Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2/Proof 2

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:

$\vdash \paren {p \land \paren {q \lor r} } \iff \paren {\paren {p \land q} \lor \paren {p \land r} } $

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

$\blacksquare$