Modus Ponendo Ponens/Sequent Form

Theorem

The Modus Ponendo Ponens can be symbolised by the sequent:

 $\ds p$ $\implies$ $\ds q$ $\ds p$  $\ds$ $\ds \vdash \ \$ $\ds q$  $\ds$

Proof 1

By the tableau method of natural deduction:

$p \implies q, p \vdash q$
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

By the tableau method of natural deduction:

$p \implies q, p \vdash q$
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: $\neg \EE$ 2, 5
7 2, 5 $q$ Rule of Explosion: $\bot \EE$ 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

$\blacksquare$

Proof by Truth Table

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.

$\blacksquare$