# Implication is Left Distributive over Conjunction/Formulation 1

## Theorem

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

This can of course be expressed as two separate theorems:

### Forward Implication

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

### Reverse Implication

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

## Proof 1

### Proof of Forward Implication

By the tableau method of natural deduction:

$p \implies \paren {q \land r} \vdash \paren {p \implies q} \land \paren {p \implies r}$
Line Pool Formula Rule Depends upon Notes
1 1 $p \implies \paren {q \land r}$ Premise (None)
2 2 $p$ Assumption (None)
3 1, 2 $q \land r$ Modus Ponendo Ponens: $\implies \mathcal E$ 1, 2
4 1, 2 $q$ Rule of Simplification: $\land \EE_1$ 3
5 1, 2 $r$ Rule of Simplification: $\land \EE_2$ 3
6 1 $p \implies q$ Rule of Implication: $\implies \II$ 2 – 4 Assumption 2 has been discharged
7 1 $p \implies r$ Rule of Implication: $\implies \II$ 2 – 5 Assumption 2 has been discharged
8 1 $\paren {p \implies q} \land \paren {p \implies r}$ Rule of Conjunction: $\land \II$ 6, 7

$\blacksquare$

### Proof of Reverse Implication

By the tableau method of natural deduction:

$\paren {p \implies q} \land \paren {p \implies r} \vdash p \implies \paren {q \land r}$
Line Pool Formula Rule Depends upon Notes
1 1 $\paren {p \implies q} \land \paren {p \implies r}$ Premise (None)
2 1 $\paren {p \land p} \implies \paren {q \land r}$ Sequent Introduction 1 Praeclarum Theorema
3 3 $p$ Assumption (None)
4 3 $p \land p$ Sequent Introduction 3 Rule of Idempotence: Conjunction
5 1, 3 $q \land r$ Modus Ponendo Ponens: $\implies \mathcal E$ 2, 4
6 1 $p \implies \paren {q \land r}$ Rule of Implication: $\implies \II$ 3 – 5 Assumption 3 has been discharged

$\blacksquare$

## Proof by Truth Table

We apply the Method of Truth Tables.

As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.

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

$\blacksquare$