$(q_1, q_2), (q_1, q_4)$ marked

$(q_3, q_2), (q_3, q_4)$ marked

$(q_5, q_2), (q_5, q_4)$ marked

Iterate:

Consider $(q_0, q_1)$:
- $\delta(q_0, 0)=q_1$, $\delta(q_1, 0)=q_3$. $(q_1, q_3)$ is unmarked.
- $\delta(q_0, 1)=q_2$, $\delta(q_1, 1)=q_4$. $(q_2, q_4)$ is unmarked (both final).
- $(q_0, q_1)$ remains unmarked.
Consider $(q_0, q_3)$:
- $\delta(q_0, 0)=q_1$, $\delta(q_3, 0)=q_0$. $(q_1, q_0)$ is unmarked.
- $\delta(q_0, 1)=q_2$, $\delta(q_3, 1)=q_4$. $(q_2, q_4)$ is unmarked.
- $(q_0, q_3)$ remains unmarked.
Consider $(q_0, q_5)$:
- $\delta(q_0, 0)=q_1$, $\delta(q_5, 0)=q_4$. $(q_1, q_4)$ is marked (non-final, final). So, mark $(q_0, q_5)$.
Consider $(q_1, q_3)$:
- $\delta(q_1, 0)=q_3$, $\delta(q_3, 0)=q_0$. $(q_3, q_0)$ is unmarked.
- $\delta(q_1, 1)=q_4$, $\delta(q_3, 1)=q_4$. $(q_4, q_4)$ trivial.
- $(q_1, q_3)$ remains unmarked.
Consider $(q_2, q_5)$:
- $\delta(q_2, 0)=q_5$, $\delta(q_5, 0)=q_4$. $(q_5, q_4)$ is marked (non-final, final). So, mark $(q_2, q_5)$.

After iterations, the unmarked pairs are: $(q_0, q_1)$, $(q_0, q_3)$, $(q_1, q_3)$, and $(q_2, q_4)$.

Equivalence classes:
- $\{q_0, q_1, q_3\}$ (since $(q_0,q_1)$, $(q_1,q_3)$, $(q_0,q_3)$ are equivalent)
- $\{q_2, q_4\}$ (since $(q_2,q_4)$ are equivalent)
- $\{q_5\}$ (distinguishable from all others)
Let $A = \{q_0, q_1, q_3\}$, $B = \{q_2, q_4\}$, $C = \{q_5\}$.
Initial State: $A$ (contains $q_0$).
Final States: $B$ (contains $q_2, q_4$).

State	$\delta(q, 0)$	$\delta(q, 1)$
$A = \{q_0, q_1, q_3\}$	$A$ (from $q_0 \to q_1$, $q_1 \to q_3$, $q_3 \to q_0$)	$B$ (from $q_0 \to q_2$, $q_1 \to q_4$, $q_3 \to q_4$)
$B = \{q_2, q_4\}$	$C$ (from $q_2 \to q_5$) / $B$ (from $q_4 \to q_4$) = $B$ or $C$? Let's recheck. $\delta(q_2, 0) = q_5 \in C$. $\delta(q_4, 0) = q_4 \in B$. This indicates $q_2$ and $q_4$ are distinguishable by input $0$. Let's re-examine the table marking for $(q_2, q_4)$. $\delta(q_2, 0) = q_5$, $\delta(q_4, 0) = q_4$. Is $(q_5, q_4)$ marked? Yes. So $(q_2, q_4)$ should be marked. This means $q_2$ and $q_4$ are not equivalent. Correction: My manual trace of Step 3 was flawed. Let's restart the marking.

Let's use a systematic table. Mark 'X' for distinguishable.

Initial: Mark $(q_i, q_j)$ if one is final and other is non-final.

Final states: $F = \{q_2, q_4\}$. Non-final: $N = \{q_0, q_1, q_3, q_5\}$.
Mark: $(q_0,q_2), (q_0,q_4), (q_1,q_2), (q_1,q_4), (q_3,q_2), (q_3,q_4), (q_5,q_2), (q_5,q_4)$.

  q0 q1 q2 q3 q4 q5
q0    .  X  .  X  .
q1       X  .  X  .
q2          X  X  X
q3             X  .
q4                X
q5

Pairs marked 'X' are distinguishable. '.' means unmarked.

Iteration 1: