# modal logic S5

The modal logic S5 is the smallest normal modal logic containing the following schemas:

• (T) $\square A\to A$, and

• (5) $\diamond A\to\square\diamond A$.

S5 is also denoted by KT5, where T and 5 correspond to the schemas T and 5 respectively.

A binary relation  $R$ on a set $W$ is said to be Euclidean iff for any $u,v,w$, $uRv$ and $uRw$ imply $vRw$. $R$ being Euclidean is first-order definable:

 $\forall u\forall v\forall w((uRv\land uRw)\to vRw).$
###### Proposition 1.

5 is valid in a frame $\mathcal{F}$ iff $\mathcal{F}$ is Euclidean.

###### Proof.

First, let $\mathcal{F}$ be a frame validating 5. Suppose $wRx$ and $wRy$. Let $M$ be a model based on $\mathcal{F}$, with $V(p)=\{x\}$. Since $\models_{x}p$, we have $\models_{w}\diamond p$, and so $\models_{w}\square\diamond p$, or $\models_{u}\diamond p$ for all $u$ such that $wRu$. In particular, $\models_{y}\diamond p$. So there is a $z$ such that $yRz$ and $\models_{z}p$. But this means $z=x$, whence $yRx$, meaning $R$ is Euclidean.

Conversely, suppose $\mathcal{F}$ is a Euclidean frame, and $M$ a model based on $\mathcal{F}$. Suppose $\models_{w}\diamond A$. Then there is a $v$ such that $wRv$ and $\models_{v}A$. Now, for any $u$ with $wRu$, we have $uRv$ since $R$ is Euclidean. So $\models_{u}\diamond A$. Since $u$ is arbitrary, $\models_{w}\square\diamond A$, and therefore $\models_{w}\diamond A\to\square\diamond A$. ∎

###### Proof.

Suppose $R$ is both reflexive and Euclidean. If $aRb$, since $aRa$, $bRa$ so $R$ is symmetric. If $aRb$ and $bRc$, then $bRa$ since $R$ has just been proven symmetric, and therefore $aRc$, or $R$ is transitive    . Conversely, suppose $R$ is an equivalence relation. If $aRb$ and $aRc$, then $bRa$ since $R$ is symmetric, so that $bRc$ since $R$ is transitive. Hence $R$ is Euclidean. ∎

This also shows that

S5 $=$ KTB4,

where B is the schema $A\to\square\diamond A$, valid in any symmetric frame (see here (http://planetmath.org/ModalLogicB)), and 4 is the schema $\square A\to\square\square A$, valid in any transitive frame (see here (http://planetmath.org/ModalLogicS4)). It is also not hard to show that

S5 $=$ KDB4 $=$ KDB5,

where $D$ is the schema $\square A\to\diamond A$, valid in any serial frame (see here (http://planetmath.org/ModalLogicD)).

As a result,

###### Proposition 2.

S5 is sound in the class of equivalence frames.

###### Proof.

Since any theorem  $A$ in S5 is deducible  from a finite sequence  consisting of tautologies  , which are valid in any frame, instances of T, which are valid in reflexive frames, instances of 5, which are valid in Euclidean frames by the proposition  above, and applications of modus ponens  and necessitation, both of which preserve validity in any frame, $A$ is valid in any frame which is both reflexive and Euclidean, and hence an equivalence frame. ∎

###### Proof.

By the discussion above, it is enough to show that the canonical frame of S5 is reflexive, symmetric, and transitive. Since S5 contains T, B, and 4, $\mathcal{F}_{\textbf{S5}}$ is reflexive, symmetric, and transitive respectively, the proofs of which can be found in the corresponding entries on T, B, and S4. ∎

Remark. Alternatively, one can also show that the canonical frame of the consistent normal logic containing 5 must be Euclidean.

###### Proof.

Let $\Lambda$ be such a logic. Suppose $uR_{\Lambda}v$ and $uR_{\Lambda}w$. We want to show that $vR_{\Lambda}w$, or $\Delta_{v}:=\{B\mid\square B\in v\}\subseteq w$. Let $A$ be any wff. If $A\notin w$, $A\notin\Delta_{u}$ since $uR_{\Lambda}v$, so $\square A\notin u$ by the definition of $\Delta_{u}$, or $\neg\square A\in u$ since $u$ is maximal, or $\diamond\neg A\in u$ by substitution theorem on $\neg\square A\leftrightarrow\diamond\neg A$, or $\square\diamond\neg A\in u$ by modus ponens on 5 and the fact that $u$ is closed under modus ponens. This means that $\diamond\neg A\in\Delta_{u}$ by the definition of $\Delta_{u}$, or $\diamond\neg A\in v$ since $uR_{\Lambda}v$, so that $\neg\square A\in v$ by the substitution theorem on $\diamond\neg A\leftrightarrow\neg\square A$, which means $\square A\notin v$ since $v$ is maximal, or $A\notin\Delta_{v}$ by the definition of $\Delta_{v}$. ∎

Title modal logic S5 ModalLogicS5 2013-03-22 19:34:04 2013-03-22 19:34:04 CWoo (3771) CWoo (3771) 11 CWoo (3771) Definition msc 03B45 msc 03B42 S5 5 Euclidean