Disjunction with Tautology
Theorem
A disjunction with a tautology:
- $p \lor \top \dashv \vdash \top$
Proof by Natural Deduction
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $p \lor \top$ | Premise | (None) | ||
| 2 | 2 | $\top$ | Assumption | (None) | ||
| 3 | 3 | $p$ | Assumption | (None) | ||
| 4 | 3 | $p \lor \neg p$ | Rule of Addition: $\lor \II_1$ | 3 | ||
| 5 | 3 | $\top$ | Law of Excluded Middle | 4 | ||
| 6 | 1 | $\top$ | Proof by Cases: $\text{PBC}$ | 1, 2 – 2, 3 – 5 | Assumptions 2 and 3 have been discharged |
$\Box$
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By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $\top$ | Premise | (None) | ||
| 2 | 1 | $p \lor \top$ | Rule of Addition: $\lor \II_2$ | 1 |
$\blacksquare$
Proof by Truth Table
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations.
$\begin{array}{|c|c||ccc|} \hline
p & \top & p & \lor & \top \\
\hline
F & T & F & T & T \\
T & T & T & T & T \\
\hline
\end{array}$
$\blacksquare$
Sources
- 2012: M. Ben-Ari: Mathematical Logic for Computer Science (3rd ed.) ... (previous) ... (next): $\S 2.3.3$