$(p∧q) ∧ r ⊢ p ∧ (q∧r)$
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?
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...