Two questions about the partial order of propositional modal logics

Question

The propositional modal logics $T$, $B$, $S4$ and $S5$ are related as follows: $T$ has the fewest theorems, $S5$ has the most, $B$ and $S4$ are intermediate between $T$ and $S5$, but they are incomparable. I have two questions about this partial order. First, is the intersection of $B$ and $S4$ equal to $T$? That is, is the set of theorems of $T$ precisely the set of theorems which are common to $B$ and $S4$? And second, is the union of $B$ and $S4$ equal to $S5$? That is, is the set of theorems of $S5$ precisely the set of theorems which belong in either $B$ or $S4$? If either of those questions have a negative answer, I would like a specific counterexample, meaning a modal logic formula that is in one of the sets but not in the other.

Answer 1

Both your questions have negative answers (although the first has an almost-positive answer).

Below I'll write $[X]$ for the deductive closure of $X$.

The defining sentence of $S_5$, namely $$\Diamond p\rightarrow\Box\Diamond p,$$ is not in $S_4$ or $B$. On the other hand, this is a near thing: the frames validating $B\cup S_4$ are exactly the frames validating $S_5$, and so the appropriate completeness theorem (actually, there are several we could use here) says that $S_5$ is contained in $[S_4\cup B]$. In particular, in the lattice of deductively closed modal theories, $S_5$ is the least upper bound of $S_4$ and $B$.

Meanwhile, $[B]\cap [S_4]$ is trivially deductively closed, but it is also larger than $T$. Keep in mind that if $B\vdash\varphi$ and $S_4\vdash\psi$ then $[B]\cap [S_4]\vdash\varphi\vee\psi$, and this lets us build lots of counterexamples. For example, most obviously we can use the sentence $$(\Box p\rightarrow\Box\Box p)\vee (\Box p\rightarrow\Diamond p),$$ which is just the disjunction of the defining sentences of $S_4$ and $B$ respectively.