# Rule of Association/Conjunction/Formulation 2

## Theorem

$\vdash \paren {p \land \paren {q \land r} } \iff \paren {\paren {p \land q} \land r}$

## Proof

By the tableau method of natural deduction:

$\vdash \paren {p \land \paren {q \land r} } \iff \paren {\paren {p \land q} \land r}$
Line Pool Formula Rule Depends upon Notes
1 1 $p \land \paren {q \land r}$ Assumption (None)
2 1 $\paren {p \land q} \land r$ Sequent Introduction 1 Rule of Association: Formulation 1
3 $\paren {p \land \paren {q \land r} } \implies \paren {\paren {p \land q} \land r}$ Rule of Implication: $\implies \mathcal I$ 1 – 2 Assumption 1 has been discharged
4 4 $\paren {p \land q} \land r$ Assumption (None)
5 4 $p \land \paren {q \land r}$ Sequent Introduction 4 Rule of Association: Formulation 1
6 $\paren {\paren {p \land q} \land r} \implies \paren {p \land \paren {q \land r} }$ Rule of Implication: $\implies \mathcal I$ 4 – 5 Assumption 4 has been discharged
7 $\paren {p \land \paren {q \land r} } \iff \paren {\paren {p \land q} \land r}$ Biconditional Introduction: $\iff \mathcal I$ 3, 6

$\blacksquare$