Implication is Left Distributive over Disjunction/Formulation 2

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Theorem

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


Proof

By the tableau method of natural deduction:

$\vdash \paren {p \implies \paren {q \lor r} } \iff \paren {\paren {p \implies q} \lor \paren {p \implies r} } $
Line Pool Formula Rule Depends upon Notes
1 1 $p \implies \paren {q \lor r}$ Assumption (None)
2 1 $\paren {p \implies q} \lor \paren {p \implies r}$ Sequent Introduction 1 Implication is Left Distributive over Disjunction: Formulation 1
3 $\paren {p \implies \paren {q \lor r} } \implies \paren {\paren {p \implies q} \lor \paren {p \implies r} }$ Rule of Implication: $\implies \mathcal I$ 1 – 2 Assumption 1 has been discharged
4 4 $\paren {p \implies q} \lor \paren {p \implies r}$ Assumption (None)
5 4 $p \implies \paren {q \lor r}$ Sequent Introduction 4 Implication is Left Distributive over Disjunction: Formulation 1
6 $\paren {\paren {p \implies q} \lor \paren {p \implies r} } \implies \paren {p \implies \paren {q \lor r} }$ Rule of Implication: $\implies \mathcal I$ 4 – 5 Assumption 4 has been discharged
7 3, 6 $\paren {p \implies q} \lor \paren {p \implies r}$ Biconditional Introduction: $\iff \mathcal I$ 16, 17 19

$\blacksquare$

Sources