Anja Petković Komel in her PhD thesis Meta-analysis of type theories with an application to the design of formal proofs studied transformations of type theories among other things.
Definition 9.3.3 on page 66 gives the notion of a conservative transformation. This is more general than a conservative extension, which is a conservative inclusion.
Let me give a quick summary. To understand the general definition, we need to know what a boundary is. Every type-theoretic judgement has a boundary and a head:
- the boundary of $\Gamma \vdash A \; \mathsf{type}$ is $\Gamma \vdash \Box \; \mathsf{type}$ and $A$ is the head.
- the boundary of $\Gamma \vdash t : A$ is $\Gamma \vdash \Box : A$ and $t$ is the head.
- the boundary of $\Gamma \vdash A \equiv B$ looks the same as the judgement itself because its head is trivial (judgemental equality is proof irrelevant), but we might write the judgement as $\Gamma \vdash A \equiv B \;\mathsf{by}\; \star$, the boundary as $\Gamma \vdash A \equiv B \;\mathsf{by}\; \Box$ and take $\star$ to be a trivial token that serves as the head.
- similarly for $\Gamma \vdash t \equiv s : A$.
A boundary $\Gamma \vdash \mathscr{b}$ may be filled with a head $e$ to give a judgement $\Gamma \vdash \mathscr{b}\boxed{e}$. For example, filling the boundary $\Gamma \vdash \Box : A$ with $t$ produces $\Gamma \vdash t : A$.
Boundaries are an important concept that is often left implicit, but they deserve to treated properly. For example, a goal in a proof assistant is precisely a well-formed boundary. The goal is solved when its head is filled to give a derivable judgement.
Given a type-theoretic transformation $f : \mathcal{T} \to \mathcal{U}$, we say that $f$ is conservative when the following holds: for any well-formed boundary $\Gamma \vdash_\mathcal{T} \mathscr{b}$ in $T$, if the boundary $f \Gamma \vdash_\mathcal{U} f \mathscr{b}$ can be filled with some $e$ to give a derivable judgement $f \Gamma \vdash_\mathcal{U} (f \mathscr{b})\boxed{e}$, then there is $e'$ such that $\Gamma \vdash_\mathcal{T} \mathscr{b}\boxed{e'}$.
Concretely, suppose $f : \mathcal{T} \to \mathcal{U}$ is an inclusion so that $f(e) = e$. Then conservativity of $f$ amounts to: if $\Gamma \vdash_\mathcal{T} \mathscr{b}$ is well-formed in $\mathcal{T}$ and $\Gamma \vdash_\mathcal{U} \mathscr{b}\boxed{e}$ is derivable in $\mathcal{U}$, then there is $e'$ in $\mathcal{T}$ such that $\mathcal{T}$ derives $\Gamma \vdash_\mathcal{T} \mathscr{b}\boxed{e'}$. In particular, when we specialize this condition to term judgements and term equations, we get:
- if $\Gamma \vdash_\mathcal{T} A \; \mathsf{type}$ and $\Gamma \vdash_\mathcal{U} e : A$ then there is $e'$ such that $\Gamma \vdash_\mathcal{T} e' : A$
- if $\Gamma \vdash_\mathcal{T} s : A$ and $\Gamma \vdash_\mathcal{T} t : A$ and $\Gamma \vdash_\mathcal{U} s \equiv t : A$ then $\Gamma \vdash_\mathcal{T} s \equiv t : A$
You should not take the above as precise definitions, because I skipped over the treatment of meta-variables, which matters. The details are in Petković Komel's dissertation. She also gives examples of conservative transformations, such as definitional extension.
The concept of a conservative transformation subsumes the usual notions of conservativity that one sees in logic books, because first-order theories are special cases of type theories.
Let us apply the concept to your specific question: if an extension adds a new element to a type, is it conservative? Concretely, suppose we have a type $A$ in theory $\mathcal{T}$ to which we add a new constant $\vdash c : A$ to obtain an extended theory $\mathcal{T}'$.
If $A$ was empty beforehand, then this is not conservative, consider the boundary $\vdash \Box : A$. Otherwise, consider a closed term $t : A$ and define a further extension $\mathcal{T}''$ which, in addition to $\vdash c : A$, also has the equation $\vdash c \equiv t : A$. Example 9.3.5 of Petković Komel's thesis shows that $\mathcal{T}''$ is a conservative extension of $\mathcal{T}$. We thus have a chain of extensions
$$\mathcal{T} \to \mathcal{T}' \to \mathcal{T}''$$
which can be used to show that $\mathcal{T}'$ is a conservative extension: given a boundary $$\Gamma \vdash_\mathcal{T} \mathscr{b}$$ and a filling $$\Gamma \vdash_{\mathcal{T}'} \mathscr{b}\boxed{e},$$ we also get
$$\Gamma \vdash_{\mathcal{T}''} \mathscr{b}\boxed{e},$$
and hence by conservativity of $\mathcal{T}''$ over $\mathcal{T}$ we find $e'$ such that
$$\Gamma \vdash_{\mathcal{T}} \mathscr{b}\boxed{e'}.$$
Working through Example 9.3.5 tells us how $e'$ is related to $e$: we just replace all occurrences of $c$ with $t$, as one would expect of a definitional extension.
The wheels are turning, the mechanism works.
You asked how the concept is useful. In chapter 10 of the dissertation, a general elaboration theorem is proved: every finitary type theory can be elaborated to a standard type theory. Conservativity plays a role in this theorem: the fact that the retrogression map (the opposite of elaboration) is conservative means that the elaboration does not reconstruct information that is not really there.
Conservativity also plays a role in the design of PAs. Suppose we use an SMT solver $S$ to solve goals in a proof assistant $P$. The situation amounts to a transformation $f : P \to S$ that converts boundaries in $P$ to boundaries in $S$. To say that $f$ is conservative means that it is safe to use the solver $S$: for if $S$ manages to fill a boundary $f \mathscr{b}$ then we know that the original boundary $\mathscr{b}$ can also be filled in $P$. Of course, we need an explicit way of converting the head that $S$ found back to a head in $P$ – but that is precisely a computable witness of conservativity. And also, note that we need not translate back entire derivations, we only need to be able to reconstruct the head.
Prop
s one can prove, or may not change which types we can prove are inhabited. So if you hear the phase conservative, you should ask yourself “conservative over what?” $\endgroup$