modal logic B
The modal logic B (for Brouwerian) is the smallest normal modal logic containing the following schemas:

•
(T) $\mathrm{\square}A\to A$, and

•
(B) $A\to \mathrm{\square}\diamond A$.
In this entry (http://planetmath.org/ModalLogicT), we show that T is valid in a frame iff the frame is reflexive^{}.
Proposition 1.
B is valid in a frame $\mathrm{F}$ iff $\mathrm{F}$ is symmetric.
Proof.
First, suppose B is valid in a frame $\mathcal{F}$, and $wRu$. Let $M$ be a model based on $\mathcal{F}$, with $V(p)=\{w\}$, $p$ a propositional variable. Since $w\in V(p)$, ${\vDash}_{w}p$, and ${\vDash}_{w}p\to \mathrm{\square}\diamond p$ by assumption^{}, ${\vDash}_{v}\diamond p$ for all $v$ such that $wRv$. In particular, ${\vDash}_{u}\diamond p$, which means there is a $t$ such that $uRt$ and ${\vDash}_{t}p$. But this means that $t\in V(p)$, so $t=w$, whence $uRw$, and $R$ is symmetric.
Conversely, let $\mathcal{F}$ be a symmetric frame, $M$ a model based on $\mathcal{F}$, and $w$ a world in $M$. Suppose ${\vDash}_{w}A$. If ${\vDash \u0338}_{w}\mathrm{\square}\diamond A$, then there is a $u$ such that $wRu$, with ${\vDash \u0338}_{u}\diamond A$. This mean for no $t$ with $uRt$, we have ${\vDash}_{t}A$. Since $R$ is symmetric, $uRw$, so ${\vDash \u0338}_{w}A$, a contradiction^{}. Therefore, ${\vDash}_{w}\mathrm{\square}\diamond A$, and ${\vDash}_{w}A\to \mathrm{\square}\diamond A$ as a result. ∎
As a result,
Proposition 2.
B is sound in the class of symmetric frames.
Proof.
Since any theorem^{} in B is deducible from a finite sequence^{} consisting of tautologies^{}, which are valid in any frame, instances of B, which are valid in symmetric frames by the proposition^{} above, and applications of modus ponens^{} and necessitation, both of which preserve validity in any frame, whence the result. ∎
In addition^{}, using the canonical model of B, we have
Proposition 3.
B is complete^{} in the class of reflexive, symmetric frames.
Proof.
Since B contains T, its canonical frame ${\mathcal{F}}_{\text{\mathbf{B}}}$ is reflexive. We next show that any consistent normal logic $\mathrm{\Lambda}$ containing the schema B is symmetric. Suppose $w{R}_{\mathrm{\Lambda}}u$. We want to show that $u{R}_{\mathrm{\Lambda}}w$, or that ${\mathrm{\Delta}}_{u}:=\{B\mid \mathrm{\square}B\in u\}\subseteq w$. It is then enough to show that if $A\notin w$, then $A\notin {\mathrm{\Delta}}_{u}$. If $A\notin w$, $\mathrm{\neg}A\in w$ because $w$ is maximal, or $\mathrm{\square}\diamond \mathrm{\neg}A\in w$ by modus ponens on B, or $\mathrm{\square}\mathrm{\neg}\mathrm{\square}A\in w$ by the substitution theorem on $A\leftrightarrow \mathrm{\neg}\mathrm{\neg}A$, or $\mathrm{\neg}\mathrm{\square}A\in {\mathrm{\Delta}}_{w}$ by the definition of ${\mathrm{\Delta}}_{w}$, or $\mathrm{\neg}\mathrm{\square}A\in u$ since $w{R}_{\mathrm{\Lambda}}u$, or $\mathrm{\square}A\notin u$, since $u$ is maximal, or $A\notin {\mathrm{\Delta}}_{u}$ by the definition of ${\mathrm{\Delta}}_{u}$. So ${R}_{\mathrm{\Lambda}}$ is symmetric, and ${R}_{\text{\mathbf{B}}}$ is both reflexive and symmetric. ∎
Title  modal logic B 

Canonical name  ModalLogicB 
Date of creation  20130322 19:34:11 
Last modified on  20130322 19:34:11 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  8 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 03B45 
Defines  B 