Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 2

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Theorem

The disjunction operator is right distributive over the conjunction operator:

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


Proof

By the tableau method of natural deduction:

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

$\blacksquare$