Rule of Material Equivalence/Formulation 1/Proof 1

Theorem

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

Proof

By the tableau method of natural deduction:

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

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$p \iff q$	Premise	(None)
2		1	$p \implies q$	Biconditional Elimination: $\iff \EE_1$	1
3		1	$q \implies p$	Biconditional Elimination: $\iff \EE_2$	1
4		1	$\paren {p \implies q} \land \paren {q \implies p}$	Rule of Conjunction: $\land \II$	2, 3

By the tableau method of natural deduction:

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

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$\paren {p \implies q} \land \paren {q \implies p}$	Premise	(None)
2		1	$p \implies q$	Rule of Simplification: $\land \EE_1$	1
3		1	$q \implies p$	Rule of Simplification: $\land \EE_2$	1
4		1	$p \iff q$	Biconditional Introduction: $\iff \II$	2, 3

$\blacksquare$