Modus Ponendo Ponens/Variant 1

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Theorem

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


Proof 1

By the tableau method of natural deduction:

$p \vdash \paren {p \implies q} \implies q$
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 $\paren {p \implies q} \implies q$ Rule of Implication: $\implies \II$ 2 – 3 Assumption 2 has been discharged

$\blacksquare$


Proof by Truth Table

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.

$\blacksquare$


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