# Conjunction and Implication

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## Theorems

### Conjunction Equivalent to Negation of Implication of Negative

#### Formulation 1

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

#### Formulation 2

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

### Implication Equivalent to Negation of Conjunction with Negative

#### Formulation 1

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

#### Formulation 2

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

### Conjunction with Negative Equivalent to Negation of Implication

#### Formulation 1

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

#### Formulation 2

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

### Modus Ponendo Tollens

#### Formulation 1

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

#### Formulation 2

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

## Law of Excluded Middle

Note that the Modus Ponendo Tollens:

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

can be proved in both directions without resorting to Law of Excluded Middle.

All the others:

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

are not reversible in intuitionistic logic.