# Modus Ponendo Tollens/Variant/Formulation 2

## Theorem

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

## Proof

By the tableau method of natural deduction:

$\vdash \paren {\neg \paren {p \land q} } \iff \paren {p \implies \neg q}$
Line Pool Formula Rule Depends upon Notes
1 1 $\neg \paren {p \land q}$ Assumption (None)
2 1 $p \implies \neg q$ Sequent Introduction 1 Modus Ponendo Tollens: Formulation 1: Forward Implication
3 $\paren {\neg \paren {p \land q} } \implies \paren {p \implies \neg q}$ Rule of Implication: $\implies \II$ 1 – 2 Assumption 1 has been discharged
4 4 $p \implies \neg q$ Assumption (None)
5 4 $\neg \paren {p \land q}$ Sequent Introduction 4 Modus Ponendo Tollens: Formulation 1: Reverse Implication
6 $\paren {p \implies \neg q} \implies \paren {\neg \paren {p \land q} }$ Rule of Implication: $\implies \II$ 4 – 5 Assumption 4 has been discharged
7 $\paren {\neg \paren {p \land q} } \iff \paren {p \implies \neg q}$ Biconditional Introduction: $\iff \II$ 3, 6

$\blacksquare$