Biconditional is Transitive/Formulation 2

Theorem

The biconditional operator is transitive:

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

Proof

By the tableau method of natural deduction:

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

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$\paren {p \iff q} \land \paren {q \iff r}$	Assumption	(None)
2		1	$p \iff q$	Rule of Simplification: $\land \EE_1$	1
3		1	$q \iff r$	Rule of Simplification: $\land \EE_2$	1
4		1	$p \iff r$	Sequent Introduction	2, 3	Biconditional is Transitive: Formulation 1
5			$\paren {\paren {p \iff q} \land \paren {q \iff r} } \implies \paren {p \iff r}$	Rule of Implication: $\implies \II$	1 – 4	Assumption 1 has been discharged

$\blacksquare$

Sources

• 1964: Donald Kalish and Richard Montague: Logic: Techniques of Formal Reasoning ... (previous) ... (next): $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Theorem $\text{T93}$