# Implication is Left Distributive over Conjunction/Formulation 2

## Theorem

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

This can be expressed as two separate theorems:

### Forward Implication

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

### Reverse Implication

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

## Proof 1

### Proof of Forward Implication

Let us use the following abbreviations

 $\displaystyle \phi$ $\text{ for }$ $\displaystyle p \implies \left({q \land r}\right)$ $\displaystyle \psi$ $\text{ for }$ $\displaystyle \left({p \implies q}\right) \land \left({p \implies r}\right)$

By the tableau method of natural deduction:

$\left({p \implies \left({q \land r}\right)}\right) \implies \left({\left({p \implies q}\right) \land \left({p \implies r}\right)}\right)$
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: $\implies \mathcal I$ 1 – 2 Assumption 1 has been discharged

Expanding the abbreviations leads us back to:

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

$\blacksquare$

### Proof of Reverse Implication

Let us use the following abbreviations

 $\displaystyle \phi$ $\text{ for }$ $\displaystyle \left({p \implies q}\right) \land \left({p \implies r}\right)$ $\displaystyle \psi$ $\text{ for }$ $\displaystyle p \implies \left({q \land r}\right)$

By the tableau method of natural deduction:

$\left({\left({p \implies q}\right) \land \left({p \implies r}\right)}\right) \implies \left({p \implies \left({q \land r}\right)}\right)$
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: $\implies \mathcal I$ 1 – 2 Assumption 1 has been discharged

Expanding the abbreviations leads us back to:

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

$\blacksquare$

## Proof 2

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.

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

$\blacksquare$