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:
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.
$\blacksquare$