Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 2

Theorem

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

This can be expressed as two separate theorems:

Forward Implication

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

Reverse Implication

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

Proof 1

By the tableau method of natural deduction:

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

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$p \land \neg q$	Assumption	(None)
2		1	$\neg \paren {p \implies q}$	Sequent Introduction	1	Conjunction with Negative is Equivalent to Negation of Conditional: Formulation 1: Forward Implication
3			$\paren {p \land \neg q} \implies \paren {\neg \paren {p \implies q} }$	Rule of Implication: $\implies \II$	1 – 2	Assumption 1 has been discharged
4		4	$\neg \paren {p \implies q}$	Assumption	(None)
5		4	$p \land \neg q$	Sequent Introduction	4	Conjunction with Negative is Equivalent to Negation of Conditional: Formulation 1:Reverse Implication
6			$\paren {\neg \paren {p \implies q} } \implies \paren {p \land \neg q}$	Rule of Implication: $\implies \II$	4 – 5	Assumption 4 has been discharged
7			$\paren {p \land \neg q} \iff \paren {\neg \paren {p \implies q} }$	Biconditional Introduction: $\iff \II$	3, 6

$\blacksquare$

Proof by Truth Table

We apply the Method of Truth Tables.

As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations.

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

$\blacksquare$