Rule of Association/Disjunction/Formulation 2/Proof 1
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Theorem
- $\vdash \paren {p \lor \paren {q \lor r} } \iff \paren {\paren {p \lor q} \lor r}$
Proof
By the tableau method of natural deduction:
| 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: $\implies \II$ | 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: $\implies \II$ | 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: $\iff \II$ | 3, 6 |
$\blacksquare$