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:
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.
$\blacksquare$
Sources
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $2$ Conditionals and Negation: Exercise $1 \ \text{(m)}$