Modus Ponendo Ponens/Variant 3

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Theorem

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


Proof 1

By the tableau method of natural deduction:

$\vdash \paren {\paren {p \implies q} \land p} \implies q$
Line Pool Formula Rule Depends upon Notes
1 1 $\paren {p \implies q} \land p$ Assumption (None)
2 1 $p \implies q$ Rule of Simplification: $\land \mathcal E_1$ 1
3 1 $p$ Rule of Simplification: $\land \mathcal E_2$ 1
4 1 $q$ Modus Ponendo Ponens: $\implies \mathcal E$ 2, 3
5 $\paren {\paren {p \implies q} \land p} \implies q$ Rule of Implication: $\implies \mathcal I$ 1 – 4 Assumption 1 has been discharged

$\blacksquare$


Proof by Truth Table

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.

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

$\blacksquare$


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