In [BN07, Subsection 4.3], the authors show that if a symmetric scheme $\mathcal{SE}$ is IND-CPA and a MAC $\mathcal{MA}$ is EUF-CMA (though they use the SUF-CMA notation, but I think it refers to the same notion?), then the symmetric scheme $\mathcal{SE'}$, constructed via the Encrypt-the-MAC technique, is IND-CCA secure.
Unfortunately, they show it the following way:
- $\mathcal{SE}$ being IND-CPA implies $\mathcal{SE'}$ being IND-CPA
- $\mathcal{MA}$ being EUF-CMA implies $\mathcal{SE'}$ being INT-CTXT
- $\mathcal{SE'}$ being INT-CPA and INT-CTXT implies it being IND-CCA
I struggle to see how a game-based proof of this would look-like. For instance, let us say we have an adversary $\mathcal{A}$ that breaks IND-CCA security of $\mathcal{SE'}$, and let us consider a reduction $\mathcal{R}$ that has access to this adversary and tries to break either IND-CPA security of $\mathcal{SE}$ or the EUF-CMA security of $\mathcal{MA}$. How would the reduction proceed, since it does not have access to a decryption oracle?
Assuming we work in the real-or-random setup, I understand the reduction can record the queries that $\mathcal{A}$ does in order to know how it should answer on these particular queries. However, how is it possible for $\mathcal{R}$ to answer decryption queries which are not stored in its database? If we consider that $\mathcal{A}$ won't try to query the decryption of some $y$ without previously having asked the encryption of $x$ which resulted in $y=\mathrm{Enc}(x)$, then what's the point of giving a decryption oracle to $\mathcal{A}$ at all? Should $\mathcal{R}$ answer with a random $r=\mathrm{Enc}^{-1}(y)$, hoping that $\mathcal{A}$ won't ever query $x$ such that $r\neq y=f(x)$? In that case doesn't this potentially break $\mathcal{A}$'s strategy to break IND-CCA?