# Modus Ponendo Ponens/Sequent Form

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## Theorem

The Modus Ponendo Ponens can be symbolised by the sequent:

- $p \implies q, p \vdash q$

## Proof 1

By the tableau method of natural deduction:

Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|

1 | 1 | $p \implies q$ | Premise | (None) | ||

2 | 2 | $p$ | Premise | (None) | ||

3 | 1, 2 | $q$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 1, 2 |

$\blacksquare$

## Proof 2

We apply the Method of Truth Tables.

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

As can be seen, when $p$ is true, and so is $p \implies q$, then $q$ is also true.

$\blacksquare$

## Proof 3

By the tableau method of natural deduction:

Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|

1 | 1 | $p \implies q$ | Premise | (None) | ||

2 | 2 | $p$ | Premise | (None) | ||

3 | 3 | $p$ | Assumption | (None) | ||

4 | 1, 3 | $q$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 1, 3 | ||

5 | 5 | $\lnot p$ | Assumption | (None) | ||

6 | 2, 5 | $\bot$ | Principle of Non-Contradiction: $\neg \mathcal E$ | 2, 5 | ||

7 | 2, 5 | $q$ | Rule of Explosion: $\bot \mathcal E$ | 6 | ||

8 | $p \lor \lnot p$ | Law of Excluded Middle | (None) | |||

9 | 1, 2 | $q$ | Proof by Cases: $\text{PBC}$ | 8, 3 – 4, 5 – 7 | Assumptions 3 and 5 have been discharged |

$\blacksquare$