Self-Distributive Law for Conditional/Forward Implication/Formulation 1
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Theorem
- $p \implies \paren {q \implies r} \vdash \paren {p \implies q} \implies \paren {p \implies r}$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $p \implies \paren {q \implies r}$ | Premise | (None) | ||
| 2 | 2 | $p \implies q$ | Assumption | (None) | ||
| 3 | 3 | $p$ | Assumption | (None) | ||
| 4 | 1, 3 | $q \implies r$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 1, 3 | ||
| 5 | 2, 3 | $q$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 2, 3 | ||
| 6 | 1, 2, 3 | $r$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 4, 5 | ||
| 7 | 1, 2 | $p \implies r$ | Rule of Implication: $\implies \II$ | 3 – 6 | Assumption 3 has been discharged | |
| 8 | 1 | $\paren {p \implies q} \implies \paren {p \implies r}$ | Rule of Implication: $\implies \II$ | 2 – 7 | Assumption 2 has been discharged |
$\blacksquare$
Sources
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $2$ Conditionals and Negation: Exercise $1 \ \text{(j)}$