# Implication is Left Distributive over Conjunction/Formulation 2

## Theorem

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

This can be expressed as two separate theorems:

### Forward Implication

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

### Reverse Implication

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

## Proof 1

### 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:

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

$\blacksquare$

## Proof by Truth Table

We apply the Method of Truth Tables.

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

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

$\blacksquare$