# Conjunction with Tautology

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## Theorem

A conjunction with a tautology:

- $p \land \top \dashv \vdash p$

## Proof by Natural Deduction

By the tableau method of natural deduction:

Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|

1 | 1 | $p \land \top$ | Premise | (None) | ||

2 | 1 | $p$ | Rule of Simplification: $\land \mathcal E_1$ | 1 |

$\Box$

By the tableau method of natural deduction:

Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|

1 | 1 | $p$ | Premise | (None) | ||

2 | $q \lor \neg q$ | Law of Excluded Middle | (None) | |||

3 | $\top$ | Law of Excluded Middle | (None) | |||

4 | 1 | $p \land \top$ | Rule of Conjunction: $\land \mathcal I$ | 1, 3 |

$\blacksquare$

## 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.

$\begin{array}{|c|ccc||c|} \hline p & p & \land & \top & \top \\ \hline F & F & F & T & T \\ T & T & T & T & T \\ \hline \end{array}$

$\blacksquare$

## Sources

- 2012: M. Ben-Ari:
*Mathematical Logic for Computer Science*(3rd ed.) ... (previous) ... (next): $\S 2.3.3$