Remark: here I'm using the indexes $0,1,2$ and $0,1$ instead of $1, 2, 3$ and $1,2$.
We have to show the $3$-indistinguishability problem is equivalent to the $2$-indistinguishability one.
$2$-indistinguishability is easier than $3$-indistinguishability.
First let consider it exists an adversary $\mathcal{A}_3$ against the
$3$-indistinguishability problem with advantage $\epsilon$.
Define $\mathcal{B}^{\mathcal{A}_3}_2$:
Receive three messages $(m_0, m_1, m_2)$ from $\mathcal{A}_3$
$x \xleftarrow{\\\$} \mathbb{Z}_3$
$(m^\prime_0, m^\prime_1) := (m_{(1+ x \mod 3)}, m_{(2+ x \mod 3)})$
Send $(m^\prime_0, m^\prime_1)$ to the challenger, and receive $c$.
Send $c$ to $\mathcal{A}_3$ and receive $b$.
If $b=x$ then return a random bit $b^\prime$ else return $(b- x \mod 3)$.
We first prove that $\mathcal{B}_2$ has probability to win $\frac{1-\epsilon}{4} +\epsilon$.
Let call $b''$ the bit chosen by the challenger.
$\mathbb{P}(\mathcal{B}_2 wins)= \mathbb{P}(b- x \mod 3 = b'')\mathbb{P}(\mathcal{B}_2 wins|
b- x \mod 3 = b'') +
\mathbb{P}(b- x \mod 3 \neq b'')\mathbb{P}(\mathcal{B}_2 wins|
b- x \mod 3 \neq b'')$
$= \epsilon \cdot 1 + (1 - \epsilon)\mathbb{P}((b=x) \wedge b^\prime =b'' |
b- x \mod 3 \neq b^{\prime\prime}) $
$= \epsilon + (1 - \epsilon)\frac{1}{2}\cdot\frac{1}{2}.qed$
Now we can look the advantage:
$|\frac{1}{2} - (\frac{1-\epsilon}{4} +\epsilon)|=|\frac{1- 3 \epsilon}{4}| = \frac{1}{12}|\frac{1}{3}- \epsilon|$.
If $|\frac{1}{3}- \epsilon|$ is non negligible, it implies $|\frac{1}{2} - (\frac{1-\epsilon}{4} +\epsilon)|$ is also non negligible.
$3$-indistinguishability is easier than $2$-indistinguishability.
Now let consider it exists an adversary $\mathcal{A}_2$ against the
$2$-indistinguishability problem with advantage $\epsilon$.
Define $\mathcal{B}^{\mathcal{A}_2}_3$:
Receive two messages $(m_0, m_1)$from $\mathcal{A}_2$
$b \xleftarrow{\\\$} \mathbb{Z}_2$
$m_2 := m_{b}$
Send $(m_0, m_1, m_2)$ to the challenger, and receive $c$.
Send $c$ to $\mathcal{A}_2$ and receive $b^\prime$.
$x \xleftarrow{\\\$} \big\{b, 2\big\}$
If $b^\prime=b$ then return x else return $b^\prime$
Whe first prove compute the probability to win for $\mathcal{B}_3$.
$\mathbb{P}(\mathcal{B}_3 wins) =
\frac{1}{3}\mathbb{P}(\mathcal{B}_3 wins|b''=2) +
\frac{1}{3}\mathbb{P}(\mathcal{B}_3 wins| b''=b)+
\frac{1}{3}\mathbb{P}(\mathcal{B}_3 wins| b''=1 - b) $
$\mathbb{P}(\mathcal{B}_3 wins) =
\frac{1}{3}\mathbb{P}(b'=b \wedge x=2|b''=2) +
\frac{1}{3}\mathbb{P}(b'=b \wedge x=b| b''=b)+
\frac{\epsilon}{3}.$
Because $x$ is picked independently of $b'$:
$\mathbb{P}(\mathcal{B}_3 wins)$
$= \frac{1}{3}\mathbb{P}(b'=b|b''=2) \cdot \mathbb{P}(x=2|b''=2) +
\frac{1}{3}\mathbb{P}(b'=b|b''=b') \cdot \mathbb{P}(x=b''|b''=b')+
\frac{\epsilon}{3} $
$\mathbb{P}(\mathcal{B}_3 wins) =
\frac{1}{3}\epsilon \cdot \frac1 2 +
\frac{1}{3}\epsilon \cdot \frac1 2+
\frac{\epsilon}{3} = \frac{2\epsilon}{3}.$
Now we compute the advantage of $\mathcal{B}_3$:
$|\frac1 3 - \frac{2\epsilon}{3} |= \frac1 6 |\frac 1 2 - \epsilon|.$
If $|\frac{1}{2}- \epsilon|$ is non negligible, it implies $|\frac{1}{3} - \frac{2\epsilon}{3}|$ is also non negligible.
We deduce these problems are equivalent.
