Let $\textit{C}$ be the modal logical schema $(\square (\square \alpha \rightarrow \alpha) \wedge \square (\square \lnot\alpha \rightarrow \lnot \alpha))\rightarrow (\square \alpha \vee \square \lnot \alpha)$.


I believe we can establish that $\textit{C}$ corresponds to the second order condition that the accessibility relation $\textit{R}$ is ancestrally directed in the following sense: If $xRy\_0$ and $xRz\_0$ and if $Y$ and $Z$ are the set of possible worlds seen by $y\_0$ and $z\_0$, respectively, in a finite number of steps, then $Y\cap Z$ is nonempty.


Let $K$, $D$ and $4$ be as usual where $D$ corresponds with the seriality and $4$ with the transitivity of the accessibility relation. The schema $.2$ is $\lozenge \square \alpha \rightarrow \square \lozenge \alpha$ which correponds with the directedness of the accessibility relation. 


Questions: How (if at all) is .2 derivable in KD4C? How (if at all) is C derivable in KD4.2? 



On directedness, transitivity and ancestral directedness
Your axiom .2 is $\mathrm{G1}$ from *Hughes&Cresswell*, so I will use this name. They (and also Wikipedia) call the corresponding frames *convergent* (another name for *directed*). Also, I am going to rewrite axiom $\mathrm{C}$ as
$(\Diamond p \land \Box(\Box p \rightarrow p) \land \Box(p \rightarrow \Diamond p)) \rightarrow \Box p$.



> 
> *$\mathrm{C}$ is not derivable in $\mathbf{KD4G1}$.*
> 
> 
> 


If it were, then it would also be derivable in $\mathbf{S4.2}$, which extends $\mathbf{KD4G1}$.
But $\mathrm{C}$ plus $\mathbf{S4.2}$ reflexivity immediately implies
$\Diamond p \rightarrow \Box p$,
then $p \rightarrow \Box p$, therefore $p \leftrightarrow \Box p$
and $\mathbf{S4.2}$ would collapse to $\mathbf{Triv}$,
which is known not to be the case.


Note that the frame condition you give for C is incorrect,
since it implies seriality when $y\_0=z\_0=y$,
while C is valid on the non-serial frame $F=(\{x,y\},\{(x,y)\})$.
Assume we fix this by weakening the definition of ancestral convergence to $y\_0\not=z\_0$.
Then we still get only a necessary, but not sufficient, condition for a C-frame.
Indeed, the frame
$F'=(\{x,y,z\},\{(x,y),(x,z),(y,y),(y,z),(z,z)\})$ is serial, transitive and convergent, therefore ancestrally convergent,
yet $\mathrm{C}$ fails to validate at $x$ for valuation $V(p)=\{y\}$.
And $F'$ validates $\mathrm{G1}$, which is another proof that $\mathrm{C}$ is not derivable in $\mathbf{KD4G1}$.



> 
> *$\mathrm{G1}$ is derivable in $\mathbf{KD4C}$.*
> 
> 
> 


Here is an outline of the proof:


a) In $\mathbf{KD4}$, derive
$(\Box p \land \Box\Box p) \leftrightarrow \Box p$ and
$\Box p \rightarrow \Box\Diamond p$ and
$\Box\Diamond p \rightarrow \Diamond p$ and
$\Box p \rightarrow \Diamond\Box p$ and
$\Diamond (p \land \Box p) \leftrightarrow \Diamond\Box p$,
all quite straightforward.


b) Substitute $p\land\Box p/p$ in $\mathrm{C}$, then use all a) and simplify it to
$(\Diamond\Box p \land \Box(\Box p \rightarrow p)) \rightarrow \Box p$.


c) Substitute $\Diamond p/p$ in the result of b), then apply 3-rd a) and simplify it to
$\Diamond\Box\Diamond p \rightarrow \Box\Diamond p$.


d) Apply $\Diamond$-monotony to 2-nd a) and derive
$\Diamond\Box p \rightarrow \Diamond\Box\Diamond p$.


e) $\mathrm{G1}$, i.e. $\Diamond\Box p \rightarrow \Box\Diamond p$, follows from c) and d).


