Implication Equivalent to Negation of Conjunction with Negative/Formulation 1/Proof

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Theorem

$p \implies q \dashv \vdash \neg \paren {p \land \neg q}$


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.

$\begin{array}{|ccc||ccccc|} \hline p & \implies & q & \neg & (p & \land & \neg & q) \\ \hline \F & \T & \F & \T & \F & \F & \T & \F \\ \F & \T & \T & \T & \F & \F & \F & \T \\ \T & \F & \F & \F & \T & \T & \T & \F \\ \T & \T & \T & \T & \T & \F & \F & \T \\ \hline \end{array}$

$\blacksquare$


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