Modus Ponendo Ponens/Variant 2

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Theorem

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


Proof 1

By the tableau method of natural deduction:

$\vdash p \implies \paren {\paren {p \implies q} \implies q} $
Line Pool Formula Rule Depends upon Notes
1 1 $p$ Assumption (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
5 $p \implies \paren {\paren {p \implies q} \implies q}$ Rule of Implication: $\implies \II$ 1 – 4 Assumption 1 has been discharged

$\blacksquare$


Proof 2

We apply the Method of Truth Tables.

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

As can be seen by inspection, the main connective is true throughout.

$\blacksquare$