Non-Equivalence as Equivalence with Negation/Formulation 1/Proof 2

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Theorem

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


Proof

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.


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

$\blacksquare$