Rule of Exportation/Forward Implication/Formulation 1
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Theorem
- $\paren {p \land q} \implies r \vdash p \implies \paren {q \implies r}$
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $\paren {p \land q} \implies r$ | Premise | (None) | ||
2 | 2 | $p$ | Assumption | (None) | ||
3 | 3 | $q$ | Assumption | (None) | ||
4 | 2, 3 | $p \land q$ | Rule of Conjunction: $\land \II$ | 2, 3 | ||
5 | 1, 2, 3 | $r$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 1, 4 | ||
6 | 1, 2 | $q \implies r$ | Rule of Implication: $\implies \II$ | 3 – 5 | Assumption 3 has been discharged | |
7 | 1 | $p \implies \paren {q \implies r}$ | Rule of Implication: $\implies \II$ | 2 – 6 | Assumption 2 has been discharged |
$\blacksquare$