# Disjunction with Tautology

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## 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: $\lor \mathcal I_1$ | 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 |

$\Box$

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: $\lor \mathcal I_2$ | 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): $\S 2.3.3$