(Disclaimer: I am a noob in cryptography so please keep that mind in your answers).
In symmetric private key cryptography, an adversary $\mathcal A$ provided with encryption and decryption oracles $\mathcal O_{\text{Enc}},\mathcal O_{\text{Dec}}$ wins the IND-CCA game if it can consistently guess which of two plaintexts $m_0,m_1\in \mathcal M$ corresponds to a given (by some oracle $\mathcal O_{LR}$) cipher $c^*=\text{Enc}_k(m_{b\in\{0,1\}})$, that is, if it can output $b'=b$ with probability greater than one half. On the other hand, $\mathcal A$ with the same oracles wins the OW-CCA game if given a cipher $c^*=\text{Enc}_k(m^*)$ it can output $m'=m^*$.
I want to show that IND-CCA implies OW-CCA. The argument should go like this: suppose that IND-CCA holds but there exists an $\mathcal A$ that breaks OW-CCA. Then $\mathcal A$ can be used to build an adversary $\mathcal B$ that breaks IND-CCA. The details, however, escape me; can anyone help with those?
Edit. The most basic construction I can think of goes as follows: call the oracle $\mathcal O_{LR}$ on the pair $m_0,m_1$ to obtain $c^*$. Each time $\mathcal B$ calls an oracle, we pass the query to $\mathcal A$'s equivalent oracle and then forward the answer to $\mathcal B$. Since we assume that $\mathcal A$ eventually wins, it will return a value $m'$ that corresponds to either $m_0$ (in which case we make it return $b'=0$) or $m_1$ (in which case it returns $b'=1$). By our assumption that IND-CCA cannot be broken, it follows that $\mathcal A$ cannot exist either.
Does this work?
