Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 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 1

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 1 $p \land \paren {q \lor r}$ Assumption (None)
2 1 $\paren {\paren {p \land q} \lor \paren {p \land r} }$ Sequent Introduction 1 Conjunction is Left Distributive over Disjunction: Formulation 1
3 $\paren {p \land \paren {q \lor r} } \implies \paren {\paren {p \land q} \lor \paren {p \land r} }$ Rule of Implication: $\implies \II$ 1 – 2 Assumption 1 has been discharged
4 4 $\paren {p \land q} \lor \paren {p \land r}$ Assumption (None)
5 4 $p \land \paren {q \lor r}$ Sequent Introduction 4 Conjunction is Left Distributive over Disjunction: Formulation 1
6 $\paren {\paren {p \land q} \lor \paren {p \land r} } \implies \paren {p \land \paren {q \lor r} }$ Rule of Implication: $\implies \II$ 4 – 5 Assumption 4 has been discharged
7 $\paren {p \land \paren {q \lor r} } \iff \paren {\paren {p \land q} \lor \paren {p \land r} }$ Biconditional Introduction: $\iff \II$ 3, 6

$\blacksquare$


Proof 2

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

$\blacksquare$


Proof by Truth Table

We apply the Method of Truth Tables to the proposition.

As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations.

$\begin{array}{|ccccc|c|ccccccc|} \hline p & \land & (q & \lor & r) & \iff & (p & \land & q) & \lor & (p & \land & r) \\ \hline \F & \F & \F & \F & \F & \T & \F & \F & \F & \F & \F & \F & \F \\ \F & \F & \F & \T & \T & \T & \F & \F & \F & \F & \F & \F & \T \\ \F & \F & \T & \T & \F & \T & \F & \F & \T & \F & \F & \F & \F \\ \F & \F & \T & \T & \T & \T & \F & \F & \T & \F & \F & \F & \T \\ \T & \F & \F & \F & \F & \T & \T & \F & \F & \F & \T & \F & \F \\ \T & \T & \F & \T & \T & \T & \T & \F & \F & \T & \T & \T & \T \\ \T & \T & \T & \T & \F & \T & \T & \T & \T & \T & \T & \F & \F \\ \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\ \hline \end{array}$

$\blacksquare$


Sources

(erroneously referring to it as one of De Morgan's Laws)