Non-Equivalence as Conjunction of Disjunction with Disjunction of Negations

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Theorem

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


Proof 1

By the tableau method of natural deduction:

$\neg \paren {p \iff q} \vdash \paren {p \lor q} \land \paren {\neg p \lor \neg q} $
Line Pool Formula Rule Depends upon Notes
1 1 $\neg \paren {p \iff q}$ Premise (None)
2 1 $\paren {p \lor q} \land \neg \paren {p \land q}$ Sequent Introduction 1 Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction
3 1 $\paren {p \lor q} \land \paren {\neg p \lor \neg q}$ Sequent Introduction 2 De Morgan's Laws: Disjunction of Negations


By the tableau method of natural deduction:

$\paren {p \lor q} \land \paren {\neg p \lor \neg q} \vdash \neg \paren {p \iff q} $
Line Pool Formula Rule Depends upon Notes
1 1 $\paren {p \lor q} \land \paren {\neg p \lor \neg q}$ Premise (None)
2 1 $\paren {p \lor q} \land \neg \paren {p \land q}$ Sequent Introduction 1 De Morgan's Laws: Disjunction of Negations
3 1 $\neg \paren {p \iff q}$ Sequent Introduction 2 Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction

$\blacksquare$


Proof 2

We apply the Method of Truth Tables.

As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.


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

$\blacksquare$