An oft-cited result about intuitionistic logic is that
Proposition 1. Classical logic can be embedded into intuitionistic logic.
This is the Godel-Gentzen double-negation translation: given a formula $\phi$, we insert $\lnot \lnot$ throughout the formula to obtain a new formula $\texttt{encode}(\phi)$, such that $$ \phi \text{ is provable in classical logic} \iff \texttt{encode}(\phi) \text{ is provable in intuitionistic logic}. \tag{1} $$
People cite this as a result that intuitionistic logic is more expressive or more general. So there is an unspoken assumption that:
Proposition 2. Intuitionistic logic does not embed into classical logic.
My question is, what is the standard definition of embedding used such that Propositions 1 and 2 are both true?
Why this is not obvious:
The condition (1) cannot be the definition of embedding used: it is easy to provide such an encoding $\texttt{encode}$ from any logic to any other non-trivial logic. Just send any provable statement to a simple provable sentence (like $1 = 1$), and any unprovable statement to simple unprovable sentence (like $0 = 1$).
There are less trivial encodings, too. Suppose we require the encoding to be computable, or that it preserves some structure (each step of a proof of $\phi$ can be translated into a step of a proof of $\texttt{encode}(\phi)$). Then I think it may be true that such a translation exists from intuitionistic logic (or any other logic) into classical logic: we just have to introduce sufficiently many axioms as part of the encoded formula to describe how the source logic works. So for a formula in intuitionistic logic $\psi$, we define $\texttt{encode}(\psi)$ in classical logic as the statement that the conjunction of all those axioms implies "$\psi$ is provable", where "is provable" is also encoded in the target language. (Some consistency assumptions may be necessary on the set of axioms to make sure it faithfully describes which statements of intuitionistic logic are provable.)