# Modus Ponendo Ponens/Variant 1

## Theorem

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

## Proof 1

By the tableau method of natural deduction:

$p \vdash \left({p \implies q}\right) \implies q$
Line Pool Formula Rule Depends upon Notes
1 1 $p$ Premise (None)
2 2 $p \implies q$ Assumption (None)
3 1, 2 $q$ Modus Ponendo Ponens: $\implies \mathcal E$ 2, 1
4 1 $\left({p \implies q}\right) \implies q$ Rule of Implication: $\implies \mathcal I$ 2 – 3 Assumption 2 has been discharged

$\blacksquare$

## Proof 2

We apply the Method of Truth Tables.

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

As can be seen by inspection, when $p$ is true, the value of the main connective is also true.

$\blacksquare$