Let $\mathbf{SBM}$ be the normal modal logic system defined as $\mathbf{T}$ plus the following two axioms:

$$\mathrm{SB}: \Box(\Diamond p \rightarrow p)\rightarrow (p \rightarrow \Box p)$$

$$\mathrm{M}: \Box\Diamond p \rightarrow \Diamond\Box p $$

If I am not mistaken, the frames for $\mathbf{SBM}$ are the same as the frames for $\mathbf{Triv}$ ($\mathbf{T}$ plus $p \leftrightarrow \Box p$), namely the frames where every world sees precisely itself. The argument (sketch) goes like this:

1) $\mathrm{SB}$ corresponds to the following frame condition: if $w$ sees $w'$ then there exists a path back from $w'$ to $w$ (i.e. $w'$ sees $w$ in a finite number of steps) such that $w$ sees all the worlds along the path, including $w$ if the reflexivity condition from $\mathbf{T}$ is added. Let's call this the visible back-path (VBP) condition. It can be proved using induction on the length of the path.

2) $\mathrm{M}$ corresponds to the McKinsey (McK) frame condition: for every partitioning of the set of worlds into two disjoint subsets, every world sees a world whose direct (immediate) successors lie all within the same partition. It can be proved by writing $\mathrm{M}$ as $\Diamond(\Box p \lor \Box \lnot p)$.

3) Consider a reflexive VBP frame that is not a frame for $\mathbf{Triv}$, i.e. there is a world $w$ that sees at least one world $w'$ distinct form itself. For every such world $w'$ define $L_w(w')$ to be the length of the shortest VBP from $w'$ to $w$. Define a partitioning of the set of worlds into two subsets such that the first one includes $w$ and all the worlds $w'$ seen by $w$ for which $L_w(w')$ is even, the second one includes all the remaining worlds. Then the McK frame condition is not satisfied at $w$.

At this point there are two possibilities, but I cannot prove either of them:

a) $p \leftrightarrow \Box p$ is a theorem of $\mathbf{SBM}$, i.e. the system $\mathbf{SBM}$ is the same as $\mathbf{Triv}$.

b) $p \leftrightarrow \Box p$ is not a theorem of $\mathbf{SBM}$, in which case $\mathbf{SBM}$ is distinct from $\mathbf{Triv}$ and therefore incomplete.

  • $\begingroup$ I've deleted an incorrect answer (I missed that the system was required to include $T$). $\endgroup$ May 26, 2014 at 22:53

1 Answer 1


I was pointed out (see here) that this is the same as the system $\mathbf{KTLM}$ presented in "A Simple Incomplete Extension of T which is the Union of Two Complete Modal Logics with f.m.p." by Roy A. Benton (2002). In the article, theorem 4.2 establishes that the frames for $\mathbf{SBM}$ are the same as the frames for $\mathbf{Triv}$, basically steps 1--3 above. Then lemma 4.5 shows that $p \rightarrow \Box p$ is not a theorem of $\mathbf{SBM}$, which implies that $\mathbf{SBM}$ is distinct from $\mathbf{Triv}$, hence incomplete.

The proof of the latter uses a "recessive" general frame $(W,R,P)$ with $W = \{0,1,2,...\}$, $R = \{(m,n):m=0 \lor m-1 \le n\}$ and $P=\{S \subseteq W:S\text{ is finite or cofinite}\}$. These admissible sets are well defined and $\mathrm{T}$ and $\mathrm{SB}$ are valid on the non-restricted frame (which is reflexive and satisfies the VBP condition). To prove that $\mathrm{M}$ is valid on the general frame we observe that for any valuation $V$, if $V(\Box \Diamond p,m)=T$ for some $m$ then $V(p,n)=T$ for arbitrarily large $n$, hence the set $\{n:V(p,n)=T\}$ must be cofinite, and by picking a large-enough $n$ seen by $m$ we have $V(\Box p,n)=T$, therefore $V(\Diamond \Box p,m)=T$. Finally, the valuation with $V(p,0)=T$ and $V(p,n)=F$ everywhere else falsifies $p \rightarrow \Box p$ at 0, hence on the general frame.


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