Conjunction with Negative Equivalent to Negation of Implication/Formulation 1

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Theorem

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


This can be expressed as two separate theorems:

Forward Implication

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

Reverse Implication

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


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.

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

$\blacksquare$


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