# Rule of Association/Disjunction/Formulation 2

## Theorem

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

### Forward Implication

$\vdash \left({p \lor \left({q \lor r}\right)}\right) \implies \left({\left({p \lor q}\right) \lor r}\right)$

### Reverse Implication

$\vdash \left({p \lor \left({q \lor r}\right)}\right) \impliedby \left({\left({p \lor q}\right) \lor r}\right)$

## Proof 1

By the tableau method of natural deduction:

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

$\blacksquare$

## Proof 2

This proof is derived in the context of the following proof system: Instance 2 of the Hilbert-style systems.

By the tableau method:

$\vdash \paren {p \lor \paren {q \lor r} } \iff \paren {\paren {p \lor q} \lor r}$
Line Pool Formula Rule Depends upon Notes
1 $\paren {p \lor \paren {q \lor r} } \implies \paren {\paren {p \lor q} \lor r}$ Rule of Association: Forward Implication
2 $\paren {\paren {p \lor q} \lor r} \implies \paren {p \lor \paren {q \lor r} }$ Rule of Association: Reverse Implication
3 $\paren {\paren {p \lor \paren {q \lor r} } \implies \paren {\paren {p \lor q} \lor r} } \land \paren {\paren {\paren {p \lor q} \lor r} \implies \paren {p \lor \paren {q \lor r} } }$ Rule $RST \, 4$ 1,2
4 $\paren {p \lor \paren {q \lor r} } \iff \paren {\paren {p \lor q} \lor r}$ Rule $RST \, 2 (3)$ 3

$\blacksquare$