De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 2

From ProofWiki
Jump to navigation Jump to search

Theorem

$\vdash \paren {\neg p \land \neg q} \iff \paren {\neg \paren {p \lor q} }$


This can be expressed as two separate theorems:

Forward Implication

$\left({\neg p \land \neg q}\right) \implies \left({\neg \left({p \lor q}\right)}\right)$

Reverse Implication

$\left({\neg \left({p \lor q}\right)}\right) \implies \left({\neg p \land \neg q}\right)$


Proof 1

By the tableau method of natural deduction:

$\vdash \left({\neg p \land \neg q}\right) \iff \left({\neg \left({p \lor q}\right)}\right)$
Line Pool Formula Rule Depends upon Notes
1 1 $\neg p \land \neg q$ Assumption (None)
2 1 $\neg \left({p \lor q}\right)$ Sequent Introduction 1 De Morgan's Laws (Logic): Disjunction of Negations: Formulation 1
3 $\left({\neg p \land \neg q}\right) \implies \left({\neg \left({p \lor q}\right)}\right)$ Rule of Implication: $\implies \mathcal I$ 1 – 2 Assumption 1 has been discharged
4 4 $\neg \left({p \lor q}\right)$ Assumption (None)
5 4 $\neg p \land \neg q$ Sequent Introduction 4 De Morgan's Laws (Logic): Disjunction of Negations: Formulation 1
6 $\left({\neg \left({p \lor q}\right)}\right) \implies \left({\neg p \land \neg q}\right)$ Rule of Implication: $\implies \mathcal I$ 4 – 5 Assumption 4 has been discharged
7 $\left({\neg p \land \neg q}\right) \iff \left({\neg \left({p \lor q}\right)}\right)$ Biconditional Introduction: $\iff \mathcal I$ 3, 6

$\blacksquare$


Proof 2

We apply the Method of Truth Tables.

As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations.

$\begin{array}{|ccccc|c|cccc|} \hline \neg & p & \land & \neg & q & \iff & \neg & (p & \lor & q) \\ \hline T & F & T & T & F & T & T & F & F & F \\ T & F & F & F & T & T & F & F & T & T \\ F & T & F & T & F & T & F & T & T & F \\ F & T & F & F & T & T & F & T & T & T \\ \hline \end{array}$

$\blacksquare$


Sources

(referring to it as one of the laws of negation)