Proving associativity in propositional logic

Question

$(p∧q) ∧ r ⊢ p ∧ (q∧r)$

In the sequent above, the only thing that happens is switching brackets between p&q and r, to q&r and separating out p. I could use the elimination rule between p&q and r, and get these separately, combine them with the introduction rule to get the conclusion.. But what step do I take to replace the brackets from p&q to q&r? Is their a rule that would help me move brackets?

Answer 1

If you want to prove it formally, you have to stay at the formal rules of the syntax, like e.g. :

The set well-formed propositional formulas is the smallest set $X$ with the properties

(i) $p_i ∈ X, i ∈ \mathbb N$,

(ii) if $ϕ,ψ ∈ X$, then $(ϕ ∧ ψ) ∈ X$, and so on.

Thus, the correct formula will be :

$((p \land q) \land r)$,

and the proof will be :

1) $(p \land q)$ --- from premise by $\land$-elim

2) $r$ --- from premise by $\land$-elim

3) $p$ --- from 1) by $\land$-elim

4) $q$ --- from 1) by $\land$-elim

5) $(q \land r)$ --- from 4) and 2) by $\land$-intro

and so on...