Conditional is Left Distributive over Disjunction/Formulation 2
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Theorem
- $\vdash \paren {p \implies \paren {q \lor r} } \iff \paren {\paren {p \implies q} \lor \paren{p \implies r} }$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $p \implies \paren {q \lor r}$ | Assumption | (None) | ||
| 2 | 1 | $\paren {p \implies q} \lor \paren {p \implies r}$ | Sequent Introduction | 1 | Conditional is Left Distributive over Disjunction: Formulation 1 | |
| 3 | $\paren {p \implies \paren {q \lor r} } \implies \paren {\paren {p \implies q} \lor \paren {p \implies r} }$ | Rule of Implication: $\implies \II$ | 1 – 2 | Assumption 1 has been discharged | ||
| 4 | 4 | $\paren {p \implies q} \lor \paren {p \implies r}$ | Assumption | (None) | ||
| 5 | 4 | $p \implies \paren {q \lor r}$ | Sequent Introduction | 4 | Conditional is Left Distributive over Disjunction: Formulation 1 | |
| 6 | $\paren {\paren {p \implies q} \lor \paren {p \implies r} } \implies \paren {p \implies \paren {q \lor r} }$ | Rule of Implication: $\implies \II$ | 4 – 5 | Assumption 4 has been discharged | ||
| 7 | 3, 6 | $\paren {p \implies q} \lor \paren {p \implies r}$ | Biconditional Introduction: $\iff \II$ | 16, 17 | 19 |
$\blacksquare$
Sources
- 1964: Donald Kalish and Richard Montague: Logic: Techniques of Formal Reasoning ... (previous) ... (next): $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Theorem $\text{T55}$