Natural deduction for equivalence

Question

We have the following rules $$A \equiv B \vdash A \to B$$ $$A \equiv B \vdash B \to A$$ $$A \to B,B \to A \vdash A \equiv B$$ Prove the theorem $$\vdash [p \equiv (q \equiv r)] \to [(p \equiv q) \equiv r]$$

Answer 1

1) $p ≡ (q ≡ r)$ --- premise

2) $r$ --- assumed [a]

3) $p$ --- assumed [b]

4) $q ≡ r$ --- from 1) and 2) by $\equiv$-Elim and $\to$-Elim

5) $p ≡ q$ --- assumed [c].

From 3), 4) and 5) using $\equiv$-Elim twice and $\to$-Elim twice:

6) $r$

7) $(p ≡ q) \to r$ --- from 5) and 6) by $\to$-Intro, discharging [c].

8) $q$ --- from 2) and using the fact that from 4) we have $r \to q$

9) $p \to q$ --- from 3) and 8) by $\to$-Intro, discharging [b]

10) $q \to r$ --- from 2) by $\to$-Intro

11) $r \to q$ --- from 8) by $\to$-Intro

12) $q ≡ r$ --- from 10) and 11)

13) $p$ --- from 12) and using the fact that from 1) we have $(q ≡ r) \to p$

14) $q \to p$ --- from 13) by $\to$-Intro

15) $(p ≡ q)$ --- from 9) and 14)

16) $r \to (p ≡ q)$ --- from 2) and 14) by $\to$-Intro, discharging [a]

$(p ≡ q) ≡ r$ --- from 7) and 15).