Non-Equivalence as Disjunction of Negated Conditionals

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Theorem

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


Proof 1

By the tableau method of natural deduction:

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

$\Box$


By the tableau method of natural deduction:

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

$\blacksquare$


Proof by Truth Table

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.


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

$\blacksquare$


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