Clavius's Law/Formulation 1
Theorem
- $\neg p \implies p \vdash p$
Proof 1
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $\neg p \implies p$ | Premise | (None) | ||
| 2 | $p \lor \neg p$ | Law of Excluded Middle | (None) | |||
| 3 | 3 | $\neg p$ | Assumption | (None) | Either $p$ is false ... | |
| 4 | 1, 3 | $p$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 1, 3 | ||
| 5 | 5 | $p$ | Assumption | (None) | ... or $p$ is true | |
| 6 | 1 | $p$ | Proof by Cases: $\text{PBC}$ | 2, 3 – 4, 5 – 5 | Assumptions 3 and 5 have been discharged |
$\blacksquare$
Proof 2
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $\neg p \implies p$ | Premise | (None) | ||
| 2 | 2 | $p \implies \bot$ | Assumption | (None) | ||
| 3 | 2 | $\neg p$ | Sequent Introduction | 2 | Negation as Implication of Bottom | |
| 4 | 1,2 | $p$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 1, 3 | ||
| 5 | 1 | $(p \implies \bot) \implies p$ | Rule of Implication: $\implies \II$ | 2 – 4 | Assumption 2 has been discharged | |
| 6 | 1 | $p$ | Sequent Introduction | 5 | Peirce's Law |
$\blacksquare$
Source of Name
This entry was named for Christopher Clavius.
Sources
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $3$ Conjunction and Disjunction: Exercise $1 \ \text{(j)}$
- 2000: Michael R.A. Huth and Mark D. Ryan: Logic in Computer Science: Modelling and reasoning about systems ... (previous) ... (next): $\S 1.2.1$: Rules for natural deduction: Exercises $1.5: \ 2 \ \text{(a)}$