Conditional is Equivalent to Negation of Conjunction with Negative/Formulation 2/Proof 1

Theorem

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

$\vdash \paren {p \implies q} \iff \paren {\neg \paren {p \land \neg q} } $
Line	Pool	Formula	Rule	Depends upon	Notes
1	1	$p \implies q$	Assumption	(None)
2	2	$p \land \neg q$	Assumption	(None)
3	2	$p$	Rule of Simplification: $\land \EE_1$	2
4	1, 2	$q$	Modus Ponendo Ponens: $\implies \mathcal E$	1, 3
5	2	$\neg q$	Rule of Simplification: $\land \EE_2$	2
6	1, 2	$\bot$	Principle of Non-Contradiction: $\neg \EE$	4, 5
7	1	$\neg \paren {p \land \neg q}$	Proof by Contradiction: $\neg \II$	2 – 6	Assumption 2 has been discharged
8		$\paren {p \implies q} \implies \paren {\neg \paren {p \land \neg q} }$	Rule of Implication: $\implies \II$	1 – 7	Assumption 1 has been discharged
9	9	$\neg \paren {p \land \neg q}$	Assumption	(None)
10	10	$p$	Assumption	(None)
11	11	$\neg q$	Assumption	(None)
12	10, 11	$p \land \neg q$	Rule of Conjunction: $\land \II$	10, 11
13	9, 10, 11	$\bot$	Principle of Non-Contradiction: $\neg \EE$	12, 9
14	9, 10	$q$	Reductio ad Absurdum	11 – 12	Assumption 11 has been discharged
15	9	$p \implies q$	Rule of Implication: $\implies \II$	10 – 14	Assumption 10 has been discharged
16		$\paren {\neg \paren {p \land \neg q} } \implies \paren {p \implies q}$	Rule of Implication: $\implies \II$	9 – 15	Assumption 9 has been discharged
17		$\paren {p \implies q} \iff \paren {\neg \paren {p \land \neg q} }$	Biconditional Introduction: $\iff \II$	8, 16

$\blacksquare$

This is one of the axioms of logic that was determined by Aristotle, and forms part of the backbone of classical (Aristotelian) logic.

This in turn invalidates this theorem from an intuitionistic perspective.

1964: Donald Kalish and Richard Montague: Logic: Techniques of Formal Reasoning ... (previous) ... (next): $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 3$: Theorem $\text{T37}$