Biconditional is Commutative/Formulation 1/Proof 2

Theorem

$p \iff q \dashv \vdash q \iff p$


Proof

By the tableau method of natural deduction:

$p \iff q \vdash q \iff p$
Line Pool Formula Rule Depends upon Notes
1 1 $p \iff q$ Premise (None)
2 1 $\paren {p \implies q} \land \paren {q \implies p}$ Sequent Introduction 1 Rule of Material Equivalence
3 1 $\paren {q \implies p} \land \paren {p \implies q}$ Sequent Introduction 2 Conjunction is Commutative
4 1 $q \iff p$ Sequent Introduction 3 Rule of Material Equivalence

$\Box$


By the tableau method of natural deduction:

$q \iff p \vdash p \iff q$
Line Pool Formula Rule Depends upon Notes
1 1 $q \iff p$ Premise (None)
2 1 $\paren {q \implies p} \land \paren {p \implies q}$ Sequent Introduction 1 Rule of Material Equivalence
3 1 $\paren {p \implies q} \land \paren {q \implies p}$ Sequent Introduction 2 Conjunction is Commutative
4 1 $p \iff q$ Sequent Introduction 3 Rule of Material Equivalence

$\blacksquare$


Sources

  • 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $4$ The Biconditional: Theorem $24$
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