De Morgan's Laws (Logic)/Disjunction/Formulation 2
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Theorem
- $\vdash \paren {p \lor q} \iff \paren {\neg \paren {\neg p \land \neg q} }$
Proof 1
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \lor q$ | Assumption | (None) | ||
2 | 1 | $\neg \paren {\neg p \land \neg q}$ | Sequent Introduction | 1 | De Morgan's Laws (Logic): Disjunction: Formulation 1 | |
3 | $\paren {p \lor q} \implies \paren {\neg \paren {\neg p \land \neg q} }$ | Rule of Implication: $\implies \II$ | 1 – 2 | Assumption 1 has been discharged | ||
4 | 4 | $\neg \paren {\neg p \land \neg q}$ | Assumption | (None) | ||
5 | 4 | $p \lor q$ | Sequent Introduction | 4 | De Morgan's Laws (Logic): Disjunction: Formulation 1 | |
6 | $\paren {\neg \paren {\neg p \land \neg q} } \implies \paren {p \lor q}$ | Rule of Implication: $\implies \II$ | 4 – 5 | Assumption 4 has been discharged | ||
7 | $\paren {p \lor q} \iff \paren {\neg \paren {\neg p \land \neg q} }$ | Biconditional Introduction: $\iff \II$ | 3, 6 |
$\blacksquare$
Proof by Truth Table
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations.
$\begin{array}{|ccc|c|cccccc|} \hline (p & \lor & q) & \iff & (\neg & (\neg & p & \land & \neg & q)) \\ \hline \F & \F & \F & \T & \F & \T & \F & \T & \T & \F \\ \F & \T & \T & \T & \T & \T & \F & \F & \F & \T \\ \T & \T & \F & \T & \T & \F & \T & \F & \T & \F \\ \T & \T & \T & \T & \T & \F & \T & \F & \F & \T \\ \hline \end{array}$
$\blacksquare$
Sources
- 1959: A.H. Basson and D.J. O'Connor: Introduction to Symbolic Logic (3rd ed.) ... (previous) ... (next): $\S 3.6$: Reference Formulae: $RF \, 15$
- 1959: A.H. Basson and D.J. O'Connor: Introduction to Symbolic Logic (3rd ed.) ... (previous) ... (next): $\S 4.7$: The Derivation of Formulae: $D \, 26$
- 1964: Donald Kalish and Richard Montague: Logic: Techniques of Formal Reasoning ... (previous) ... (next): $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Theorem $\text{T64}$
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): or