# p-morphism

Let $\mathcal{F}_{1}=(W_{1},R_{1})$ and $\mathcal{F}_{2}=(W_{2},R_{2})$ be Kripke frames. A p-morphism from $\mathcal{F}_{1}$ to $\mathcal{F}_{2}$ is a function $f:W_{1}\to W_{2}$ such that

• if $uR_{1}w$, then $f(u)R_{2}f(w)$,

• if $sR_{2}t$ and $s=f(u)$ for some $u\in W_{1}$, then $uR_{1}w$ and $t=f(w)$ for some $w\in W_{1}$,

We write $f:\mathcal{F}_{1}\to\mathcal{F}_{2}$ to denote that $f$ is a p-morphism from $\mathcal{F}_{1}$ to $\mathcal{F}_{2}$.

Let $M_{1}=(\mathcal{F}_{1},V_{1})$ and $M_{2}=(\mathcal{F}_{2},V_{2})$ be Kripke models of modal propositional logic  PL${}_{M}$. A p-morphism from $M_{1}$ to $M_{2}$ is a p-morphism $f:\mathcal{F}_{1}\to\mathcal{F}_{2}$ such that

• $M_{1}\models_{w}p$ iff $M_{2}\models_{f(w)}p$ for any propositional variable $p$.

###### Proposition 1.

For any wff $A$, $M_{1}\models_{w}A$ iff $M_{2}\models_{f(w)}A$.

###### Proof.

Induct on the number $n$ of logical connectives in $A$. When $n=0$, $A$ is either $\perp$ or a propositional variable. The case when $A$ is $\perp$ is obvious, and the other case is definition. Next, suppose $A$ is $B\to C$, then $M_{1}\models_{w}A$ iff $M_{1}\not\models_{w}B$ or $M_{1}\models_{w}C$ iff $M_{2}\not\models_{f(w)}B$ or $M_{2}\models_{f(w)}C$ iff $M_{2}\models_{f(w)}A$. Finally, suppose $A$ is $\square B$, and $M_{1}\models_{w}A$. To show $M_{2}\models_{f(w)}A$, let $t$ be such that $f(w)R_{2}t$. Then there is a $u$ such that $t=f(u)$ and $wR_{1}u$, so that $M_{1}\models_{u}B$. By induction  , $M_{2}\models_{f(u)}B$, or $M_{2}\models_{t}B$. Hence $M_{2}\models_{f(w)}A$. Conversely, suppose $M_{2}\models_{f(w)}A$. To show $M_{1}\models_{w}A$, let $u$ be such that $wR_{1}u$. So $f(w)R_{2}f(u)$, and therefore $M_{2}\models_{f(u)}B$. By induction, $M_{1}\models_{u}B$, whence $M_{1}\models_{w}A$. ∎

###### Proposition 2.

If a p-morphism $f:\mathcal{F}_{1}\to\mathcal{F}_{2}$ is one-to-one, then $\mathcal{F}_{2}\models A$ implies $\mathcal{F}_{1}\models A$ for any wff $A$.

###### Proof.

Suppose $\mathcal{F}_{2}\models A$. Let $M=(W_{1},R_{1},V_{1})$ be any model based on $\mathcal{F}_{1}$ and $w$ any world in $W_{1}$. We want to show that $M\models_{w}A$.

Define a Kripke model $M^{\prime}:=(W_{2},R_{2},V_{2})$ as follows: for any propositional variable $p$, let $V_{2}(p):=\{s\in W_{2}\mid f^{-1}(s)\cap V_{1}(p)\neq\varnothing\}$. Then $M\models_{w}p$ iff $w\in V_{1}(p)$ iff $f^{-1}(f(w))=\{w\}\subseteq V_{1}(p)$ (since $f$ is one-to-one) iff $f^{-1}(f(w))\cap V_{1}(p)\neq\varnothing$ iff $f(w)\in V_{2}(p)$ iff $M^{\prime}\models_{f(w)}p$. This shows that $f$ is a p-morphism from $M$ to $M^{\prime}$.

###### Proposition 3.

If a p-morphism $f:\mathcal{F}_{1}\to\mathcal{F}_{2}$ is onto, then $\mathcal{F}_{1}\models A$ implies $\mathcal{F}_{2}\models A$ for any wff $A$.

###### Proof.

Suppose $\mathcal{F}_{1}\models A$. Let $M=(W_{2},R_{2},V_{2})$ be any model based on $\mathcal{F}_{2}$ and $s$ any world in $W_{2}$. We want to show that $M\models_{s}A$.

Define a Kripke model $M^{\prime}:=(W_{1},R_{1},V_{1})$ as follows: for any propositional variable $p$, let $V_{1}(p):=\{w\in W_{1}\mid f(w)\in V_{2}(p)\}$. Then $w\in V_{1}(p)$ iff $f(w)\in V_{2}(p)$, so $f$ is a p-morphism from $M^{\prime}$ to $M$, and by assumption $M^{\prime}\models A$ for any wff $A$.

Now, let $w\in W_{1}$ be a world such that $f(w)=s$. Since $M^{\prime}\models A$, $M^{\prime}\models_{w}A$ in particular, and therefore $M\models_{f(w)}A$ or $M\models_{s}A$ by the last proposition. ∎

###### Corollary 1.

If $f:\mathcal{F}_{1}\to\mathcal{F}_{2}$ is bijective   , then $\mathcal{F}_{1}\models A$ iff $\mathcal{F}_{2}\models A$ for any wff $A$.

A frame $\mathcal{F}^{\prime}$ is said to be a p-morphic image of a frame $\mathcal{F}$ if there is an onto p-morphism $f:\mathcal{F}\to\mathcal{F}^{\prime}$. Let $\mathcal{C}$ be the class of all frames validating a wff. Then by the third proposition above, $\mathcal{C}$ is closed under p-morphic images: if a frame is in $\mathcal{C}$, so is any of its p-morphic images. Using this property, we can show the following: if $\mathcal{C}$ is the class of all frames validating a wff $A$, then $\mathcal{C}$ can not be

• the class of all asymetric frames

• the class of all anti-symmetric frames

###### Proof.

Let $\mathcal{F}_{1}=(\mathbb{N},<)$ and $\mathcal{F}_{2}=(\{0\},R)$, where $0R0$. Notice that $\mathcal{F}_{1}$ is in both the class of irreflexive frames and the class of asymetric frames, but $\mathcal{F}_{2}$ is in neither. Let $f:\mathbb{N}\to\{0\}$ be the obvious surjection. Clearly, $m implies $f(m)Rf(n)$. Also, if $f(m)R0$, then $f(m)Rf(m+1)$. So $f$ is a p-morphism. Suppose $\mathcal{C}$ is either the class of all irreflexive frames or the class of all asymetric frames. If $A$ is validated by $\mathcal{C}$, $A$ is validated by $\mathcal{F}_{1}$ in particular (since $\mathcal{F}_{1}$ is in $\mathcal{C}$), so that $A$ is validated by $\mathcal{F}_{2}$ as well, which means $\mathcal{F}_{2}$ is $\mathcal{C}$ too, a contradiction   . Therefore, no such an $A$ exists.

Next, let $\mathcal{F}_{3}=(\mathbb{N},S)$, where $nS(n+1)$ for all $n\in\mathbb{N}$ and $\mathcal{F}_{4}=(\{0,1\},R)$, where $R=\{(0,1),(1,0)\}$. Let $\mathcal{C}$ be the class of all anti-symmetric frames. Then $\mathcal{F}_{3}$ is in $\mathcal{C}$ but $\mathcal{F}_{4}$ is not. Let $f:\mathcal{F}_{3}\to\mathcal{F}_{4}$ be given by $f(n)=0$ if $n$ is even and $f(n)=1$ if $n$ is odd. If $aSb$, then $f(a)$ and $f(b)$ differ by $1$, so $f(a)Rf(b)$. On the other hand, if $f(a)Rx$, then $x$ is either $0$ or $1$, depending on whether $a$ odd or even. Pick $b=a+1$, so $aSb$ and $f(b)=x$. This shows that $f$ is a p-morphism. By the same argument as in the last paragraph, no wff $A$ is validated by precisely the members of $\mathcal{C}$. ∎

Title p-morphism Pmorphism 2013-03-22 19:34:54 2013-03-22 19:34:54 CWoo (3771) CWoo (3771) 12 CWoo (3771) Definition msc 03B45 Bisimulation