Biconditional is Transitive/Formulation 1/Proof 1

Theorem

$p \iff q, q \iff r \vdash p \iff r$

Proof

By the tableau method of natural deduction:

$p \iff q, q \iff r \vdash p \iff r$

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$p \iff q$	Premise	(None)
2		2	$q \iff r$	Premise	(None)
3		1	$p \implies q$	Biconditional Elimination: $\iff \EE_1$	1
4		2	$q \implies r$	Biconditional Elimination: $\iff \EE_1$	2
5		1, 2	$p \implies r$	Sequent Introduction	1, 2	Hypothetical Syllogism: Formulation 1
6		1	$q \implies p$	Biconditional Elimination: $\iff \EE_2$	1
7		2	$r \implies q$	Biconditional Elimination: $\iff \EE_2$	2
8		1, 2	$r \implies p$	Sequent Introduction	7, 6	Hypothetical Syllogism: Formulation 1
9		1, 2	$p \iff r$	Biconditional Introduction: $\iff \II$	5, 8

$\blacksquare$

Sources

• 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $4$ The Biconditional: Theorem $26$