Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2/Proof 2
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Theorem
- $\vdash \paren {p \land \paren {q \lor r} } \iff \paren {\paren {p \land q} \lor \paren {p \land r} }$
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | $\paren {p \land \paren {q \lor r} } \implies \paren {\paren {p \land q} \lor \paren {p \land r} }$ | Theorem Introduction | (None) | Conjunction Distributes over Disjunction: Forward Implication | ||
2 | $\paren {\paren {p \land q} \lor \paren {p \land r} } \implies \paren {p \land \paren {q \lor r} }$ | Theorem Introduction | (None) | Conjunction Distributes over Disjunction: Reverse Implication | ||
3 | $\paren {p \land \paren {q \lor r} } \iff \paren {\paren {p \land q} \lor \paren {p \land r} }$ | Biconditional Introduction: $\iff \II$ | 1, 2 |
$\blacksquare$