# Rule of Material Implication/Formulation 2

## Theorem

$\vdash \left({p \implies q}\right) \iff \left({\neg p \lor q}\right)$

This can be expressed as two separate theorems:

### Forward Implication

$\vdash \left({p \implies q}\right) \implies \left({\neg p \lor q}\right)$

### Reverse Implication

$\vdash \left({\neg p \lor q}\right) \implies \left({p \implies q}\right)$

## Proof 1

By the tableau method of natural deduction:

$\left({p \implies q}\right) \iff \left({\neg p \lor q}\right)$
Line Pool Formula Rule Depends upon Notes
1 1 $p \implies q$ Assumption (None)
2 1 $\neg p \lor q$ Sequent Introduction 1 Rule of Material Implication: Formulation 1
3 $\left({p \implies q}\right) \implies \left({\neg p \lor q}\right)$ Rule of Implication: $\implies \mathcal I$ 1 – 2 Assumption 1 has been discharged
4 4 $\neg p \lor q$ Assumption (None)
5 4 $p \implies q$ Sequent Introduction 4 Rule of Material Implication: Formulation 1
6 $\left({\neg p \lor q}\right) \implies \left({p \implies q}\right)$ Rule of Implication: $\implies \mathcal I$ 4 – 5 Assumption 4 has been discharged
7 $\left({p \implies q}\right) \iff \left({\neg p \lor q}\right)$ Biconditional Introduction: $\iff \mathcal I$ 3, 6

$\blacksquare$

#### Law of the Excluded Middle

This proof depends on the Law of the Excluded Middle, by way of Rule of Material Implication/Formulation 1/Forward Implication.

This is one of the axioms of logic that was determined by Aristotle, and forms part of the backbone of classical (Aristotelian) logic.

However, the intuitionist school rejects the Law of the Excluded Middle as a valid logical axiom. This in turn invalidates this proof from an intuitionistic perspective.

## Proof 2

We apply the Method of Truth Tables.

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

$\begin{array}{|ccc|c|cccc|} \hline (p & \implies & q) & \iff & (\neg & p & \lor & q) \\ \hline F & T & F & T & T & F & T & F \\ F & T & T & T & T & F & T & T \\ T & F & F & F & T & T & F & F \\ T & T & T & F & T & T & T & T \\ \hline \end{array}$

$\blacksquare$