De Morgan's Laws (Logic)/Conjunction/Formulation 1/Reverse Implication
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Theorem
- $\neg \left({\neg p \lor \neg q}\right) \vdash p \land q$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $\neg \left({\neg p \lor \neg q}\right)$ | Premise | (None) | ||
| 2 | 2 | $\neg \left ({p \land q}\right)$ | Assumption | (None) | ||
| 3 | 2 | $\neg p \lor \neg q$ | Sequent Introduction | 2 | De Morgan's Laws: Disjunction of Negations | |
| 4 | 1, 2 | $\bot$ | Principle of Non-Contradiction: $\neg \EE$ | 3, 1 | ||
| 5 | 1 | $p \land q$ | Reductio ad Absurdum | 2 – 4 | Assumption 2 has been discharged |
$\blacksquare$
Law of the Excluded Middle
This theorem depends on the Law of the Excluded Middle, by way of Reductio ad Absurdum.
This is one of the logical axioms that was determined by Aristotle, and forms part of the backbone of classical (Aristotelian) logic.
However, the intuitionist school rejects the Law of the Excluded Middle as a valid logical axiom.
This in turn invalidates this theorem from an intuitionistic perspective.
Sources
- 2000: Michael R.A. Huth and Mark D. Ryan: Logic in Computer Science: Modelling and reasoning about systems ... (previous) ... (next): $\S 1.2.5$: An aside: proof by contradiction: Exercises $1.6: \ 2 \ \text{(b)}$