Rule of Addition/Sequent Form/Formulation 1/Form 1
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Theorem
\(\ds p\) | \(\) | \(\ds \) | ||||||||||||
\(\ds \vdash \ \ \) | \(\ds p \lor q\) | \(\) | \(\ds \) |
Proof 1
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p$ | Premise | (None) | ||
2 | 1 | $p \lor q$ | Rule of Addition: $\lor \II_1$ | 1 |
$\blacksquare$
Proof by Truth Table
We apply the Method of Truth Tables.
$\begin{array}{|c||ccc|} \hline p & p & \lor & q \\ \hline \F & \F & \F & \F \\ \F & \F & \T & \T \\ \T & \T & \T & \F \\ \T & \T & \T & \T \\ \hline \end{array}$
As can be seen, when $p$ is true so is $p \lor q$.
$\blacksquare$
Sources
- 1973: Irving M. Copi: Symbolic Logic (4th ed.) ... (previous) ... (next): $2$ Arguments Containing Compound Statements: $2.3$: Argument Forms and Truth Tables: Exercise $\text{I} \ \mathbf a.$
- 1973: Irving M. Copi: Symbolic Logic (4th ed.) ... (previous) ... (next): $3$: The Method of Deduction: $3.1$: Formal Proof of Validity: Rules of Inference: $9.$