phi_2$

So $y \models \phi_1 \rightarrow \phi_2$

Answer 2

Recall that $w \models \phi_1 \to \phi_2$ if and only if for all $v \in W$, if $wRv$ and $v \models \phi_1$ then $v \models \phi_2$. This is the part of the conditions on $\models$ (or of $V$, whichever one is defined first in your treatment of the subject).

Now suppose $w \models \phi_1 \to \phi_2$ and $v \in W$, and also $w R v$.

Suppose we have $s \in W$ such that $vRs$ and $s \models \phi_1$. Then by transitivity, $w R s$. Then $s \models \phi_2$.

Thus, we see that $v \models \phi_1 \to \phi_2$.