Conjunction with Negative Equivalent to Negation of Implication/Formulation 2

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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

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


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 Equivalent to Negation of Implication: Formulation 1
3 $\paren {p \land \neg q} \implies \paren {\neg \paren {p \implies q} }$ Rule of Implication: $\implies \mathcal I$ 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 Equivalent to Negation of Implication: Formulation 1
6 $\paren {\neg \paren {p \implies q} } \implies \paren {p \land \neg q}$ Rule of Implication: $\implies \mathcal I$ 4 – 5 Assumption 4 has been discharged
7 $\paren {p \land \neg q} \iff \paren {\neg \paren {p \implies q} }$ Biconditional Introduction: $\iff \mathcal I$ 3, 6

$\blacksquare$


Proof 2

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.

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

$\blacksquare$