Tautological Antecedent
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Theorem
A conditional with a tautology as antecedent:
- $\top \implies p \dashv \vdash p$
Proof by Natural Deduction
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $\top \implies p$ | Premise | (None) | ||
2 | $\top$ | Rule of Top-Introduction: $\top \II$ | (None) | |||
3 | 1 | $p$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 1, 2 |
$\Box$
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p$ | Premise | (None) | ||
2 | 2 | $\top$ | Assumption | (None) | ||
3 | 1 | $\top \implies p$ | Rule of Implication: $\implies \II$ | 2 – 1 | Assumption 2 has been discharged |
$\blacksquare$
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.
$\begin{array}{|c|ccc||c|ccc|} \hline p & \top & \implies & p \\ \hline F & T & F & F \\ T & T & T & T \\ \hline \end{array}$
$\blacksquare$
Also see
Sources
- 2012: M. Ben-Ari: Mathematical Logic for Computer Science (3rd ed.) ... (previous) ... (next): $\S 2.3.3$