Conditional is Left Distributive over Conjunction/Reverse Implication/Formulation 2/Proof
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Theorem
- $\vdash \paren {\paren {p \implies q} \land \paren {p \implies r} } \implies \paren {p \implies \paren {q \land r} }$
Proof
Let us use the following abbreviations
| \(\ds \phi\) | \(\text {for}\) | \(\ds \paren {p \implies q} \land \paren {p \implies r}\) | ||||||||||||
| \(\ds \psi\) | \(\text {for}\) | \(\ds p \implies \paren {q \land r}\) |
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $\phi$ | Assumption | (None) | ||
| 2 | 1 | $\psi$ | Sequent Introduction | 1 | Conditional is Left Distributive over Conjunction: Formulation 1 | |
| 3 | $\phi \implies \psi$ | Rule of Implication: $\implies \II$ | 1 – 2 | Assumption 1 has been discharged |
Expanding the abbreviations leads us back to:
- $\paren {\paren {p \implies q} \land \paren {p \implies r} } \implies \paren {p \implies \paren {q \land r} }$
$\blacksquare$