Prove $\vdash (A_1 ↔ A_2) \vee (A_2 ↔ A_3) \vee (A_3 ↔ A_1) $ using natural deduction.

Long comment

I think that the proof is quite long: we can try with a simpler (but similar) example :

$\vdash (A_1 \to A_2) \lor (A_2 \to A_3) \lor (A_3 \to A_1)$.

In this case, the proof is straightforward, using EM :

1) $\vdash A_1 \lor \lnot A_1$

2) $A_1$ --- assumed [a] for $\lor$-elim

3) $A_3 \to A_1$ --- from 2) by $\to$-intro

4) $(A_2 \to A_3) \lor (A_3 \to A_1)$ --- from 3) by $\lor$-intro

5) $(A_1 \to A_2) \lor (A_2 \to A_3) \lor (A_3 \to A_1)$ --- from 4) by $\lor$-intro

6) $\lnot A_1$ --- assumed [b] for $\lor$-elim

7) $A_1$ --- assumed [c]

8) $A_2$ --- from 6) and 7)

9) $A_1 \to A_2$ --- from 7) and 8) by $\to$-intro, discharging [c]

10) $(A_1 \to A_2) \lor (A_2 \to A_3)$ --- from 9) by $\lor$-intro

11) $(A_1 \to A_2) \lor (A_2 \to A_3) \lor (A_3 \to A_1)$ --- from 10) by $\lor$-intro

Now we can conclude by $\lor$-elim from 1), discharging [a] and [b].