# Implication is Left Distributive over Conjunction/Formulation 2/Proof 1

## Theorem

$\vdash \left({p \implies \left({q \land r}\right)}\right) \iff \left({\left({p \implies q}\right) \land \left({p \implies r}\right)}\right)$

## Proof

### Proof of Forward Implication

Let us use the following abbreviations

 $\ds \phi$ $\text {for}$ $\ds p \implies \paren {q \land r}$ $\ds \psi$ $\text {for}$ $\ds \paren {p \implies q} \land \paren {p \implies r}$

By the tableau method of natural deduction:

$\paren {p \implies \paren {q \land r} } \implies \paren {\paren {p \implies q} \land \paren {p \implies r} }$
Line Pool Formula Rule Depends upon Notes
1 1 $\phi$ Assumption (None)
2 1 $\psi$ Sequent Introduction 1 Implication is Left Distributive over Conjunction: Formulation 1
3 $\phi \implies \psi$ Rule of Implication: $\implies \II$ 1 – 2 Assumption 1 has been discharged

Expanding the abbreviations leads us back to:

$\paren {p \implies \paren {q \land r} } \implies \paren {\paren {p \implies q} \land \paren {p \implies r} }$

$\blacksquare$

### Proof of Reverse Implication

Let us use the following abbreviations

 $\ds \phi$ $\text {for}$ $\ds \paren {p \implies q} \land \paren {p \implies r}$ $\ds \psi$ $\text {for}$ $\ds p \implies \paren {q \land r}$

By the tableau method of natural deduction:

$\paren {\paren {p \implies q} \land \paren {p \implies r} } \implies \paren {p \implies \paren {q \land r} }$
Line Pool Formula Rule Depends upon Notes
1 1 $\phi$ Assumption (None)
2 1 $\psi$ Sequent Introduction 1 Implication is Left Distributive over Conjunction: Formulation 1
3 $\phi \implies \psi$ Rule of Implication: $\implies \II$ 1 – 2 Assumption 1 has been discharged

Expanding the abbreviations leads us back to:

$\paren {\paren {p \implies q} \land \paren {p \implies r} } \implies \paren {p \implies \paren {q \land r} }$

$\blacksquare$