Biconditional is Commutative/Formulation 2

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Theorems

The biconditional operator is commutative:

$\vdash \left({p \iff q}\right) \iff \left({q \iff p}\right)$


Proof

By the tableau method of natural deduction:

$\vdash \left({p \iff q}\right) \iff \left({q \iff p}\right)$
Line Pool Formula Rule Depends upon Notes
1 1 $\left({p \iff q}\right)$ Assumption (None)
2 1 $\left({q \iff p}\right)$ Sequent Introduction 1 Biconditional is Commutative: Formulation 1
3 1 $\left({p \iff q}\right) \implies \left({q \iff p}\right)$ Rule of Implication: $\implies \mathcal I$ 1 – 2 Assumption 1 has been discharged
4 4 $\left({p \iff q}\right)$ Assumption (None)
5 4 $\left({q \iff p}\right)$ Sequent Introduction 4 Biconditional is Commutative: Formulation 1
6 4 $\left({p \iff q}\right) \implies \left({q \iff p}\right)$ Rule of Implication: $\implies \mathcal I$ 4 – 5 Assumption 4 has been discharged
7 $\left({p \iff q}\right) \iff \left({q \iff p}\right)$ Biconditional Introduction: $\iff \mathcal I$ 3, 6

$\blacksquare$


Sources