True Statement is implied by Every Statement/Formulation 1
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Theorem
\(\ds p\) | \(\) | \(\ds \) | ||||||||||||
\(\ds \vdash \ \ \) | \(\ds q \implies p\) | \(\) | \(\ds \) |
Proof 1
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p$ | Premise | (None) | ||
2 | 1 | $\neg q \lor p$ | Rule of Addition: $\lor \II_2$ | 1 | ||
3 | 1 | $q \implies p$ | Sequent Introduction | 1 | Rule of Material Implication |
$\blacksquare$
Proof by Truth Table
We apply the Method of Truth Tables.
As can be seen by inspection, where the truth value in the relevant column on the left hand side is $\T$, that under the one on the right hand side is also $\T$:
$\begin{array}{|c||ccc|} \hline p & q & \implies & p \\ \hline \F & \F & \T & \F \\ \F & \T & \F & \F \\ \T & \F & \T & \T \\ \T & \T & \T & \T \\ \hline \end{array}$
$\blacksquare$
Sources
- 1973: Irving M. Copi: Symbolic Logic (4th ed.) ... (previous) ... (next): $2$ Arguments Containing Compound Statements: $2.3$: Argument Forms and Truth Tables: Exercise $\text{II} \ \mathbf 6.$
- 1996: H. Jerome Keisler and Joel Robbin: Mathematical Logic and Computability ... (previous) ... (next): $\S 1.14$: Exercise $17 \ \text{(ii)}$