De Morgan's Laws (Logic)/Disjunction/Formulation 2/Proof 1
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Theorem
- $\vdash \paren {p \lor q} \iff \paren {\neg \paren {\neg p \land \neg q} }$
Proof
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 |
$\blacksquare$