I am wondering whether a rule like the following is consistent with decidable conversion and type-checking for dependent type theory:
$$ \frac{f\, g : (x:\mathsf{bool}) \to C~x\quad f~\mathsf{tt} \equiv g~\mathsf{tt} \quad f~\mathsf{ff} \equiv g~\mathsf{ff}}{f\equiv g} $$
That is, if two functions with domain $\mathsf{bool}$ agree definitionally on $\mathsf{tt}$ and $\mathsf{ff}$, then they are convertible. An analogous rule for functions on general inductive types like $\mathbb{N}$ is certainly bad, but it seems not too ridiculous to me to wonder whether the special case of $\mathsf{bool}$ (and other non-recursive inductive types, like sum-types and enumerated types) might be possible.
One thing that makes me somewhat hopeful is that this rule would make $(x:\mathsf{bool}) \to C~x$ behave very much like $C~\mathsf{tt} \times C~\mathsf{ff}$, and indeed be definitionally isomorphic to it if the latter also has an $\eta$-rule. Since such a rule for products is possible, I thought maybe it could also be possible for the function-type. And implementation-wise, I could imagine a bidirectional conversion-checking algorithm noticing when the type is a function with domain $\mathsf{bool}$ and deciding to apply this rule instead of the ordinary $\eta$-rule for functions.